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A ray of light incident at an angle \(\theta\) on a refracting face of a prism emerges from the other face normally. If the angle of the prism is \(5^{\circ}\) and the prism is made of a material of refractive index 1.5 , the angle of incidence is
(a) Key concept:
In thin prisms, the distance between the refracting surfaces is ineligible and the angle of prism (A) is very small. Since \(A=r_1+r_2\), therefore if \(A\) is small then both \(r_1\), and \(r_2\) are also small, and the same is true for \(i_1\) and \(\mathrm{i}_2\).
According to the question, ray emerges from other surface of prism normally,
\(\therefore\) Angle of incidence at second face,
\(
r^{\prime}=0^{\circ}
\)
Now, \(r+r^{\prime}=A\)
\(
\Rightarrow r=A-r^{\prime}=5^{\circ}-0^{\circ}=5^{\circ}
\)
Using snell’s law
\(
\begin{aligned}
& \mu=\frac{\sin i}{\sin r} \\
& \text { or } \sin i=\mu \sin r=1.5 \times \sin 5^{\circ} \\
& =0.131 \\
& \Rightarrow \theta=i=\sin ^{-1}(0.131) \\
& =7.5^{\circ}
\end{aligned}
\)
A short pulse of white light is incident from air to a glass slab at normal incidence. After travelling through the slab, the first colour to emerge is
(d) Key Concept: When light ray goes from one medium to other medium, the frequency of light remains unchanged.
As velocity of wave is given by the relation \(v=f \lambda\). When light ray goes from one medium to other medium, the frequency of light remains unchanged. Hence \(v \propto \lambda\) or greater the wavelength, greater the speed.
The light of red colour is of highest wavelength and therefore of highest speed. Therefore, after travelling through the slab, the red colour emerges first.
An object approaches a convergent lens from the left of the lens with a uniform speed \(5 \mathrm{~m} / \mathrm{s}\) and stops at the focus. The image
(c) In our problem the object approaches a convergent lens from the left of the lens with a uniform speed of 5 \(\mathrm{m} / \mathrm{s}\), hence the image will move away from the lens with a non-uniform acceleration.
Explanation:
How to solve?
Analyze the behavior of the image distance and magnification as the object distance approaches the focal length.
Step 1: Analyze the image distance
As the object approaches the focus, \(u\) approaches \(f\).
Using the lens formula, \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\).
As \(u\) approaches \(f, \frac{1}{v}\) approaches 0 , so \(v\) approaches infinity.
The image moves away from the lens.
Step 2: Analyze the magnification
Magnification \({M}=\frac{v}{u}\).
As \(v\) approaches infinity and \(u\) approaches \(f\), the magnification increases.
The image size increases.
Step 3: Analyze the speed of the image
The speed of the image is not constant.
When the object is far from the focus, the image moves slowly.
As the object gets closer to the focus, the image moves faster.
The image moves with non-uniform acceleration.
Solution
The image moves away from the lens with a non-uniform acceleration.
You are given four sources of light each one providing a light of a single colour – red, blue, green and yellow. Suppose the angle of refraction for a beam of yellow light corresponding to a particular angle of incidence at the interface of two media is \(90^{\circ}\). Which of the following statements is correct if the source of yellow light is replaced with that of other lights without changing the angle of incidence?
(c) Key concept: According to Cauchy relationship.
\(
\lambda \propto \frac{1}{\mu}
\)
Smaller the wavelength higher the refractive index and consequently smaller the critical angle.
We know \(v=f \lambda\), the frequency of wave remains unchanged with medium hence \(v \propto \lambda\).
The critical angle, \(\sin C=\frac{1}{\mu}\)
Also, velocity of light, \(v \propto \frac{1}{\mu}\)
According to VIBGYOR, among all given sources of light, the blue light have smallest wavelength. As \(\lambda_{\text {blue }}<\lambda_{\text {yellow }}\) hence \(v_{\text {blue }}<v_{\text {yellow }}\), it means \(\mu_{\text {blue }}>\mu_{\text {yellow }}\).
It means critical angle for blue is less than yellow colour, the critical angle is least which facilitates total internal reflection for the beam of blue light.
The radius of curvature of the curved surface of a plano-convex lens is 20 cm. If the refractive index of the material of the lens be 1.5, it will
Key Concept: By lens maker’s formula for plano-convex lens, focal length is given by \(f=\frac{R}{\mu-1}\). This is always positive for \(\mu>1\) or optically denser medium of material of lens placed in air.
(c) Here, \(R=20 \mathrm{~cm}, \mu=1.5\), on substituting the values in \(f=\frac{R}{\mu-1}=\frac{20}{1.5-1}=40 \mathrm{~cm}\) of converging nature as \(f>0\). Therefore, lens act as a convex lens irrespective of the side on which the object lies.
Explanation:
\(
\begin{aligned}
&\text { If object lies on curved side then } \mathrm{R}_1=+20 \mathrm{~cm} \text { and } \mathrm{R}_2=\infty, \mu_1=1, \mu_2=1.5\\
&\begin{aligned}
& \frac{1}{f}=\left(\mu_2-\mu_1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{\infty}\right)=\frac{0.5}{20}=\frac{5}{200}=\frac{1}{40}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\mathrm{F}=+40 \mathrm{~cm}\\
&\text { If object lies on plane side } R_1=\infty \text { and } R_2=-20 \mathrm{~cm}, \mu_1=1, \mu_2=1.5\\
&\begin{aligned}
& \frac{1}{f}=\left(\mu_2-\mu_1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \frac{1}{f}=(1.5-1)\left(\frac{1}{\infty}-\left(-\frac{1}{20}\right)\right)=0.5 \times\left(+\frac{1}{20}\right)
\end{aligned}
\end{aligned}
\)
\(
\frac{1}{f}=\frac{5}{200}
\)
\(
\mathrm{f}=+40 \mathrm{~cm}
\)
So, lens will always act as a convex lens irrespective of the side on which objects lie.
The phenomena involved in the reflection of radiowaves by ionosphere is similar to
(b) Radio waves are reflected by a layer of atmosphere called the Ionosphere, so they can reach distant parts of the Earth. The reflection of radio waves by ionosphere is due to total internal reflection. It is the same as total internal reflection of light in air during a mirage, i.e., angle of incidence is greater than critical angle.
Important point: The ionized part of the Earth’s atmosphere is known as the ionosphere. Ultraviolet light from the sun collides with atoms in this region knocking electrons loose. This creates ions, or atoms with missing electrons. This is what gives the lonosphere its name- and it is the free electrons that cause the reflection and absorption of ratio waves.
The direction of ray of light incident on a concave mirror is shown by PQ while directions in which the ray would travel after reflection is shown by four rays marked \(1,2,3\) and 4 (Fig 9.1). Which of the four rays correctly shows the direction of reflected ray?
(b) The ray PQ of light passes through focus F and incident on the concave mirror, after reflection, should become parallel to the principal axis and shown by ray 2 in the figure.
Important points:
We can locate the image of any extended object graphically by drawing any two of the following four special rays:
1. A ray initially parallel to the principal axis is reflected through the focus of the mirror (1).
2. A ray passing through the center of curvature is reflected back along itself (3).
3. A ray initially passing through the focus is reflected parallel to the principal axis (2).
4. A ray incident at the pole is reflected symmetrically.
The optical density of turpentine is higher than that of water while its mass density is lower. Figure below. shows a layer of turpentine floating over water in a container. For which one of the four rays incident on turpentine in Figure below, the path shown is correct?
(b) Here, light ray goes from (optically) rarer medium air to optically denser medium turpentine, then it bends towards the normal, i.e., \(\theta 1>\theta 2\) whereas when it goes from to optically denser medium turpentine to rarer medium water, then it bends away the normal.
A car is moving with at a constant speed of \(60 \mathrm{~km} \mathrm{~h}^{-1}\) on a straight road. Looking at the rear view mirror, the driver finds that the car following him is at a distance of 100 m and is approaching with a speed of \(5 \mathrm{~km} \mathrm{~h}^{-1}\). In order to keep track of the car in the rear, the driver begins to glance alternatively at the rear and side mirror of his car after every 2 s till the other car overtakes. If the two cars were maintaining their speeds, which of the following statement (s) is/are correct?
(d) Object moving along the principat axis: On differentiating the mirror formula with respect to time, we get \(\frac{d v}{d t}=-\left(\frac{v}{u}\right)^2 \frac{d u}{d t}=-\left(\frac{f}{u-f}\right)^2 \cdot \frac{d u}{d t}\) where \(d v / d t\) is the velocity of image along the principal axis and \(d u / d t\) is the velocity of object along the principal axis. Negative sign implies that the image, in case of mirror, always moves in the direction opposite to that of the object.
As the distance between the cars decreases, the speed of the image of the car would appear to increase.
There are certain material developed in laboratories which have a negative refractive index (Fig. 9.3). A ray incident from air (medium 1) into such a medium (medium 2) shall follow a path given by
(a) The materials with negative refractive index responds to Snell’s law just opposite way. If incident ray from air (Medium 1) incident on those material, the ray refract or bend same side of the normal as in option (a).
Consider an extended object immersed in water contained in a plane trough. When seen from close to the edge of the trough the object looks distorted because
(a, b, c) When light from the submerged object before reaching to the observer gets, refracted from water surface, the rays bend away from normal and the angle subtended by the image of the object at the eye is smaller than the actual angle subtended by the object in air. Also the apparent depth of the .points close to the edge are nearer the surface of the water compared to the points away from the edge.
As we move towards right, the angle of incident increases and becomes equal to critical angle. Hence some of the points of the object far away from the edge may not be visible because of total internal reflection.
A rectangular block of glass ABCD has a refractive index 1.6. A pin is placed midway on the face AB (Fig. 9.4). When observed from the face A D, the pin shall
(a, d) As long as angle of incidence on AD of the ray emanating from pin is less than the critical angle, the pin shall appear to be near A.
When angle of incidence on AD of the ray emanating from pins is greater than the critical angle, the light suffers from total internal reflection and cannot be seen through AD.
An astronomical refractive telescope has an objective of focal length 20 m and an eyepiece of focal length 2 cm.
(a, b, c) Key concept: The magnifying power \(m\) is the ratio of the angle \(\beta\) subtended at the eye by the final image to the angle \(\alpha\) which the object subtends at the lens or the eye. Hence, in normal adjustment
\(
m \approx \frac{\beta}{\alpha} \approx \frac{h}{f_e} \frac{f_o}{h}=\frac{f_o}{f_e}
\)
In this case, the length of the telescope tube is \(f_o+f_e\).
The length of the telescope tube is \(t_0+t_e=20+(0.02)=20.02 \mathrm{~m}\)
Also,
\(
m=20 / 0.02=1000
\)
Also, the image formed is inverted.
A point source of light is placed in front of a plane mirror.
(a) The image of a point source in plane mirror will be a virtual point image, behind the mirror. So the reflected rays should meet at this point when produced backwards.
Explanation:
When a point source of light is placed in front of a plane mirror, the image formed is a virtual, upright, and same-size image that appears to be located behind the mirror at a distance equal to the object’s distance from the mirror.
Virtual Image:
The light rays from the source do not actually converge behind the mirror; instead, they appear to originate from that point.
Upright and Same Size:
The image retains the same orientation and size as the object.
Distance Behind the Mirror:
The distance between the image and the mirror is the same as the distance between the object and the mirror.
Law of Reflection:
This phenomenon is based on the law of reflection, where the angle of incidence equals the angle of reflection. Therefore, the reflected rays, when extended backward, intersect at the location of the virtual image.
Total internal reflection can take place only if
(b) The correct answer is light goes from optically denser medium to rarer medium. For total internal reflection to occur, light must travel from a medium with a higher refractive index to a medium with a lower refractive index. According to Snell’s Law, \(\frac{\sin i}{\sin r}=\frac{\mu_2}{\mu_1}\)
where \(\mathrm{r}=90^{\circ}\) for particular incidence angle called critical angle. When the incidence angle is equal to or greater than \(\mathrm{i}_{\mathrm{c}}\), then total internal reflection occurs. It takes place when ray of light travels from optically denser medium \(\left(\mu_1>\mu_2\right)\) to optically rarer medium.
Explanation:
Total internal reflection happens when the angle of incidence is greater than the critical angle, and the light is traveling from a denser medium to a rarer medium.
Denser to Rarer:
If light travels from a denser medium (like glass or water) to a rarer medium (like air), it bends away from the normal (an imaginary line perpendicular to the surface).
Critical Angle:
As the angle of incidence increases, the angle of refraction also increases. When the angle of incidence reaches a certain critical angle, the angle of refraction becomes 90 degrees, and the light travels along the surface between the two media.
Total Internal Reflection:
If the angle of incidence is increased further beyond the critical angle, the light is no longer refracted but is entirely reflected back into the denser medium. This is total internal reflection.
In image formation from spherical mirrors, only paraxial rays are considered because they
(c) In image formation by spherical mirrors, only paraxial rays are considered because they form a nearly point image of a point source.
Explanation:
Paraxial rays:
These are light rays that are close to the principal axis of the mirror and make a small angle with it.
Why paraxial rays are preferred:
When paraxial rays are used, they converge to a single point after reflection, forming a clear, focused image of a point object.
Aberrations:
If rays far from the principal axis (marginal rays) are considered, they don’t all converge at the same point, leading to blurring and distortion in the image, a phenomenon called spherical aberration.
Geometrical simplicity:
While it’s true that paraxial rays are geometrically easier to handle, the primary reason for using them is to avoid aberrations and achieve a sharper image.
A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm . The image will form at
(d) The mirror formula is \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\).
For a convex mirror, the focal length \(f\) is positive.
Object distance \(u\) is taken as negative by convention (following sign convention).
Object distance: \(u=-30 \mathrm{~cm}\)
Focal length of convex mirror: \(f=+30 \mathrm{~cm}\)
\(
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{+30}=\frac{1}{v}+\frac{1}{-30}
\end{aligned}
\)
\(
\frac{1}{v}=\frac{1}{30}-\frac{1}{-30}
\)
\(
v=+15 \mathrm{~cm}
\)
The image will form 15 cm behind the mirror.
The figure below shows two rays \(A\) and \(B\) being reflected by a mirror and going as \(A^{\prime}\) and \(B^{\prime}\). The mirror
(a) The image shows that incident rays A and B are parallel and are falling on the object. Therefore, rays \(A^{\prime}\) and \(B^{\prime}\) are the reflected rays. It is given that \(A^{\prime}\) and \(B^{\prime}\) are still parallel after reflection. So, this situation is possible only in the case of a plane mirror. In the case of a convex mirror, if the incident ray is parallel to the principal axis, then after reflection, it will diverge or appear to pass through focus. Whereas in the case of a concave mirror, if the incident ray is parallel to the principal axis, then after reflection, it will converge and pass-through focus, thus forming a real image of an object.
The image formed by a concave mirror
(c) The correct answer is: is certainly real if the object is virtual.
Explanation: A concave mirror can form both real and virtual images depending on the position of the object relative to the mirror. If the object is virtual (behind the mirror), the image formed by a concave mirror is always real. If the object is real (in front of the mirror), the image can be either real or virtual depending on the object’s distance from the mirror’s focal point.
Why other options are incorrect:
(a) is always real:
This is incorrect because concave mirrors can form virtual images when the object is placed between the mirror and the focal point.
(b) is always virtual:
This is incorrect because concave mirrors can form real images when the object is placed beyond the focal point.
(d) is certainly virtual if the object is real:
This is incorrect because if the object is real and placed beyond the focal point, the image formed is real. If the object is real and placed between the focal point and the mirror, the image is virtual.
Figure below shows three transparent media of refractive indices \(\mu_1, \mu_2\) and \(\mu_3\). A point object \(O\) is placed in the medium \(\mu_2\). If the entire medium on the right of the spherical surface has refractive index \(\mu_1\), the image forms at \(O^{\prime}\). If this entire medium has refractive index \(\mu_3\), the image forms at \(O^{\prime \prime}\). In the situation shown,
(d) A point object \(O\) is in medium \(\mu_2\).
If the entire medium to the right of the spherical surface is \(\mu_1\), the image forms at \(O^{\prime}\).
If the entire medium to the right of the spherical surface is \(\mu_3\), the image forms at \(O^{\prime \prime}\).
Rays from object \(O\) above the principal axis enter the medium with refractive index \(\mu_3\).
Rays from object \(O\) below the principal axis enter the medium with refractive index \(\mu_1\).
The rays entering \(\mu_3\) will form an image at \(O^{\prime \prime}\). The rays entering \(\mu_1\) will form an image at \(O^{\prime}\).
Since different parts of the spherical surface refract light into different media, two distinct images are formed.
Two images form, one at \(O^{\prime}\) and the other at \(O^{\prime \prime}\).
Four modifications are suggested in the lens formula to include the effect of the thickness \(t\) of the lens. Which one is likely to be correct?
(c) The likely correct modification to the lens formula to include the effect of the lens thickness \(t\) is \(\frac{1}{v-t}-\frac{1}{u+t}=\frac{1}{f}\).
Explanation:
The standard thin lens formula is \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\), where \(v\) is the image distance, \(u\) is the object distance, and \(f\) is the focal length. The thickness \(t\) of the lens affects the light path, effectively shifting the positions of the image and object.
Option (c) accounts for this shift by adding the lens thickness \(t\) to both the object and image distances. When the lens is thin, \(t\) approaches zero, and the equation reduces to the standard thin lens formula.
A double convex lens has two surfaces of equal radii \(R\) and refractive index \(m=1 \cdot 5\). We have,
(b)
\(
\begin{aligned}
& \frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) \\
& \mu=1.5, \mathrm{R}_1=\mathrm{R}, \mathrm{R}_2=-\mathrm{R}, \mathrm{f}=+\mathrm{f} \\
& \frac{1}{\mathrm{f}}=(1.5-1)\left(\frac{1}{\mathrm{R}}+\frac{1}{\mathrm{R}}\right) \Rightarrow \frac{1}{\mathrm{f}}=0.5 \times \frac{2}{\mathrm{R}}=\frac{1}{2} \times \frac{2}{\mathrm{R}} \\
& \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{R}} \Rightarrow \mathrm{f}=+\mathrm{R}
\end{aligned}
\)
A point source of light is placed at a distance of \(2 f\) from a converging lens of focal length \(f\). The intensity on the other side of the lens is maximum at a distance
(c) Since the object is placed at \(2 f\), the image of the object will be formed at distance of \(2 f\) from a converging lens.
It can also be shown from the lens formula:
\(
\frac{1}{v}-\frac{1}{u}=\frac{1}{f}
\)
Here, \(u=-2 f\) and \(f=f\)
On putting the respective values we get:
\(
\begin{aligned}
& \frac{1}{v}-\frac{1}{-2 f}=\frac{1}{f} \\
& \Rightarrow \frac{1}{v}=\frac{1}{f}-\frac{1}{2 f} \\
& =\frac{1}{2 f}
\end{aligned}
\)
Therefore, image distance \(v=2 f\)
When a point source of light is placed at 2F of a convex lens, we get all its light rays will be convergent at \(2 f\) on the other side and we get intensity oat \(2 f\) will be maximum.
A parallel beam of light is incident on a converging lens parallel to its principal axis. As one moves away from the lens on the other side on its principal axis, the intensity of light
(d)
Understanding the Setup:
A parallel beam of light is incident on a converging lens (convex lens) parallel to its principal axis.
The lens will refract the light rays, causing them to converge.
Identifying the Focal Point:
The light rays that are parallel to the principal axis converge at a point known as the focal point (F) of the lens, which is located on the principal axis at a distance equal to the focal length (f) from the lens.
Behavior of Light After Passing Through the Lens:
As the parallel rays of light pass through the lens, they converge towards the focal point. At this point, the intensity of the light is at its maximum because all the light rays are concentrated in a small area.
Moving Away from the Lens:
As you move away from the lens along the principal axis, initially, you are approaching the focal point. Therefore, the intensity of the light will increase as you get closer to the focal point.
Once you pass the focal point and continue moving away from the lens, the light rays begin to diverge. As they spread out, the intensity of the light decreases because the same amount of light is now distributed over a larger area.
Conclusion:
The intensity of light first increases as you approach the focal point and then decreases as you move further away from the focal point.
Therefore, the correct answer to the question is that the intensity of light “first increases then decreases.”
Final Answer: The intensity of light first increases then decreases as one moves away from the lens on the other side along the principal axis.
A symmetric double convex lens is cut in two equal parts by a plane perpendicular to the principal axis. If the power of the original lens was 4 D, the power of a cut-lens will be
(a)
Biconvex lens is cut perpendicular to the principal axis, it will become a plano-convex lens. For length of biconvex lens,
\(
\begin{aligned}
& \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \frac{1}{f}=(n-1) \frac{2}{R} \quad\left(\because R_1=R, R_2=-R\right)
\end{aligned}
\)
\(
\mathrm{f}=\frac{\mathrm{R}}{2(\mathrm{n}-1)} \dots(i)
\)
\(
\begin{aligned}
&\text { For plano-convex lens, }\\
&\begin{aligned}
& \frac{1}{f_1}=(n-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\
& f_1=\frac{R}{(n-1)} \quad \ldots \ldots \ldots . . \text { (ii) }
\end{aligned}
\end{aligned}
\)
On comparing eqs. (i) and (ii), we see the focal length become double.
\(
\mathrm{f}_1=2f
\)
As power of lens, \(\mathrm{P} \propto \frac{1}{\text { focal length }}\)
Hence, power will become half.
New power \(=\frac{4 D}{2}=2 \mathrm{D}\)
A symmetric double convex lens is cut in two equal parts by a plane containing the principal axis. If the power of the original lens was 4 D, the power of a divided lens will be
(c) The original lens is a symmetric double convex lens.
The original lens is cut into two equal parts by a plane containing the principal axis.
The power of the original lens is 4 D.
When a symmetric double convex lens is cut along a plane containing the principal axis, the radii of curvature of the surfaces remain unchanged.
The focal length of a lens depends on the radii of curvature and the refractive index of the material.
Since these parameters are unchanged, the focal length of each half remains the same as the original lens.
The power P of a lens is the reciprocal of its focal length f , i.e., \({P}=\frac{1}{{f}}\).
Since the focal length remains the same, the power of each divided part will also remain the same.
The original power was 4 D. Therefore, the power of each divided lens is also 4 D.
Two concave lenses \(L_1\) and \(L_2\) are kept in contact with each other. If the space between the two lenses is filled with a material of smaller refractive index, the magnitude of the focal length of the combination
(c) The focal length of the combination will increase. For finding out the combination of lens we have the formula:
\(
\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2}
\)
Where, \(F\) is the focal length for the combination
\(d\) is the separation between two lenses
Here, \(d=0\)
\(
\begin{aligned}
& \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2} \\
& F=\frac{f_1 f_2}{f_1+f_2}
\end{aligned}
\)
Hence, the focal length will increase.
A thin lens is made with a material having refractive index \(\mu=1.5\). Both the sides are convex. It is dipped in water ( \(\mu=1.33\) ). It will behave like
(a) consider a lens of focal length I with radius of curvature of both the surfaces \(\mathrm{R}_1\) and \(\mathrm{R}_2\)
\(
\frac{1}{f}=(1.5-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=0.5\left(\frac{1}{R_1}+\frac{1}{R_2}\right)
\)
if the lens is placed in water then the focal length \(f^{\prime}\) is given by
\(
\frac{1}{f^{\prime}}=\left(\frac{1.5}{1.33}-1\right)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)=\frac{0.17}{1.33}\left(\frac{2}{f}\right)
\)
\(f^{\prime}=3.91 f\)
Since, the sign of \(f^{\prime}\) is the same as that of \(f\), so the lens will behave as a converging lens.
A convex lens is made of a material having refractive index \(1 \cdot 2\). Both the surfaces of the lens are convex. If it is dipped into water ( \(\mu=1.33\) ), it will behave like
(b) The refractive index of the convex lens material is \(\mu_l=1.2\). The refractive index of water is \(\mu_w=1.33\).
Compare the refractive index of the lens material \(\left(\mu_l\right)\) with the refractive index of the water ( \(\mu_w\) ).
We have \(\mu_l=1.2\) and \(\mu_w=1.33\).
Thus, \(\mu_l<\mu_w\).
When a lens is immersed in a medium with a higher refractive index than its own material, its behavior reverses.
A convex lens, which is normally convergent in air, will behave as a divergent lens in water.
The convex lens will behave like a divergent lens.
A point object \(O\) is placed on the principal axis of a convex lens of focal length \(f=20 \mathrm{~cm}\) at a distance of 40 cm to the left of it. The diameter of the lens is 10 cm. An eye is placed 60 cm to right of the lens and a distance \(h\) below the principal axis. The maximum value of \(h\) to see the image is
(b) The focal length of the convex lens is \(f=20 \mathrm{~cm}\).
The object distance is \(u=40 \mathrm{~cm}\).
The diameter of the lens is \(D=10 \mathrm{~cm}\).
The eye is placed at a distance of 60 cm to the right of the lens.
Use the lens formula: \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\).
Substitute the given values: \(\frac{1}{v}-\frac{1}{-40 \mathrm{~cm}}=\frac{1}{20 \mathrm{~cm}}\).
Solve for \(v: \frac{1}{v}=\frac{1}{20 \mathrm{~cm}}-\frac{1}{40 \mathrm{~cm}}=\frac{2-1}{40 \mathrm{~cm}}=\frac{1}{40 \mathrm{~cm}}\). The image distance is \(v=40 \mathrm{~cm}\).
The radius of the lens is \(R=\frac{D}{2}=\frac{10 \mathrm{~cm}}{2}=5 \mathrm{~cm}\).
This means that the image is formed at the center of curvature. Therefore, we can draw the ray diagram as follows.
\(
\begin{aligned}
&\text { Now in the figure, } \triangle C E D \text { and } \triangle C A B \text { are symmetric. So we have }\\
&\frac{A B}{D E}=\frac{B C}{D C} \Rightarrow \frac{5}{h}=\frac{40}{20} \Rightarrow \frac{5}{h}=2 \Rightarrow h=\frac{5}{2} \Rightarrow h=2.5 \mathrm{~cm}
\end{aligned}
\)
The rays of different colours fail to converge at a point after going through a converging lens. This defect is called
(d) The correct answer is chromatic aberration. Since refractive index for different wave length of light is different. Hence the different colours of light forms images at different position. This phenomenon is called chromatic aberration.
Explanation: Chromatic aberration occurs when different wavelengths of light (colors) focus at different points within a lens, resulting in a blurry image with color fringes, especially around the edges of objects.
Why other options are incorrect:
(a) Spherical aberration:
This defect occurs when light rays passing through the edges of a lens focus at a different point than rays passing through the center, causing a blurry image. It’s not related to different colors.
(b) Distortion:
Distortion refers to a geometric distortion of the image, where straight lines appear curved or objects appear stretched or compressed.
(c) Coma:
Coma is a type of optical aberration that causes light rays to converge unevenly around the lens’s axis, resulting in a blurred image with comet-like tails. It’s not related to color differences.
If the light moving in a straight line bends by a small but fixed angle, it may be a case of
(a, b) Understanding the Scenario:
A ray of light is traveling straight and then bends by a small but fixed angle. This indicates that there is a change in direction of the light ray.
Analyzing Reflection:
When light hits a reflective surface (like a mirror), it can change direction. According to the law of reflection, the angle of incidence is equal to the angle of reflection. If the surface is flat, the bending of the ray can be considered a fixed angle.
Conclusion: Reflection can cause the ray to bend by a fixed angle.
Analyzing Refraction:
Refraction occurs when light passes from one medium to another (for example, from air to glass). The light ray bends at the interface according to Snell’s law, which relates the angles of incidence and refraction to the indices of refraction of the two media.
If the angle of incidence is small and the indices of refraction are known, the angle of refraction can also be determined, thus making it a fixed angle.
Conclusion: Refraction can also cause the ray to bend by a fixed angle.
Analyzing Dispersion:
Dispersion occurs when light passes through a prism and different wavelengths (colors) of light bend by different amounts. This results in a spectrum of colors and not a single fixed angle.
Conclusion: Dispersion does not cause a fixed angle of bending.
Analyzing Diffraction:
Diffraction occurs when light passes through a small aperture or around an obstacle, causing the light to spread out and create interference patterns. This results in multiple rays and does not produce a single fixed angle of bending.
Conclusion: Diffraction does not result in a fixed angle of bending.
Final Conclusion: Based on the analysis, the bending of a ray of light by a small but fixed angle can be attributed to either reflection or refraction. Therefore, the correct answer to the question is that the bending may be due to reflection or refraction.
Mark the correct options.
(b) If the final rays are converging, we have a real image.This is because a real image is formed by converging reflected/refracted rays from a mirror/lens.
(a) If the incident rays are converging, we have a virtual object.
Incorrect: A real object emits light rays that diverge. If rays are converging before hitting a lens or mirror, they must have originated from a virtual object (a point behind the lens or mirror where light appears to diverge from). A virtual object is the point of convergence of light rays that would appear to originate from behind the lens or mirror if they were extended.
(c) The image of a virtual object is called a virtual image
Incorrect. The image of a virtual object can be real or virtual, depending on the optical system. There’s no rule that says it must be virtual.
(d) If the image is virtual, the corresponding object is called a virtual object
Incorrect. A virtual image can be formed by a real object. So this is not necessarily true.
Which of the following (referred to a spherical mirror) do (does) not depend on whether the rays are paraxial or not?
(a, c, d) Paraxial rays are the light rays close to the principal axis. The focus of the spherical mirror for paraxial rays are different from the focus for marginal rays. So, the focus depends on whether the rays are paraxial or marginal. The pole, radius of curvature and the principal axis of a spherical mirror do not depend on paraxial or marginal rays.
Note: The point where paraxial rays (light rays close to the principal axis) converge after reflection is indeed called the focus or principal focus. This is a key concept in understanding how mirrors form images.
The image of an extended object, placed perpendicular to the principal axis of a mirror, will be erect if
(c, d) The virtual image of a real object and the real image of a virtual object are always erect.
Explanation: A real image occurs when the rays converge. A real image is always formed below the principal axis, so these are inverted whereas a virtual image is always formed above the principal axis so these are always erect.
A convex lens forms a real image of a point object placed on its principal axis. If the upper half of the lens is painted black,
(c, d) If the upper half portion of the convex lens is painted, then only the intensity of the image will decrease, as the amount of light passing through the lens will decrease. Also, there will be no shift in the position of the image because all the parameters remain the same.
Consider three converging lenses \(L_1, L_2\) and \(L_3\) having identical geometrical construction. The index of refraction of \(L_1\) and \(L_2\) are \(\mu_1\) and \(\mu_2\) respectively. The upper half of the lens \(L_3\) has a refractive index \(\mu_1\) and the lower half has \(\mu_2\) (figure below). A point object \(O\) is imaged at \(O_1\) by the lens \(L_1\) and at \(O_2\) by the lens \(L_2\) placed in same position. If \(L_3\) is placed at the same place,
(a, b) Both upper part and lower part form their own images. As they have different refractiveindices two different images are formed.
Explanation:
The lens \(L_1\) converges the light at point \(O_1\) and lens \(L_2\) converges the light at \(O_2\). As the upper half of lens \(L_3\) has a refractive index equal to that of \(\mathrm{L}_1\), it will converge the light at \(O_1\) and thus the image will be formed at \(\mathrm{O}_1\). Also, the lower half of lens \(L_3\) has a refractive index equal to that of lens \(L_2\), it will converge the light at \(O_2\) and thus the image will be formed at \(\mathrm{O}_2\).
A screen is placed at a distance 40 cm away from an illuminated object. A converging lens is placed between the source and the screen and it is attempted to form the image of the source on the screen. If no position could be found, the focal length of the lens
(b) Let the distance between the source and the lens be x, then the distance between the lens and the screen is \(40-x\)
\(
\frac{1}{v}-\frac{1}{u}=\frac{1}{f}
\)
\(
\frac{1}{(40-x)}+\frac{1}{x}=\frac{1}{f}
\)
\(
x^2-40 x+40 f=0; \quad x=\frac{40 \pm \sqrt{1600-160 f}}{2}
\)
for \(x\) to take real values \((1600-160 f)>0\)
\(\mathrm{f}<10 \mathrm{~cm}\)
For the image to be formed on the screen the focal length should be less than 10 cm, but since no such positions were found hence the lens used has focal length greater than 10 cm.
An object is placed on the principal axis of a concave mirror of focal length 10 cm at a distance of 8.0 cm from the pole. What is the position and the nature of the image?
(b)
\(
\begin{aligned}
\frac{1}{f} & =\frac{1}{u}+\frac{1}{v} \\
\frac{1}{v} & =\frac{1}{f}-\frac{1}{u} \\
& =\frac{1}{-10 \mathrm{~cm}}-\frac{1}{-8 \cdot 0 \mathrm{~cm}} \\
& =\frac{1}{40 \mathrm{~cm}} \\
v & =40 \mathrm{~cm}
\end{aligned}
\)
The positive sign shows that the image is formed at 40 cm from the pole on the other side of the mirror (figure above). As the image is formed beyond the mirror, the reflected rays do not intersect, the image is thus virtual.
A rod of length 10 cm lies along the principal axis of a concave mirror of focal length 10 cm in such a way that the end closer to the pole is 20 cm away from it. What is the length of the image?
(d)
The situation is shown in figure above. The radius of curvature of the mirror is \(r=2 f=20 \mathrm{~cm}\). Thus, the nearer end \(B\) of the \(\operatorname{rod} A B\) is at the centre of the curvature and hence, its image will be formed at \(B\) itself. We shall now locate the image of \(A\).
Here \(u=-30 \mathrm{~cm}\) and \(f=-10 \mathrm{~cm}\). We have
\(
\begin{aligned}
& \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\
& \frac{1}{v}=\frac{1}{f}-\frac{1}{u} \\
& \quad=\frac{1}{-10 \mathrm{~cm}}-\frac{1}{-30 \mathrm{~cm}} \\
& v=-15 \mathrm{~cm}
\end{aligned}
\)
Thus, the image of \(A\) is formed at 15 cm from the pole. The length of the image is, therefore, 5.0 cm.
At what distance from a convex mirror of focal length 2.5 m should a boy stand so that his image has a height equal to half the original height? The principal axis is perpendicular to the height.
(c)
\(
\begin{aligned}
&\text { We have, }\\
&\begin{aligned}
m & =-\frac{v}{u}=\frac{1}{2} \\
v & =-\frac{u}{2} \\
\frac{1}{u} & +\frac{1}{v}=\frac{1}{f} \\
\frac{1}{u} & +\frac{1}{-u / 2}=\frac{1}{2.5 \mathrm{~m}} \\
-\frac{1}{u} & =\frac{1}{2.5 \mathrm{~m}} \\
u & =-2.5 \mathrm{~m}
\end{aligned}
\end{aligned}
\)
Thus, he should stand at a distance of 2.5 m from the mirror.
A 2.0 cm high object is placed on the principal axis of a concave mirror at a distance of 12 cm from the pole. If the image is inverted, real and 5.0 cm high, What is the location of the image and the focal length of the mirror?
(a) The magnification is \(m=-\frac{v}{u}\)
\(
\begin{aligned}
\frac{-5 \cdot 0 \mathrm{~cm}}{2 \cdot 0 \mathrm{~cm}} & =\frac{-v}{-12 \mathrm{~cm}} \\
v & =-30 \mathrm{~cm} .
\end{aligned}
\)
The image is formed at 30 cm from the pole on the side of the object. We have,
\(
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
= & \frac{1}{-30 \mathrm{~cm}}+\frac{1}{-12 \mathrm{~cm}}=-\frac{7}{60 \mathrm{~cm}}
\end{aligned}
\)
or, \(f=-\frac{60 \mathrm{~cm}}{7}=-8.6 \mathrm{~cm}\).
The focal length of the mirror is 8.6 cm.
Consider the situation shown in the figure below. What is the maximum angle \(\theta\) for which the light suffers total internal reflection at the vertical surface?
(a) The critical angle for this case is
\(
\begin{gathered}
\theta^{\prime \prime}=\sin ^{-1} \frac{1}{1 \cdot 25}=\sin ^{-1} \frac{4}{5} \\
\sin \theta^{\prime \prime}=\frac{4}{5}
\end{gathered}
\)
Since \(\theta^{\prime \prime}=\frac{\pi}{2}-\theta^{\prime}\), we have \(\sin \theta^{\prime}=\cos \theta^{\prime \prime}=3 / 5\).
From Snell’s law,
\(
\frac{\sin \theta}{\sin \theta^{\prime}}=1.25
\)
\(
\begin{aligned}
\sin \theta & =1.25 \times \sin \theta^{\prime} \\
& =1.25 \times \frac{3}{5}=\frac{3}{4}
\end{aligned}
\)
\(
\theta=\sin ^{-1} \frac{3}{4}
\)
If \(\theta^{\prime \prime}\) is greater than the critical angle, \(\theta\) will be smaller than this value. Thus, the maximum value of \(\theta\), for which total reflection takes place at the vertical surface, is \(\sin ^{-1}(3 / 4)\).
Light falls on a plane reflecting surface. For what angle of incidence is the reflected ray normal to the incident ray?
(b)
Here, \(X Y\) is a plane reflecting surface, \(O A\) and \(A B\) are the incident and reflected rays, respectively
\(i=\) angle of incidence
\(r=\) angle of reflection
\(
\begin{array}{lr}
\text { As, } & A B \perp O A, \\
\therefore & i+r=90^{\circ}
\end{array}
\)
Also, according to laws of reflection,
\(
\begin{aligned}
r & =i \\
i+i & =90^{\circ} \Rightarrow 2 i=90^{\circ} \\
i & =\frac{90^{\circ}}{2}=45^{\circ}
\end{aligned}
\)
Which of the following is correct for the image formed by a plane mirror?
(c) The correct answer is Virtual and laterally inverted.
Explanation:
A plane mirror always forms a virtual image. This means the image appears to be behind the mirror, and the light rays do not actually converge there. The image is also laterally inverted, meaning left and right sides are swapped.
A plane mirror reflects a pencil of light to form a real image, then the pencil of light incident on the mirror is
(b) When convergent beam is incident on a plane mirror, then mirror forms real image, for a virtual object.
Explanation: A plane mirror typically creates virtual images, but if the light rays are already converging before hitting the mirror, they can converge further after reflection, forming a real image.
A ray of light is incident on a plane mirror at an angle of \(30^{\circ}\). The deviation produced in the ray after reflection is
\(
\begin{aligned}
&\text { (d) Deviation by a plane mirror, }\\
&\delta=180^{\circ}-2 \theta=180^{\circ}-60^{\circ}=120^{\circ}
\end{aligned}
\)
If the reflected ray is rotated by an angle of \(4 \theta\) in anti-clockwise direction, then the mirror was rotated by
(a) The reflected ray is rotated by an angle of \(4 \theta\) in the anti-clockwise direction.
When a plane mirror is rotated by an angle \(\alpha\), the reflected ray rotates by \(2 \alpha\) in the same direction.
The angle of rotation of the reflected ray is twice the angle of rotation of the mirror.
Let \(\phi_{\text {reflected }}\) be the angle of rotation of the reflected ray.
Let \(\phi_{\text {mirror }}\) be the angle of rotation of the mirror.
\(
\phi_{\text {reflected }}=2 \times \phi_{\text {mirror }}
\)
\(
\begin{aligned}
&\text { Given } \phi_{\text {reflected }}=4 \theta \text {. }\\
&\begin{aligned}
& 4 \theta=2 \times \phi_{\text {mirror }} \\
& \phi_{\text {mirror }}=\frac{4 \theta}{2} \\
& \phi_{\text {mirror }}=2 \theta
\end{aligned}
\end{aligned}
\)
The direction of rotation of the mirror is the same as the direction of rotation of the reflected ray.
Since the reflected ray rotates anti-clockwise, the mirror also rotates anticlockwise.
The mirror was rotated by \(2 \theta\) in the anti-clockwise direction.
A man is 180 cm tall and his eyes are 10 cm below the top of his head. In order to see this entire height right from top to head, he uses a plane mirror kept at a distance of 1 m from him. The minimum length of the plane mirror is
(b) According to ray diagram shown in the figure
\(
\text { Length of mirror }=\frac{1}{2}(10+170)=90 \mathrm{~cm}
\)
An object is moving towards a stationary plane mirror with a speed of \(2 \mathrm{~m} / \mathrm{s}\). Velocity of the image w.r.t. the object is
(b)
When the object is kept in front of the mirror which is stationary and plane and that object is moving with the speed of \(2 \mathrm{~ms}^{-1}\) in left of the mirror.
For this we have to draw a diagram that will explain the situation presented in the question.
Here, in the above figure we get to know that the object and image are identical and is known as the first law of reflection, size and shape remains same there will be no change. Second thing is that the speed of will remain same in magnitude but different in the direction.
So, let us say \(v_o\) be the velocity of the object and \(v_i\) be the velocity of the image. Thus according to given condition the left of the mirror represents similarity to the number line that means left side will be positive numbered side and right side will be negative numbered side.
Thus the direction being opposite the velocities are given by:
\(
v_o=-2 m s^{-1} \text { and } v_i=2 m s^{-1}
\)
Negative because moving in opposite directions.
Thus the speed of the with respect to the object \(=v_i-v_o\)
\(
\begin{aligned}
& =2 m s^{-1}-\left(-2 m s^{-1}\right) \\
& =4 m s^{-1}
\end{aligned}
\)
To get three images of a single object, one should have two plane mirrors at an angle of
\(
\begin{aligned}
&\text { (c) As number of images, } n=3\\
&\begin{array}{ll}
\text { So, } & n=\frac{360}{\theta}-1 \\
\therefore & 3=\frac{360}{\theta}-1 \\
\therefore & \theta=90^{\circ}
\end{array}
\end{aligned}
\)
An object \((O)\) is placed between two parallel plane mirrors as shown in figure. Distance between 4th image is
\(
\begin{aligned}
&\text { (b) Distance between 4th images }\\
&I_u I_u^{\prime}=2 n(x+y)=2 \times 4[3+1]=32 m
\end{aligned}
\)
A concave mirror of focal length \(f\) (in air) is immersed in water \((\mu=4 / 3)\). The focal length of the mirror in water will be
(a) On immersing a mirror in water, the focal length of the mirror remains unchanged.
Explanation:
The focal length of the concave mirror in air is \(f\).
The mirror is immersed in water with a refractive index \(\mu=\frac{4}{3}\).
The focal length of a mirror depends solely on its radius of curvature.
It is determined by the mirror’s physical shape and material.
Mirrors form images through reflection, not refraction.
The law of reflection is independent of the medium’s refractive index.
Since the reflection process is unaffected by the medium, the focal length of the mirror remains unchanged.
An object is placed 40 cm from a concave mirror of focal length 20 cm. The image formed is
(a) The object distance from the concave mirror is \(u=40 \mathrm{~cm}\)
The focal length of the concave mirror is \(f=20 \mathrm{~cm}\)
For a concave mirror, the focal length is considered negative, so \(f=-20 \mathrm{~cm}\)
The mirror formula is \(\frac{1}{v}+\frac{1}{u}=\frac{1}{f}\)
Magnification \(m=-\frac{v}{u}\)
Use the mirror formula: \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\)
Substitute the values: \(\frac{1}{v}=\frac{1}{-20}-\frac{1}{-40}\)
\(
v=-40 \mathrm{~cm}
\)
Since \(v\) is negative, the image is formed on the same side as the object. Therefore, the image is real and inverted.
Use the magnification formula: \(m=-\frac{v}{u}\). Substitute the values: \(m=-\frac{-40}{-40}\). Calculate \(m: m=-1\).
Since the absolute value of magnification \(|m|=1\), the image is the same size as the object.
The image formed is real, inverted, and the same size as the object.
A point object is placed at a distance of 30 cm from a convex mirror of focal length 30 cm. The image will form at
(d) Given, \(u=-30 \mathrm{~cm}\) and \(f=30 \mathrm{~cm}\)
\(
\frac{1}{v}+\frac{1}{-30}=\frac{1}{30} \Rightarrow \frac{1}{v}=\frac{1}{30}+\frac{1}{30}=\frac{1}{15}
\)
Distance from the mirror, \(v=+15 \mathrm{~cm}\)
The positive sign shows that the image is formed behind the mirror.
An object is placed at a distance of 30 cm from a concave mirror and its real image is formed at a distance of 30 cm from the mirror. The focal length of the mirror is
(a) Given, \(u=-30 \mathrm{~cm}\)
\(
v=-30 \mathrm{~cm}
\)
Using relation, \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
\(
\frac{1}{-30}+\frac{1}{-30}=\frac{1}{f}
\)
Focal length of the mirror, \(f=-15 \mathrm{~cm}\)
A convex mirror of focal length \(f\) forms an image which is \(1 / n\) times the object. The distance of the object from the mirror is
\(
\begin{array}{ll}
\text { (a) Magnification, } & m=+\frac{1}{n}=-\frac{v}{u} \\
\therefore & v=-\frac{u}{n}
\end{array}
\)
From mirror formula, \(\frac{1}{f}=\frac{1}{\left(-\frac{u}{n}\right)}+\frac{1}{u}\)
\(\therefore\) The distance of the object from the mirror, \(u=-(n-1) f\)
The focal length of a concave mirror is 50 cm. Where should be the object placed, so that its image is two times and inverted?
(a) In case of concave mirror, \(m=-2\)
Also,
\(
m=\frac{f}{f-u}
\)
\(
\Rightarrow \quad-2=\frac{-50}{-50-u}
\)
\(\therefore\) Distance of the object from the mirror, \(u=-75 \mathrm{~cm}\)
The object should be placed at a distance of 75 cm from the concave mirror.
The image formed by a convex mirror of focal length 30 cm is a quarter of the size of the object. The distance of the object from the mirror is
\(
\begin{aligned}
&\begin{aligned}
& \text { (b) Magnification, } m=\frac{f}{f-u} \\
& \Rightarrow \quad\left(+\frac{1}{4}\right)=\frac{+30}{+30-u}
\end{aligned}\\
&\Rightarrow \text { The distance of the object from the mirror, } u=-90 \mathrm{~cm}
\end{aligned}
\)
The distance of the object from the mirror is 90 cm.
A concave mirror gives an image three times as large as the object placed at a distance of 20 cm from it. For the image to be real, the focal length should be
(b) Magnification,
\(
m=\frac{f}{f-u} \Rightarrow-3=\frac{f}{f-(-20)}
\)
\(\Rightarrow \quad\) The focal length of mirror, \(f=-15 \mathrm{~cm}\)
The focal length of the concave mirror is 15 cm.
An object of size 7.5 cm is placed in front of a convex mirror of radius of curvature 25 cm at a distance of 40 cm. The size of the image should be
(b) Magnification, \(m=\frac{I}{O}=\frac{f}{f-u}\)
\(
\begin{aligned}
&\Rightarrow \quad \frac{I}{7.5}=\frac{(R / 2)}{(R / 2)-u}=\frac{(25 / 2)}{(25 / 2)-(-40)}\\
&\therefore \text { The size of the image, } I=1.78 \mathrm{~cm}
\end{aligned}
\)
A point object is placed at a distance of 10 cm and its real image is formed at a distance of 20 cm from a concave mirror. If the object is moved by 0.1 cm towards the mirror, the image will shift by about
(c) From the relation,
\(
\begin{aligned}
\frac{1}{v}+\frac{1}{u} & =\frac{1}{f} \\
\text { We have, } \quad \frac{-d v}{v^2}-\frac{d u}{u^2} & =0
\end{aligned}
\)
\(
\begin{gathered}
d v=\frac{v^2}{u^2}(-d u) \\
=-\left(\frac{20}{10}\right)^2(0.1)=-0.4 \mathrm{~cm} \text { away from the mirror }
\end{gathered}
\)
The image will shift by about 0.4 cm towards the mirror.
A light wave has a frequency of \(4 \times 10^{14} \mathrm{~Hz}\) and a wavelength of \(5 \times 10^{-7} \mathrm{~m}\) in a medium. The refractive index of the medium is
\(
\begin{aligned}
&\text { (a) Refractive index of the medium, }\\
&\mu=\frac{c}{v}=\frac{c}{n \lambda}=\frac{3 \times 10^8}{4 \times 10^{14} \times 5 \times 10^{-7}}=1.5
\end{aligned}
\)
\(\mu_1\) and \(\mu_2\) are the refractive indices of two mediums and \(v_1\) and \(v_2\) are the velocities of light in these two mediums respectively. Then, the relation connecting these quantities is
(d) The refractive index of medium 1 is \(\mu_1=\frac{c}{v_1}\).
The refractive index of medium 2 is \(\mu_2=\frac{c}{v_2}\).
Since both expressions equal \(c\), we have \(\mu_1 v_1=\mu_2 v_2\).
The correct relation connecting the refractive indices and velocities of light in two mediums is \(\mu_1 v_1=\mu_2 v_2\).
Absolute refractive indices of glass and water are \(\frac{3}{2}\) and \(\frac{4}{3}\). The ratio of velocities of light in glass and water will be
\(
\text { (c) As, } \quad \mu=\frac{c}{v} \Rightarrow v \propto \frac{1}{\mu}
\)
\(
\begin{aligned}
&\Rightarrow \text { Ratio of velocities of light }\\
&\frac{v_g}{v_w}=\frac{\mu_w}{\mu_g}=\frac{4 / 3}{3 / 2}=\frac{8}{9}
\end{aligned}
\)
The refractive index of a certain glass is 1.5 for light whose wavelength in vacuum is \(6000 Å\). The wavelength of this light when it passes through glass is
\(
\text { (a) Wavelength of light, } \lambda^{\prime}=\frac{\lambda}{\mu}=\frac{6000}{1.5}=4000 Å
\)
When a ray of light falls on a given plate at an angle of incidence \(60^{\circ}\), the reflected and refracted rays are found to be normal to each other. The refractive index of the material of the plate is
(c)
\(
\begin{aligned}
&\text { Refractive index of the material, }\\
&\mu=\frac{\sin i}{\sin r}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\sqrt{3}=1.732
\end{aligned}
\)
The refractive indices of glass and water with respect to air are \(3 / 2\) and \(4 / 3\), respectively. The refractive index of glass with respect to water will be
\(
\begin{aligned}
&\text { (b) Refractive index of glass with respect to water, }\\
&{ }_w \mu_g=\frac{{ }_a \mu_g}{{ }_a \mu_w}=\frac{3 / 2}{4 / 3}=\frac{9}{8}
\end{aligned}
\)
If \({ }_i \mu_j\) represents refractive index when a light ray goes from medium \(i\) to medium \(j\), then the product \({ }_2 \mu_1 \times{ }_3 \mu_2 \times{ }_4 \mu_3\) is equal to
\(
\text { (c) }{ }_2 \mu_1 \times{ }_3 \mu_2 \times{ }_4 \mu_3=\frac{\mu_1}{\mu_2} \times \frac{\mu_2}{\mu_3} \times \frac{\mu_3}{\mu_4}=\frac{\mu_1}{\mu_4}={ }_4 \mu_1=\frac{1}{{ }_1 \mu_4}
\)
When light is refracted into a medium from vacuum
(c) The frequency of a light wave, as it travels from one medium to another, always remains unchanged, while wavelength decreases.
Decrease in the wavelength of light entering a medium of refractive index \(\mu\) is given by
\(
\lambda_M=\frac{\lambda}{\mu}
\)
where \(\lambda_M=\) wavelength in medium
\(
\lambda=\text { wavelength in vacuum }
\)
\(\mu=\) refractive index
A spot is placed on the bottom of a slab made of transparent material of refractive index 1.5. The spot is viewed vertically from the top when it seems to be raised by 2 cm. Then, the height of the slab is
(c)
\(
\text { Let the real depth (height of the slab) be } h \mathrm{~cm} \text {. }
\)
\(
\text { The apparent depth is given as } h-2 \mathrm{~cm} \text { because the spot appears to be raised by } 2 \mathrm{~cm} \text {. }
\)
\(
\text { Apparent depth }=\frac{\text { Real depth }}{\text { Refractive index }}
\)
\(
h-2=\frac{h}{1.5}
\)
\(
h=6 \mathrm{~cm}
\)
The height of the slab is 6 cm.
An air bubble inside a glass slab \((\mu=1.5)\) appears 6 cm when viewed from one side and 4 cm when viewed from the opposite side. The thickness of the slab is
(c) The relationship between the actual depth d actual and the apparent depth can be expressed using the formula:
\(
d_{\text {actual }}=d_{\text {apparent }} \times \mu
\)
where \(\mu\) is the refractive index of the medium (in this case, glass).
Using the formula for the first side:
\(
d_{1_{\text {actual }}}=d_{1_{\text {apparent }}} \times \mu
\)
Substituting the values:
\(
d_{1_{\text {actual }}}=6 \mathrm{~cm} \times 1.5=9 \mathrm{~cm}
\)
Using the formula for the opposite side:
\(
d_{2_{\text {actual }}}=d_{2_{\text {apparent }}} \times \mu
\)
Substituting the values:
\(
d_{2_{a c t u a l}}=4 \mathrm{~cm} \times 1.5=6 \mathrm{~cm}
\)
The total thickness ( \(T\) ) of the glass slab is the sum of the actual depths from both sides:
\(
T=d_{1_{\text {actual }}}+d_{2_{\text {actual }}}
\)
Substituting the values:
\(
T=9 \mathrm{~cm}+6 \mathrm{~cm}=15 \mathrm{~cm}
\)
An under water swimmer is at a depth of 12 m below the surface of water. A bird is at a height of 18 m from the surface of water, directly above his eyes. For the swimmer the bird appears to be at a distance from the surface of water equal to (Refractive index of water is \(4 / 3\) )
(a) The real height of the bird from the surface of water is \(h=18 \mathrm{~m}\). The refractive index of water is \(\mu=\frac{4}{3}\).
When light travels from a rarer medium (air) to a denser medium (water), the apparent depth/height is related to the real depth/height by the refractive index.
The apparent height \(h^{\prime}\) is given by \(h^{\prime}=\mu \times h\).
Here, \(\mu\) is the refractive index of the denser medium (water) with respect to the rarer medium (air).
\(
h^{\prime}=\frac{4}{3} \times 18 \mathrm{~m}=24 \mathrm{~m}
\)
The bird appears to be at a distance of 24 m from the surface of the water for the swimmer.
A vessel of depth \(2 d \mathrm{~cm}\) is half filled with a liquid of refractive index \(\mu_1\) and the upper half with a liquid of refractive index \(\mu_2\). The apparent depth of the vessel when seen perpendicularly from above is
\(
\text { (b) Apparent depth, } h^{\prime}=\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}=\frac{d}{\mu_1}+\frac{d}{\mu_2}=d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)
\)
Explanation:
The total depth of the vessel is \(2 d \mathrm{~cm}\).
The lower half is filled with a liquid of refractive index \(\mu_1\).
The upper half is filled with a liquid of refractive index \(\mu_2\).
\(
\text { Apparent Depth }=\frac{\text { Real Depth }}{\text { Refractive Index }}
\)
The total depth is \(2 d \mathrm{~cm}\).
Each liquid fills half the vessel, so the real depth of each layer is \(\frac{2 d}{2}=d \mathrm{~cm}\).
The real depth of the lower layer is \(d\).
The refractive index of the lower layer is \(\mu_1\).
The apparent depth of the lower layer is \(h_1=\frac{d}{\mu_1}\).
The real depth of the upper layer is \(d\).
The refractive index of the upper layer is \(\mu_2\).
The apparent depth of the upper layer is \(h_2=\frac{d}{\mu_2}\).
The total apparent depth is the sum of the apparent depths of both layers.
Total apparent depth \(H=h_1+h_2\).
\(
\begin{aligned}
& \boldsymbol{H}=\frac{d}{\mu_1}+\frac{d}{\mu_2} \\
& \boldsymbol{H}=d\left(\frac{1}{\mu_1}+\frac{1}{\mu_2}\right)
\end{aligned}
\)
Three immiscible transparent liquids with refractive indices \(3 / 2,4 / 3\) and \(6 / 5\) are arranged one on top of another. The depths of the liquids are \(3 \mathrm{~cm}, 4 \mathrm{~cm}\) and 6 cm, respectively. The apparent depth of the vessel is
(a)
\(
\begin{aligned}
d_{\mathrm{app}} & =\frac{d_1}{\mu_1}+\frac{d_2}{\mu_2}+\frac{d_3}{\mu_3}=\frac{3}{3 / 2}+\frac{4}{4 / 3}+\frac{6}{6 / 5} \\
= & 2+3+5=10 \mathrm{~cm}
\end{aligned}
\)
A glass slab is immersed in water. What will be the critical angle for a light ray at glass-water interface? [Take, \({ }_a n_g=1.50,{ }_a n_w=1.33\) and \(\sin ^{-1}(0.887)=62.5\) ]
(c) The refractive index of glass relative to water is given by
\(
{ }_w n_g=\frac{{ }_a n_g}{{ }_a n_w}=\frac{1.50}{1.33}
\)
If \(C\) be the critical angle at glass water interface, then
\(
\begin{aligned}
& \sin C=\frac{1}{{ }_w n_g}=\frac{1.33}{1.50}=0.8867 \\
& C=\sin ^{-1}(0.8867)=62.5^{\circ}
\end{aligned}
\)
The wavelength of light in two liquids \(x\) and \(y\) is \(3500 Å\) and \(7000 Å\) respectively, then the critical angle of \(x\) relative to \(y\) will be
\(
\begin{aligned}
&\text { (c) Critical angle, }\\
&\begin{aligned}
& & \sin C & =\frac{n_2}{n_1}=\frac{\lambda_1}{\lambda_2} \Rightarrow \sin C=\frac{3500}{7000}=\frac{1}{2} \\
\therefore & C =30^{\circ} &
\end{aligned}
\end{aligned}
\)
White light is incident on the interface of glass and air as shown in the figure. If green light is just totally internally reflected, then the emerging ray in air contains
\(
\begin{aligned}
&\text { (a) Critical angle, }\\
&\begin{aligned}
\sin C & =\frac{1}{\mu} \\
C & =\sin ^{-1}\left(\frac{1}{\mu}\right)
\end{aligned}
\end{aligned}
\)
As \(\mu\) decreases with increase in \(\lambda\). Also, yellow, orange and red have higher wavelength than green, so \(\mu\) will be less for these rays. Thus, critical angle for these rays will be high, hence if green is just totally internally reflected, then yellow, orange and red rays will emerge out.
The critical angle of a prism is \(30^{\circ}\). The velocity of light in the medium is
(a) We have, \(\quad \sin \theta_C=\frac{1}{\mu}\)
\(
\mu=\frac{1}{\sin 30^{\circ}}=2
\)
\(
\begin{aligned}
&\text { The velocity of }\\
&\text { light, } \begin{aligned}
v & =\frac{c}{\mu}=\frac{3.0 \times 10^8}{2} \\
& =1.5 \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
\)
A ray of light travelling in a transparent medium falls on a surface separating the medium from air at an angle of incidence of \(45^{\circ}\). The ray undergoes total internal reflection. If \(n\) is the refractive index of the medium with respect to air, select the possible value of \(n\) from the following.
\(
\begin{aligned}
& \text { (d) We have, } \sin i>\sin \theta_C \\
& \qquad \begin{aligned}
\sin 45^{\circ} & >\frac{1}{n} \text { or } \frac{1}{\sqrt{2}}
\end{aligned}>\frac{1}{n} \\
& n>\sqrt{2} \text { or } n>1.414
\end{aligned}
\)
Therefore, possible values of n can be 1.5.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required, if the focal length is to be 20 cm?
(c)
\(
\begin{aligned}
\frac{1}{f} & =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right), \text { we get } \\
\frac{1}{20} & =(1.55-1)\left[\frac{1}{R}-\left(\frac{1}{-R}\right)\right]=0.55\left[\frac{1}{R}+\frac{1}{R}\right]=0.55 \times \frac{2}{R} \\
R & =1.10 \times 20=22 \mathrm{~cm}
\end{aligned}
\)
What is the refractive index of the material of a plano-convex lens, if the radius of curvature of the convex surface is 20 cm and focal length of the lens is 60 cm?
(a)
Given, \(R_1=\infty, R_2=-20 \mathrm{~cm}, f=60 \mathrm{~cm}, \mu=\)?
From Lens maker’s formula, we have
\(
\begin{aligned}
& \quad \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \Rightarrow \quad \frac{1}{60}=(\mu-1)\left(\frac{1}{\infty}-\frac{1}{-20}\right)=\frac{\mu-1}{20} \\
& \therefore \text { Refractive index, } \mu=1+\frac{1}{3}=\frac{4}{3}
\end{aligned}
\)
A convex lens has a focal length of 20 cm. It is used to form an image of an object placed 15 cm from lens. The image is
(d) Focal length \(f=+20 \mathrm{~cm}\) (positive for convex lens).
Object distance \(u=-15 \mathrm{~cm}\) (negative as per convention).
\(
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\
& \frac{1}{20}=\frac{1}{v}-\frac{1}{-15}
\end{aligned}
\)
\(
v=-60 \mathrm{~cm}
\)
Since \(v\) is negative, the image is virtual.
For a virtual image formed by a convex lens, the image is always erect.
\(
\begin{aligned}
m & =-\frac{v}{u} \\
m & =-\frac{-60}{-15} \\
m & =-4
\end{aligned}
\)
Since \(|m|>1\), the image is enlarged.
The image formed is virtual, erect, and enlarged.
Note: When object lies between focus and pole of convex lens, image formed is virtual, erect and enlarged in size.
In the figure, an air lens of radii of curvature 10 cm \(\left(R_1=R_2=10 \mathrm{~cm}\right)\) is cut in a cylinder of glass \((\mu=2 / 3)\). The focal length and the nature of the lens is
(a) Focal length of air lens in glass is given as
\(
\begin{aligned}
\frac{1}{f} & =\left({ }_g \mu_a-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& =\left(\frac{2}{3}-1\right)\left[\frac{1}{10}-\left(-\frac{1}{10}\right)\right] \Rightarrow f=-15 \mathrm{~cm}
\end{aligned}
\)
So, nature of lens is concave type.
An air bubble is contained inside water. It behaves as a
(a) An air bubble contained inside water behaves as a concave lens. This is because when light passes from a higher refractive index (water) to a lower refractive index (air), it diverges, which is the behavior of a concave lens.
A plano-convex lens is made of glass of refractive index 1.5. The radius of curvature of its convex surface is \(R\). Its focal length is
\(
\begin{aligned}
&\text { (c) Focal length of plano-convex lens, }\\
&f=\frac{R}{\mu-1}=\frac{R}{(15-1)}=2 R
\end{aligned}
\)
At what distance from a convex lens of focal length 30 cm an object should be placed, so that the size of image be \((1 / 4)\) th of the object?
\(
\begin{aligned}
&\text { (d) Image is real, because virtual image is enlarged. }\\
&\therefore \quad \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \text { or } \frac{1}{x / 4}+\frac{1}{x}=\frac{1}{30} \quad(\because u=-x)
\end{aligned}
\)
\(
\begin{aligned}
\frac{5}{x} & =\frac{1}{30} \\
x & =150 \mathrm{~cm}
\end{aligned}
\)
Distance of an object from a concave lens of focal length 20 cm is 40 cm. Then linear magnification of the image is
(b) Use the lens formula: \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\).
Substitute the values: \(\frac{1}{v}=\frac{1}{-20}+\frac{1}{-40}\).
Find a common denominator: \(\frac{1}{v}=\frac{-2}{40}+\frac{-1}{40}\).
Simplify: \(\frac{1}{v}=\frac{-3}{40}\).
Solve for \(v: v=-\frac{40}{3} \mathrm{~cm}\).
Use the magnification formula: \(m=\frac{v}{u}\).
Substitute the values: \(m=\frac{-\frac{40}{3}}{-40}\).
Simplify: \(m=\frac{1}{3}\).
The linear magnification of the image is \(\frac{1}{3}\), which is less than 1.
In order to obtain a real image of magnification 2 using converging lens of focal length 20 cm, where should an object be placed?
(d) For real image from converging lens,
\(
m=-2
\)
\(
\begin{aligned}
& \therefore \text { Magnification, } m=\frac{f}{f+u} \Rightarrow-2=\frac{f}{u+f}=\frac{20}{u+20} \\
& \Rightarrow \quad u=-30 \mathrm{~cm}
\end{aligned}
\)
An object is placed at 10 cm from a lens and real image is formed with magnification of 0.5, then the lens is
(b) We have \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \Rightarrow \frac{1}{+5}-\frac{1}{-10}=\frac{1}{f}\left(|v|=\frac{|u|}{2}\right)\)
Focal length, \(f=\frac{10}{3} \mathrm{~cm}\)
Positive sign implies that the lens is convex.
The real image which is exactly equal to the size of an object is to be obtained on a screen with the help of a convex lens of focal length 15 cm. For this, what must be the distance between the object and screen?
(d) To obtain a real image that is exactly equal to the size of the object using a convex lens, the object must be placed at the center of curvature of the lens. The center of curvature is at a distance of twice the focal length from the lens.
Given the focal length of the convex lens, \(f=15 \mathrm{~cm}\).
For the image to be the same size as the object, the object must be placed at the center of curvature, which is at a distance of \(2 f\) from the lens.
Calculate the distance of the center of curvature from the lens: \(2 f=2 \times 15 \mathrm{~cm}=30 \mathrm{~cm}\).
The distance between the object and the screen is twice the distance of the center of curvature, which is \(2 \times 30 \mathrm{~cm}=60 \mathrm{~cm}\).
The distance between the object and the screen must be 60 cm.
A plano-convex lens of curvature of 30 cm and refractive index 1.5 produces a real image of an object kept 90 cm from it. What is the magnification?
(d)
\(
\begin{aligned}
&\text { We have, } \frac{1}{v}-\frac{1}{-90}=\frac{1}{f} \text { and }(1.5-1)\left(\frac{1}{30}\right)=\frac{1}{f}\\
&\begin{array}{ll}
\therefore & v=180 \mathrm{~cm} \\
\text { or } & |m|=\left|\frac{v}{u}\right|=\frac{180}{90}=2.0
\end{array}
\end{aligned}
\)
The minimum distance between an object and its real image formed by a convex lens is
(d) From the lens formula, \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\).
So, \(v=\frac{u f}{u+f}\).
The distance \(\boldsymbol{D}=\boldsymbol{v}-\boldsymbol{u}\).
Substitute \(v: D=\frac{u f}{u+f}-u\).
Simplify: \(D=\frac{u f-u(u+f)}{u+f}=\frac{u f-u^2-u f}{u+f}=\frac{-u^2}{u+f}\).
For a real object and real image, \(u\) is negative and \(v\) is positive. Let \(u^{\prime}=-u\) (object distance magnitude).
Then \(\boldsymbol{D}=\boldsymbol{u}^{\prime}+\boldsymbol{v}\).
Using the lens formula with \(u^{\prime}\) as a positive value: \(\frac{1}{v}-\frac{1}{-u^{\prime}}=\frac{1}{f}\).
\(
\begin{aligned}
& \frac{1}{v}+\frac{1}{u^{\prime}}=\frac{1}{f} \\
& v=\frac{u^{\prime} f}{u^{\prime}-f}
\end{aligned}
\)
The distance \(D=u^{\prime}+v=u^{\prime}+\frac{u^{\prime} f}{u^{\prime}-f}\).
\(
D=\frac{u^{\prime}\left(u^{\prime}-f\right)+u^{\prime} f}{u^{\prime}-f}=\frac{u^2-u^{\prime} f+u^{\prime} f}{u^{\prime}-f}=\frac{u^2}{u^{\prime}-f} .
\)
For a real image, \(u^{\prime}>f\).
\(
\frac{d D}{d u^{\prime}}=\frac{2 u^{\prime}\left(u^{\prime}-f\right)-u^2(1)}{\left(u^{\prime}-f\right)^2}
\)
Set \(\frac{d D}{d u^{\prime}}=0\) to find the minimum.
\(
\begin{aligned}
& 2 u^{\prime}\left(u^{\prime}-f\right)-u^2=0 . \\
& 2 u^2-2 u^{\prime} f-u^2=0 . \\
& u^2-2 u^{\prime} f=0 . \\
& u^{\prime}\left(u^{\prime}-2 f\right)=0 .
\end{aligned}
\)
Since \(u^{\prime} \neq 0\), we have \(u^{\prime}=2 f\).
\(
\begin{aligned}
&\text { Substitute } u^{\prime}=2 f \text { into the expression for } D \text {. }\\
&D_{\min }=\frac{(2 f)^2}{2 f-f}=\frac{4 f^2}{f}=4 f
\end{aligned}
\)
The minimum distance between an object and its real image formed by a convex lens is \(4 f\).
A convex lens of refractive index \(3 / 2 \) has a power of 2.5 D. If it is placed in a liquid of refractive index 2 , the new power of the lens is
\(
\begin{aligned}
&\text { (d) We have, }\\
&\begin{aligned}
& P=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\
& 2.5=\left(\frac{3}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \dots(i)\\
& P^{\prime}=\left(\frac{3 / 2}{2}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \dots(ii) \\
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Solving Eqs. (i) and (ii), we get }\\
&\therefore \quad P^{\prime}=-1.25 \mathrm{D}
\end{aligned}
\)
Two thin lenses, one of focal length +60 cm and the other of focal length -20 cm are put in contact. The combined focal length is
\(
\begin{aligned}
&\text { (d) Focal length of combination, }\\
&\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{60}+\frac{1}{(-20)} \Rightarrow F=-30 \mathrm{~cm}
\end{aligned}
\)
A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of combination is
\(
\begin{aligned}
&\text { (a) Focal length of the combination, }\\
&\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}
\end{aligned}
\)
\(
\begin{aligned}
\frac{1}{F} & =\frac{1}{(+40)}+\frac{1}{(-25)} \\
F & =-\frac{200}{3} \mathrm{~cm} \\
\text { Power, } P & =\frac{100}{F}=\frac{100}{(-200 / 3)}=-1.5 \mathrm{D}
\end{aligned}
\)
Two similar plano-convex lenses are combined together in three different ways as shown in the adjoining figure. The ratio of the focal lengths in three cases will be
(b) In each case, there is a combination of two plano-convex lenses placed close to each other. Focal length of combination \(\frac{1}{F}=\frac{1}{2 f}+\frac{1}{2 f}\) is same in all cases, so ratio of focal length in three cases is \(1: 1: 1\).
A biconvex lens has a focal length \(f\). It is cut into two parts along a line perpendicular to principal axis. The focal length of each part will be
(d)
\(
\begin{aligned}
&\text { For a biconvex lens, assume } R_1=R \text { and } R_2=-R \text {. }\\
&\begin{aligned}
& \frac{1}{f}=(\mu-1)\left(\frac{1}{R}-\frac{1}{-R}\right) \\
& \frac{1}{f}=(\mu-1)\left(\frac{2}{R}\right)
\end{aligned}
\end{aligned}
\)
When cut perpendicular to the principal axis, each half becomes a plano-convex lens.
For a plano-convex lens, one surface is flat, so \(R_2=\infty\).
Let the new focal length be \(f^{\prime}\).
\(
\begin{aligned}
& \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\
& \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{1}{R}\right)
\end{aligned}
\)
From Step 1, \(\frac{1}{f}=(\mu-1) \frac{2}{R}\).
From Step 2, \(\frac{1}{f^{\prime}}=(\mu-1) \frac{1}{R}\).
Therefore, \(\frac{1}{f^{\prime}}=\frac{1}{2 f}\).
This implies \(f^{\prime}=2 f\).
The focal length of each part will be \(2 f\).
A converging lens is used to form an image on a screen. When the upper half of the lens is covered by an opaque screen, then
(d) When upper half of the lens is covered, image is formed by the rays coming from lower half of the lens. Or image will be formed by less number of rays. Therefore, intensity of the image will decrease. But complete image will be formed.
In a plano-convex lens the radius of curvature of the convex lens is 10 cm. If the plane side is polished, then the focal length will be (Refractive index \(=1.5\) )
\(
\begin{aligned}
&\text { (b) Focal length, } F=\frac{R}{2(\mu-1)}\\
&\Rightarrow \quad F=\frac{10}{2(15-1)}=10 \mathrm{~cm}
\end{aligned}
\)
A ray of light is incident on an equilateral glass prism placed on a horizontal table. For minimum deviation which of the following is true?
(b) In minimum deviation position, refracted ray inside an isosceles prism is parallel to the base of the prism.
A ray of light is incident at an angle of \(60^{\circ}\) on one face of a prism of angle \(30^{\circ}\). The ray emerging out of the prism makes an angle of \(30^{\circ}\) with the incident ray. The emergent ray is
(a) Deviation of ray in prism is given by
\(
\delta=i+e-A
\)
So,
\(
e=\delta+A-i=30^{\circ}+30^{\circ}-60^{\circ}=0^{\circ}
\)
So, emergent ray will make an angle of \(90^{\circ}[latex] with the face through which it emerges (Since the angle of emergence is [latex]0^{\circ}\), the emergent ray is perpendicular to the surface.)
When light rays are incident on a prism at an angle of \(45^{\circ}\), the minimum deviation is obtained. If refractive index of the material of prism is \(\sqrt{2}\), then the angle of prism will be
\(
\text { (d) Refractive index of prism, } \mu=\frac{\sin \frac{A+\delta_m}{2}}{\sin \frac{A}{2}}
\)
\(
\begin{array}{lrl}
\text { As, } & i=45^{\circ}=\frac{A+\delta_m}{2} \\
\text { So, } & \frac{\sin 45^{\circ}}{\sin (A / 2)} & =\sqrt{2} \\
\Rightarrow & \frac{1}{2} & =\sin \frac{A}{2} \Rightarrow A=60^{\circ}
\end{array}
\)
A ray of light passes through an equilateral glass prism in such a manner that the angle of incidence is equal to the angle of emergence and each of these angles is equal to \(3 / 4\) of the angle of the prism. The angle of deviation is
(d) Given, \(i=e=\frac{3}{4} A=\frac{3}{4} \times 60^{\circ}=45^{\circ}\)
In the position of minimum deviation,
\(
\begin{aligned}
2 i & =A+\delta_m \\
\delta_m & =2 i-A \\
& =2 \times 45^{\circ}-60=30^{\circ}
\end{aligned}
\)
In a thin prism of glass (refractive index 1.5 ), which of the following relations between the angle of minimum deviations \(\delta_m\) and angle of prism \(r\) will be correct?
\(
\begin{aligned}
&\text { (a) Angle of minimum deviation in prism, }\\
&\begin{array}{rlr}
\delta_m & =(\mu-1) A=(\mu-1)(2 r) & (\because r=A / 2) \\
& =(1.5-1)(2 r)=0.5 \times 2 r=r &
\end{array}
\end{aligned}
\)
The angle of minimum deviation for a prism of refractive index 1.5 is equal to the angle of the prism. The angle of prism is (Given, \(\cos 41^{\circ}=0.75\) )
(d) Refractive index of prism, \(\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}\)
Substituting \(A=\delta_m\) and \(\mu=1.5\), we get
\(
\begin{aligned}
1.5 & =\frac{\sin A}{\sin \left(\frac{A}{2}\right)} \\
\frac{1.5}{2} & =\cos \frac{A}{2} \\
0.75 & =\cos \frac{A}{2}
\end{aligned}
\)
\(
\text { We get, } \quad A=82^{\circ}
\)
Dispersive power depends upon
(b) the material of the prism. Dispersive power is a property of the material of the prism and is defined as the ratio of the difference in the refractive indices for two different wavelengths of light to the refractive index for a central wavelength. It is independent of the angle of the prism or the height of the prism.
A thin prism \(P_1\) with angle \(6^{\circ}\) and made from glass of refractive index 1.54 is combined with another thin prism \(P_2\) of refractive index 1.72 to produce dispersion without deviation. The angle of prism \(P_2\) will be
(b)
\(
\begin{aligned}
&\text { To produce dispersion without deviation, }\\
&\begin{aligned}
& \frac{A^{\prime}}{A}=\frac{\left(\mu_y-1\right)}{\left(\mu_y^{\prime}-1\right)} \Rightarrow \frac{A^{\prime}}{6^{\circ}}=\frac{(1.54-1)}{(1.72-1)} \\
& A^{\prime}=4.5^{\circ}=4^{\circ} 30^{\prime}
\end{aligned}
\end{aligned}
\)
The focal length of a normal eye lens is about
(b) The focal length of a normal eye lens is about 2 cm.
An object is placed at a distance \(u\) from a simple microscope of focal length \(f\). The angular magnification obtained depends
(a) The angular magnification ( \(M\) ) of a simple microscope is given by the formula: \(M=1+D / f\), where \(D\) is the least distance of distinct vision and \(f\) is the focal length of the lens.
Magnifying power of a simple microscope is (when final image is formed at \(D=25 \mathrm{~cm}\) from eye)
(b) When final image is formed at \(D=25 \mathrm{~cm}\) from eye.
In this situation, \(v=-D\)
From lens formula,
\(
\frac{1}{v}-\frac{1}{u}=\frac{1}{f}
\)
We have
\(
\begin{aligned}
\frac{1}{-D}-\frac{1}{(-u)} & =\frac{1}{f} \\
\frac{D}{u} & =1+\frac{D}{f}
\end{aligned}
\)
So, magnifying power \(=\frac{D}{u}=\left(1+\frac{D}{f}\right)\)
In a compound microscope, the intermediate image is
(c) real, inverted and magnified.
Explanation:
In a compound microscope, the objective lens forms an intermediate image that is:
Real: The image is formed by the actual convergence of light rays.
Inverted: The image is upside down compared to the original object.
Magnified: The image is larger than the object.
A compound microscope has two lenses. The magnifying power of one is 5 and the combined magnifying power is 100. The magnifying power of the other lens is
\(
\begin{aligned}
&\text { (b) Magnifying power, }\\
&\begin{aligned}
m & =m_o \times m_e \\
100 & =5 \times m_e \\
m_e & =20
\end{aligned}
\end{aligned}
\)
The length of the compound microscope is 14 cm. The magnifying power for relaxed eye is 25. If the focal length of eye lens is 5 cm, then the object distance for objective lens will be
(a) Length of compound microscope,
\(
\begin{array}{ll}
& L_{\infty}=v_o+f_e \\
\Rightarrow & 14=v_o+5 \\
\therefore & v_o=9 \mathrm{~cm}
\end{array}
\)
Magnifying power of microscope for relaxed eye,
\(
M=\frac{v_o}{u_o} \cdot \frac{D}{f_e} \text { or } 25=\frac{9}{u_o} \cdot \frac{25}{5}
\)
or distance of object for objective, \(u_o=\frac{9}{5}=1.8 \mathrm{~cm}\)
If the focal length of objective and eye lens are 1.2 cm and 3 cm respectively and the object is put 1.25 cm away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope is
(b) When final image is formed at infinity,then magnifying power,
\(
M_{\infty}=-\frac{v_o}{u_o} \times \frac{D}{f_e}
\)
\(
\begin{aligned}
&\text { From }\\
&\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o},
\end{aligned}
\)
\(
\begin{aligned}
\frac{1}{+1.2} & =\frac{1}{v_o}-\frac{1}{(-1.25)} \\
v_o & =30 \mathrm{~cm} \\
\left|M_{\infty}\right| & =\frac{30}{1.25} \times \frac{25}{3}=200
\end{aligned}
\)
The focal length of objective and eye lens of a microscope are 4 cm and 8 cm, respectively. If the least distance of distinct vision is 24 cm and object distance is 4.5 cm from the objective lens, then the magnifying power of the microscope will be
(b) For objective lens,
\(
\begin{gathered}
\frac{1}{f_o}=\frac{1}{v_o}-\frac{1}{u_o} \\
\frac{1}{4}=\frac{1}{v_o}-\frac{1}{(-4.5)} \Rightarrow\left|v_o\right|=36 \mathrm{~cm}
\end{gathered}
\)
\(\therefore\) Magnifying power, \(|M|=\frac{v_o}{u_o}\left(1+\frac{D}{f_e}\right)=\frac{36}{4.5}\left(1+\frac{24}{8}\right)=32\)
If the telescope is reversed, i.e. seen from the objective side, then
(a) If the telescope is reversed i.e., seen from the objective side, then object will appear very small, because in this case magnification becomes reciprocal of initial magnification.
The aperture of a telescope is made large, because to
(a) In a telescope, large aperture of objective helps in improving the brightness (image) by gathering more light from distant object.
In an astronomical telescope, the focal length of the objective lens is 100 cm and of eyepiece is 2 cm. The magnifying power of the telescope for the normal eye is
\(
\begin{aligned}
&\text { (a) Magnification of astronomical telescope for normal eye, }\\
&m=-\frac{f_0}{f_e}=-\frac{100}{2}=-50
\end{aligned}
\)
The focal lengths of the objective and eye lenses of a telescope are respectively, 200 cm and 5 cm. The maximum magnifying power of the telescope will be
\(
\begin{aligned}
&\text { (b) Magnifying power of astronomical telescope, }\\
&m=-\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=-\frac{200}{5}\left(1+\frac{5}{25}\right)=-48
\end{aligned}
\)
The number of lenses in a terrestrial telescope is
(b) In terrestrial telescope, there are three lenses, objective, eye-piece and third is erecting lens, it is placed at \(2 f\) from intermediate image.
Magnifying power of a Galilean telescope is given by
(a) The magnifying power \((M)\) of a Galilean telescope is given by the formula:
\(
M=\frac{f_{\mathrm{o}}}{f_{\mathrm{e}}}\left(1-\frac{f_{\mathrm{e}}}{D}\right)
\)
Here, D is the distance between the two lenses.
In Galilean telescope, the final image formed is
(b) If in an astronomical telescope, the convergent eyepiece is replaced by a divergent lens which is placed in such a way that rays from objective are directed towards its focus, final image will be erect, enlarged and virtual and hence, it is called Gallilean telescope.
Reflecting telescope consists of
(b) A concave mirror of large aperture.
Explanation: Reflecting telescopes use concave mirrors to collect and focus light, whereas convex mirrors disperse light. The “large aperture” part is important because a larger mirror can gather more light, leading to better image quality.
Resolving power of a microscope is given by
(c) The resolving power of a microscope is given by the formula:
\(
\frac{2 \mu \sin \theta}{\lambda}
\)
Where:
\(\boldsymbol{\mu}\) is the refractive index of the medium between the objective lens and the specimen.
\(\sin \theta\) is the sine of half the angle of the cone of light that enters the objective lens from the specimen (related to the numerical aperture).
\(\lambda\) is the wavelength of light used for illumination.
The resolving power of telescope whose lens has a diameter of 1.22 m for a wavelength of \(5000 Å\) is
\(
\begin{aligned}
&\text { (b) Resolving power of telescope }\\
&=\frac{d}{1.22 \lambda}=\frac{1.22}{1.22 \times 5000 \times 10^{-10}}=2 \times 10^6
\end{aligned}
\)
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