Past NEET Entrance Papers

Hint

(b)

In prism, \(r_1+c=A\)
\(
\begin{aligned}
& r_1=90^{\circ}-c \dots(i)\\
& \sin c=\frac{1}{\mu} \Rightarrow \cos c=\frac{\sqrt{\mu^2-1}}{\mu}
\end{aligned}
\)
\(\Rightarrow\) Apply Snell’s law, on incidence surface
\(
\begin{array}{r}
1 \cdot \sin 30^{\circ}=\mu \sin \left(r_1\right) \Rightarrow 1 \times \frac{1}{2}=\mu \times \sin \left(90^{\circ}-c\right) \\
\frac{1}{2}=\mu \times \frac{\sqrt{\mu^2-1}}{\mu}
\end{array}
\)
On squaring \(\frac{1}{4}=\mu^2-1\)
\(
\Rightarrow \mu^2=\frac{5}{4} \Rightarrow \mu=\frac{\sqrt{5}}{2}
\)

Hint

(a) The magnifying power of a telescope when viewing distant objects can be calculated using the formula:
\(
\text { Magnifying Power }(M)=\frac{\text { Focal Length of Objective Lens }}{\text { Focal Length of Eyepiece Lens }} =\frac{\left(f_o\right)}{\left(f_e\right)}
\)
Given in the problem:
Focal Length of Objective Lens, \(f_o=140 \mathrm{~cm}\)
Focal Length of Eyepiece Lens, \(f_e=5.0 \mathrm{~cm}\)
Substituting these values into the formula gives:
\(
M=\frac{140 \mathrm{~cm}}{5.0 \mathrm{~cm}}=28
\)

Hint

(a)

When light undergoes minimum deviation through a prism, the angle of incidence \((i)\) equals the angle of emergence ( \(i^{\prime}\) ). This means that \(i=i^{\prime}\).

Explanation
When light undergoes minimum deviation, the refracted ray is parallel to the base of the prism.
The light ray is symmetrical about the prism’s axis of symmetry.
The angles of refraction are equal ( \(\boldsymbol{r}_1=\boldsymbol{r}_2\) ).
The formula for minimum deviation can be derived using Snell’s law and the geometry of the prism.

The angle of minimum deviation depends on the refractive index of the prism material. The refractive index of the prism material is different for different colors of light. This is because different colors of light travel at different speeds within the prism.

The angle of minimum deviation decreases as the wavelength of the incident light increases. This means that the angle of deviation for red light is the least.

Hint

(b) \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
\(
\frac{1}{v}=-\frac{1}{u}+\frac{1}{f}
\)
Compare with \(y=m x+c\)
\(
\text { slope } m=-1
\)

Graph (b) has negative slope.

Hint

(b)

Key Concept:
The speed of light in a medium is given by \(v=\frac{\boldsymbol{c}}{\boldsymbol{n}}\), where \(\boldsymbol{c}\) is the speed of light in vacuum and \(n\) is the refractive index of the medium.
Time is calculated as distance divided by speed: \(t=\frac{d}{v}\).

Step 1: Calculate the speed of light in the glass
The speed of light in the glass is given by:
\(v=\frac{c}{n}\)
\(v=\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{1.5}\)
\(v=2 \times 10^8 \mathrm{~m} / \mathrm{s}\)

Step 2: Calculate the time taken for light to pass through the glass slab
The time taken is given by:
\(t=\frac{d}{v}\)
\(t=\frac{5 \times 10^{-3} \mathrm{~m}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}\)
\(t=2.5 \times 10^{-11} \mathrm{~s}\)

Solution: The time taken by sunlight to pass through the glass slab is \(2.5 \times 10^{-11}\) seconds.

Hint

(c) Both Statement I and Statement II are correct; hence, the answer is (c) Both Statement I and Statement II are correct.

Explanation:
Statement I: To form a clear image, light rays need to reflect or refract in a predictable manner, which is only possible with regular reflection or refraction.

Statement II: The color we perceive from an object is determined by the wavelengths of light that are reflected from it, which depends on the composition of the incident light.

Key points to remember:
Regular reffection: When light rays bounce off a smooth surface in parallel, creating a clear image.
Diffuse reflection: When light rays scatter in different directions due to an uneven surface, not forming a clear image.
Refraction: Bending of light when it passes from one medium to another.

Hint

(a)

Step 1: Determine the speed of light in air ( \(v_1\) )
The speed of light in air can be calculated using the formula:
\(
v_1=\frac{x}{t_1}
\)
where \(x\) is the distance traveled in air and \(t_1\) is the time taken.

Step 2: Determine the speed of light in the denser medium (\(v_2\))
The speed of light in the denser medium can be calculated using:
\(
v_2=\frac{10 x}{t_2}
\)
where \(10 x\) is the distance traveled in the denser medium and \(t_2\) is the time taken.

Step 3: Relate the speeds to the refractive indices
The refractive index \(n\) of a medium is defined as the ratio of the speed of light in vacuum (or air) to the speed of light in the medium. Thus, we have:
\(n_r=\frac{c}{v_1} \quad\) (for air)
\(n_d=\frac{c}{v_2} \quad\) (for the denser medium)

Step 4: Use the critical angle formula
The critical angle \(\theta_c\) can be calculated using Snell’s law, which states:
\(
\sin \left(\theta_c\right)=\frac{n_r}{n_d}
\)
Substituting the expressions for the refractive indices:
\(
\sin \left(\theta_c\right)=\frac{v_d}{v_r}
\)

Step 5: Substitute the values of \(v_1\) and \(v_2\)
Using the values we calculated for \(v_1\) and \(v_2\) :
\(
\sin \left(\theta_c\right)=\frac{v_2}{v_1}=\frac{\frac{10 x}{t_2}}{\frac{x}{t_1}}
\)

Step 6: Simplify the equation
This simplifies to:
\(
\sin \left(\theta_c\right)=\frac{10 t_1}{t_2}
\)

Step 7: Calculate the critical angle
To find the critical angle, we take the inverse sine:
\(
\theta_c=\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)
\)

Hint

(d) This will be a combination of 3 lenses placed in air
\(
\begin{gathered}
\frac{1}{f_1}=(1.6-1) \frac{1}{\infty}-\frac{1}{20}=\frac{-0.6}{20} \mathrm{~cm} \\
\frac{1}{f_2}=(1.5-1) \frac{1}{20}-\frac{1}{-20}=\frac{0.5}{10} \mathrm{~cm} \\
\frac{1}{f_3}=(1.6-1) \frac{1}{(-20)}-\frac{1}{\infty}=\frac{-0.6}{20} \mathrm{~cm}
\end{gathered}
\)
The combined focal length is given by
\(
\begin{gathered}
\frac{1}{f_{\mathrm{eq}}}=\frac{1}{f_1}+\frac{1}{f_2}+\frac{1}{f_3}=\frac{-0.6}{20}+\frac{0.5}{10}+\frac{-0.6}{20} \\
\frac{1}{f_{\mathrm{eq}}}=\frac{-0.1}{10} \Rightarrow f_{\mathrm{eq}}=-100 \mathrm{~cm}
\end{gathered}
\)

Hint

(a)

\(
\begin{aligned}
& \frac{1}{f_{e q}}=\frac{1}{f_1}+\frac{1}{f_2} \\
& \frac{1}{f_{e q}}=\frac{1}{f}-\frac{1}{f} \\
& \mathrm{f}_{\mathrm{eq}}=\infty
\end{aligned}
\)

Hint

(c) Focal length of a lens is given by,
\(\frac{1}{f}=(\mu-1) \frac{1}{R_1}-\frac{1}{R_2}\), when it is kept in air.
Where \(\mu\) is the refractive index of the lens medium, and \(R_1\) and \(R_2\) are radii of curvature of the curved surfaces.
Since lens is made of three materials so three \(\mu\) and hence three images.

Hint

(a) Ray is falling horizontally, i.e., perpendicular to vertical surface so angle of incidence \(i_1=0\) ; hence, angle of refraction \(r_1=0\)
We know, for any prism \(r_1+r_2=A\).
So, \(r_2=6^{\circ}\)

Again from Snell’s law
\(
\frac{\sin i_2}{\sin r_2}=\mu
\)
\(
\begin{aligned}
&\frac{{i}_2}{{r}_2}=\mu\\
&\text { as, } {i}_2 \text { and } {r}_2 \text { are small, } {i}_2=\mu r_2=1.5 \times 6=9^{\circ}
\end{aligned}
\)

Hint

(a) Angle of incidence is critical angle for green.
Now \(\sin ^{\circ} \mathrm{C}=\frac{1}{\mu} \quad \mu \propto \frac{1}{\lambda}\)
So \(\sin ^{\circ} \mathrm{C} \propto \lambda\)
Wavelengths for colours above green have critical angle more than the given angle, so they will be transmitted.
So yellow, orange and red have more wavelength than green, so they will came out in air.

Hint

(c) When of an equal size image is formed the magnifi cation is unity.
\(
\text { i.e., } v=u
\)
image distance \(=\) object distance
Here, \(v = x\)
but according to lens formula,
\(
\begin{gathered}
\frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\
\frac{2}{v}=\frac{1}{f}
\end{gathered}
\)
\(f=\frac{v}{2}=\frac{x}{2}=\) focal length of the lens is \({x} / 2\).

Hint

(v)

\(
\mu=\frac{C}{\mathrm{~V}}=\frac{\frac{1}{\sqrt{\mu_0 \sigma_0}}}{\frac{1}{\sqrt{1.5 \mu_0 \times 2 \varepsilon_0}}}=\sqrt{3}
\)

Hint

(c) Lens maker’s formula: \(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}+\frac{1}{R_2}\right)\) for a biconvex lens.
Power of a lens: \(P=\frac{1}{f}\), where \(f\) is the focal length in meters.
Step 1: Calculate the focal length using the lens maker’s formula.
Substitute the given values into the lens maker’s formula:
\(\frac{1}{f}=(1.5-1)\left(\frac{1}{0.2}+\frac{1}{0.2}\right)\)

Simplify the equation:
\(\frac{1}{f}=0.5\left(\frac{2}{0.2}\right)\)
\(\frac{1}{f}=0.5 \cdot 10\)
\(\frac{1}{f}=5\)
Calculate the focal length:
\(f=\frac{1}{5} \mathrm{~m}=0.2 \mathrm{~m}\)

Step 2: Calculate the power of the lens.
Use the formula for the power of a lens:
\(P=\frac{1}{f}\)
Substitute the value of \(f\) :
\(P=\frac{1}{0.2}\)
\(P=5 \mathrm{D}\)
Solution: The power of the lens is 5 diopters.

Hint

(c)

\(
\begin{aligned}
&\mu=\frac{C}{u} \Rightarrow u \propto \frac{1}{\mu}\\
&\text { Critical angle }\\
&\begin{aligned}
& {Sin i_c}=\frac{\mu_R}{\mu_D}=\frac{u_D}{u_R}=\frac{1.5}{2}=\frac{3}{4} \\
& i_c=\sin ^{-1}\left(\frac{3}{4}\right) \\
& \sin {i}_c=\frac{\mu_R}{\mu_D}=\frac{u_D}{u_R} \\
& i_c=\sin ^{-1}\left(\frac{3}{4}\right)
\end{aligned}
\end{aligned}
\)

Hint

(d) Power of a lens in diopters is the inverse of its focal length in meters: \(P=\frac{1}{f}\) For a system of lenses, the total power is the sum of the individual powers:
\(
P_{\text {total }}=P_1+P_2+P_3
\)
Calculate the power of each lens and then sum them to find the total power.
Step 1: Calculate the power of the concave lens
Convert the focal length to meters: \(f_1=-25 \mathrm{~cm}=-0.25 \mathrm{~m}\)
Calculate the power: \(P_1=\frac{1}{f_1}=\frac{1}{-0.25}=-4 \mathrm{D}\)

Step 2: Calculate the power of one convex lens
Convert the focal length to meters: \(f_2=40 \mathrm{~cm}=0.4 \mathrm{~m}\)
Calculate the power: \(P_2=\frac{1}{f_2}=\frac{1}{0.4}=2.5 \mathrm{D}\)

Step 3: Calculate the power of the other convex lens
Since both convex lenses have the same focal length, their power is the same:
\(
P_3=P_2=2.5 \mathrm{D}
\)

Step 4: Calculate the total power of the system
Sum the powers of all three lenses: \(\boldsymbol{P}_{\text {total }}=\boldsymbol{P}_1+\boldsymbol{P}_2+\boldsymbol{P}_3=-4+2.5+2.5=1 \mathrm{D}\)

Hint

(c) The speed of light in a medium with refractive index \({n}\) is given by \(v=\frac{c}{n}\). Time is calculated as distance divided by speed: time \(=\frac{\text { distance }}{\text { speed }}\).
Calculate the time it takes for the light to travel through the glass slab and back, and this will be the time delay between the two beams.

Step 1: Calculate the speed of light in the glass slab.
The speed of light in the glass is given by:
\(v=\frac{c}{n}\)
\(v=\frac{3 \times 10^8 \mathrm{~m} / \mathrm{s}}{1.5}\)
\(v=2 \times 10^8 \mathrm{~m} / \mathrm{s}\)

Step 2: Calculate the distance traveled by the light inside the glass slab.
The light travels through the slab and back, so the total distance is twice the thickness of the slab:
\(d=2 t\)
\(d=2 \times 1 \times 10^{-2} \mathrm{~m}\)
\(d=2 \times 10^{-2} \mathrm{~m}\)

Step 3: Calculate the time delay.
The time delay is the time it takes for the light to travel the distance \(d\) at speed \(v\) :
\(\Delta t=\frac{d}{v}\)
\(\Delta t=\frac{2 \times 10^{-2} \mathrm{~m}}{2 \times 10^8 \mathrm{~m} / \mathrm{s}}\)
\(\Delta t=10^{-10} \mathrm{~s}\)

Solution: The time delay between the two beams is \(10^{-10}\) seconds.

Hint

(c) secondary rainbow is due to double internal reflection and is formed above the primary one.

Explanation: A primary rainbow is formed due to one internal reflection within a raindrop, while a secondary rainbow is formed due to two internal reflections, hence it appears above the primary rainbow.

Why other options are incorrect:
1. primary rainbow is due to double internal reflection and is formed above the secondary one:
This is incorrect because the primary rainbow only involves one internal reflection. Additionally, the secondary rainbow is formed above the primary rainbow.
2. primary rainbow is due to double internal reflection and is formed below the secondary one:
This is incorrect for the same reasons as option 1. The primary rainbow is due to a single internal reflection, not double.
4. secondary rainbow is due to a single internal reflection and is formed above the primary one:
This is incorrect. The secondary rainbow is caused by two internal reflections. While it is formed above the primary rainbow, the reason is due to the additional reflection, not a single one.

Hint

(a) For normal adjustment, the distance between the objective and eyepiece is \(f_o+f_e\).
Magnification of a telescope is given by \(M=-\frac{f_o}{f_e}\).
The image formed by an astronomical telescope is inverted.
The aperture of the objective is larger than that of the eyepiece.
Calculate the distance between the lenses and the magnification, then evaluate the statements.

Step 1: Calculate the distance between the objective and eyepiece
The distance is the sum of the focal lengths:
\(d=f_o+f_e\)
\(d=20 \mathrm{~m}+0.02 \mathrm{~m}\)
\(d=20.02 \mathrm{~m}\)

Step 2: Calculate the magnification
The magnification is the ratio of the focal lengths:
\(M=-\frac{f_o}{f_e}\)
\(M=-\frac{20 \mathrm{~m}}{0.02 \mathrm{~m}}\)
\(\quad M=-1000\)

Step 3: Evaluate the statements
(a) The distance between the objective and eyepiece is 20.02 m . This is correct.
(b) The magnification of the telescope is -1000 . This is correct.
(c) The image of the planet is erect and diminished. This is incorrect, the image is inverted.
(d) The aperture of the eyepiece is smaller than that of the objective. This is correct.

Solution: The correct statements are (a), (b), and (d).

Hint

(c) Given \(\mathrm{i}=60^{\circ}\) and \(\mu=\sqrt{3}\)
\(\Rightarrow\) Here, angle of incidence \(\Rightarrow \mathrm{i}=\tan ^{-1}(\mu)\)
Hence, reflected and refracted rays would be perpendicular to each other.

Explanation:

Find the angle of refraction using Snell’s Law.
Snell’s Law: \(\boldsymbol{n}_1 \sin (\boldsymbol{i})=\boldsymbol{n}_{\mathbf{2}} \sin (\boldsymbol{r})\)
Air’s refractive index is approximately 1.
Glass’s refractive index is \(\sqrt{3}\).
\(
\begin{aligned}
& 1 \cdot \sin \left(60^{\circ}\right)=\sqrt{3} \cdot \sin (r) \\
& \sin \left(60^{\circ}\right)=\frac{\sqrt{3}}{2} \\
& \frac{\sqrt{3}}{2}=\sqrt{3} \sin (r) \\
& \sin (r)=\frac{\frac{\sqrt{3}}{2}}{\sqrt{3}} \\
& \sin (r)=\frac{1}{2} \\
& r=30^{\circ}
\end{aligned}
\)
Calculate the angle between the reflected and refracted rays.
The angle of reflection is equal to the angle of incidence, which is \(60^{\circ}\).
The angle between the incident ray and the normal is \(60^{\circ}\).
The angle between the reflected ray and the normal is also \(60^{\circ}\).
The angle between the refracted ray and the normal is \(30^{\circ}\).
The angle between the reflected and refracted rays is \(60^{\circ}+30^{\circ}=90^{\circ}\).
Solution
The angle between the refracted and reflected rays is \(90^{\circ}\).

Hint

(b) Parallel beam of light after refraction from convex lens converge at the focus of convex lens. In question it is given light after refraction pass through concave lens becomes parallel. Therefore light refracted from convex lens virtually meet at focus of concave lens.

According to above ray diagram, \(d=f_A-f_B\)
\(
\begin{aligned}
& =20-5 \\
& =15 \mathrm{~cm}
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
&\text { from Snell’s law }\\
&\begin{aligned}
& \sqrt{3} \sin 30^{\circ}=1 \times \sin r \\
& \Rightarrow r=60^{\circ}
\end{aligned}
\end{aligned}
\)

Hint

(d) With larger aperture of objective lens, the light gathering power in telescope is high. Also, the resolving power or the ability to observe two objects distinctly also depends on the diameter of the objective. Thus objective of large diameter is preferred. Also, with large diameters fainter objects can be observed. Hence it also contributes to the better quality and visibility of images.

Hence, all options are correct.

Hint

(a) \(
\begin{aligned}
&\text { Using lens formula for first refraction from convex lens }\\
&\begin{aligned}
& \frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f} \\
& v_1=?, \mathrm{u}=-60 \mathrm{~cm}, \mathrm{f}=30 \mathrm{~cm} \\
& \frac{1}{v_1}+\frac{1}{60}=\frac{1}{30} \Rightarrow v_1=60 \mathrm{~cm}
\end{aligned}
\end{aligned}
\)

At \(I_1\) here is first image by lens
The plane mirror will produce an image at distance 20 cm to left of it.
For second refraction from convex lens,
\(
\begin{aligned}
& \mathrm{u}=-20 \mathrm{~cm}, \mathrm{v}=?, \mathrm{f}=30 \mathrm{~cm} \\
& \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}+\frac{1}{20}=\frac{1}{30} \\
& \Rightarrow \frac{1}{v}=\frac{1}{30}-\frac{1}{20} \Rightarrow v=-60 \mathrm{~cm}
\end{aligned}
\)
Thus the final image is virtual and at a distance,
\(60-40=20 \mathrm{~cm}\) from plane mirror

Hint

(b) With a spherometer, you can measure the radius of curvature of the curved surface of a plane-convex lens, even if its material and focal length are unknown.

Explanation: A spherometer is specifically designed to measure the curvature of a spherical surface, which is exactly what the curved side of a planoconvex lens has.

Key points:
What a spherometer measures: Radius of curvature
What it cannot directly measure: Focal length or refractive index of the lens material

Hint

(a)

\(
\begin{aligned}
& u=-1.5 f \\
& \frac{1}{u}+\frac{1}{v}=\frac{1}{f} \\
& =\frac{1}{-1.5 f}+\frac{1}{v}=\frac{1}{-f} \\
& \Rightarrow \frac{1}{v}=-\frac{1}{f}+\frac{1}{1.5 f} \\
& \frac{1}{v}=\frac{-1.5+1}{1.5 f} \\
& =\frac{-0.5}{1.5 f} \\
& v=-3 f
\end{aligned}
\)

Hint

(b)

\(
\begin{aligned}
& \sin \theta_C=\frac{1}{\mu} \\
& \mu=\frac{1}{\sin \theta_c} \\
& =\frac{1}{\sin 45^{\circ}} \\
& =\frac{1}{(1 / \sqrt{2})}=\sqrt{2} \\
& \mu=\frac{C}{V} \Rightarrow V=\frac{C}{\mu} \\
& =\frac{3 \times 10^8}{\sqrt{2}} m / \sec
\end{aligned}
\)

Hint

(d)

\(
\begin{aligned}
& P=\frac{100}{f} \\
& \Rightarrow f=\frac{100}{P} \\
& =\frac{100}{10}=10 \mathrm{~cm} \\
& f=\frac{R}{2(\mu-1)} \text { (for equiconvex lens) } \\
& 10=\frac{10}{2(\mu-1)} \\
& (\mu-1)=\frac{1}{2} \\
& \Rightarrow \mu=\frac{1}{2}+1 \\
& =\frac{3}{2}
\end{aligned}
\)

Hint

(b) Since the light emerges normally from the other surface, the angle of, emergence
\(
e=0
\)
For the triangular prism, we know
\(
r_1+r_2=A
\)
But, \(\mathrm{e}=\mathrm{r}_2=0\)
So, \(r_1=A\)
For surface 1,
Snell’s Law is \(\sin i=\mu\). \(\sin r_1\)
\(
\sin i=\mu \cdot \sin A
\)
For small angles \(\theta=\sin \theta\)
So, \({i}=\mu . \mathrm{A}\)

Hint

(b)

\(
\begin{aligned}
&\text { Refractive index } \mu \text { is equal to tangent of Brewster’s angle } \mathrm{i}_{\mathrm{b}}\\
&\begin{aligned}
& \mu=\tan \mathrm{i}_{\mathrm{b}} \\
& 1<\mu<\infty \\
& 1<\tan \mathrm{i}_{\mathrm{b}}<\infty \\
& \tan ^{-1}(1)<\mathrm{i}_{\mathrm{b}}<\tan ^{-1}(\infty) \\
& 45^{\circ}<\mathrm{i}_{\mathrm{b}}<90^{\circ}
\end{aligned}
\end{aligned}
\)

Hint

(d)

For the rainbow to be observed the observer’s back must be facing the sun and it must be raining on the opposite horizon to the sun. When the light ray strikes the surface of the water droplet, it gets refracted because it’s going from a rarer medium to a denser medium. This refracted light is further separated into its observable components due to dispersion because different colors travel at different speeds in a denser medium.

Now if these dispersed rays strike the internal surface of the droplet at an angle greater than the critical angle ( \(48^{\circ}\) for water), they are reflected due to total internal reflection. Now, these reflected rays come out of the water droplet again due to refraction. And since different colors travel at different speeds, the red color emerges on top and the violet on the bottom. This is called the primary rainbow.

There’s also a possibility of the secondary rainbow. It occurs when there are two total internal reflections at the inner surface of the droplet. In the secondary rainbow, the violet appears on the top and the red on the bottom due to two reflections.

Also, the intensity of the secondary rainbow is less than that of the primary rainbow.
A Rainbow can’t be observed when the observer faces the sun.

Hint

(b) \(90^{\circ}\).
Explanation: When the angle of incidence is equal to the critical angle, the refracted ray grazes the boundary between the two media, meaning the angle of refraction is \(90^{\circ}\). This is the condition for total internal reflection to occur.

Why other options are incorrect:
(a) equal to angle of incidence:

This is not true because the angle of refraction is \(90^{\circ}\) when the angle of incidence is equal to the critical angle, not equal to the angle of incidence.
(c) \(180^{\circ}\) :

In total internal reflection, the light ray is reflected back into the same medium, not transmitted through or refracted at \(180^{\circ}\).
(d) \(0^{\circ}\) :

A \(0^{\circ}\) angle of refraction would mean the light ray is traveling straight through the medium, not being reflected back, which is not the case in total internal reflection.

Hint

(c)

Equivalent focal length in air,
\(
\frac{1}{F_1}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}
\)
When glycerin is filled inside, it behaves like a concave lens of focal length (-\(f\))
\(
\Longrightarrow \frac{1}{F_2}=\frac{1}{f}+\frac{1}{f}-\frac{1}{f}
\)
\(
=\frac{1}{f}
\)
\(
\Longrightarrow \frac{F_1}{F_2}=\frac{1}{2}
\)

Hint

(d) When an equiconvex lens is cut into two symmetrical halves along the principal axis, then there will be no change in focal length of the lens.
\(\therefore\) Power of lens, \(P=\frac{1}{f}\)
So, the power of each part will be \(P\).

Hint

(c) Given,
Double convex lens of focal length, \(f=25 \mathrm{~cm}\)
Refractive index of the material, \(\mu=1.5\)
As per the question, the radius of curvature of one surface is twice that of the other.
i.e. \(R_1=R\) and \(R_2=-2 R\)
(according to sign conventions)
Using the lens maker formula
\(
\begin{aligned}
& \frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \text { ‘ we have: } \\
& \frac{1}{25}=(1.5-1)\left[\frac{1}{R}-\left(\frac{1}{-2 R}\right)\right] \\
& \Rightarrow \frac{1}{25}=0.5\left[\frac{3 R}{2}\right] \\
& \Rightarrow \frac{1}{25}=\left[\frac{3 R}{4}\right] \\
& \Rightarrow R=18.75 \mathrm{~cm}
\end{aligned}
\)
\(
R_1=18.75 \mathrm{~cm} \text { and } R_2=37.5 \mathrm{~cm}
\)
Hence, the required radii of the double convex lens are 18.75 cm and 37.5 cm .

Hint

(b) Red light of the visible spectrum is having a maximum wavelength of about 650 nm .

Hint

(b)

For retracing the path shown in figure, light ray should be incident normally on the silvered face. Applying Snell’s law at point \(M\),
\(
\begin{aligned}
& \frac{\sin i}{\sin 30^{\circ}}=\frac{\sqrt{2}}{1} \Rightarrow \sin i=\sqrt{2} \times \frac{1}{2} \\
& \sin i=\frac{1}{\sqrt{2}} \text { i.e., } i=45^{\circ}
\end{aligned}
\)

Hint

(b)

Using mirror formula,
\(
\begin{aligned}
& \frac{1}{f}=\frac{1}{v_1}+\frac{1}{u_1} ;-\frac{1}{15}=\frac{1}{v_1}-\frac{1}{40} \Rightarrow \frac{1}{v_1}=\frac{1}{-15}+\frac{1}{40} \\
& v_1=-24 \mathrm{~cm}
\end{aligned}
\)
When object is displaced by 20 cm towards mirror. Now,
\(
\begin{aligned}
& u_2=-20 \mathrm{~cm} \\
& \frac{1}{f}=\frac{1}{v_2}+\frac{1}{u_2} ; \frac{1}{-15}=\frac{1}{v_2}-\frac{1}{20} \Rightarrow \frac{1}{v_2}=\frac{1}{20}-\frac{1}{15} \\
& v_2=-60 \mathrm{~cm}
\end{aligned}
\)
So, the image will be shift away from mirror by \((60-24) \mathrm{cm}=36 \mathrm{~cm}\).

Hint

(c) Angular magnification of an astronomical refracting telescope is given as
\(
M=\frac{f_o}{f_e}
\)
where, \(f_o\) and \(f_e\) are the focal length of objective and eye-piece, respectively.
From the given relation, it is clear that for large magnification either \(f_o\) has to be larger or \(f_e\) has to be small.
Angular resolution of an astronomical refracting telescope is given as
\(
R=\frac{D}{1.22 \lambda}
\)
where, \(D\) is the diameter of the objective.
Thus, to have large resolution, the diameter of the objective should be large.
So, objective should have large focal length \(\left(f_o\right)\) and large diameter \(D\).

Hint

(d)

(d) When mirror is rotated by \(\theta\) angle reflected ray will be rotated by \(2 \theta\).
For small angle \(\theta\),
\(
\begin{aligned}
& \tan 2 \theta \approx 2 \theta=\frac{y}{x} \\
& \therefore \quad \theta=\frac{y}{2 x}
\end{aligned}
\)

Hint

(a) The condition for dispersion without deviation is given as \((\mu-1) A=\left(\mu^{\prime}-1\right) A^{\prime}\)
Given \(\mu=1.42, A=10^{\circ}, \mu^{\prime}=1.7, A^{\prime}=\) ?
\(
\begin{aligned}
& \text { Given } \mu=1.42, A=10^{\circ}, \mu^{\prime}=1.7, A^{\prime}=\text { ? } \\
& \therefore \quad(1.42-1) \times 10=(1.7-1) A^{\prime} \\
& \quad(0.42) \times 10=0.7 \times A^{\prime} \\
& \text { or } \quad A^{\prime}=\frac{0.42 \times 10}{0.7}=6^{\circ}
\end{aligned}
\)

Hint

(d)

Here, \(\mu_g=\frac{3}{2}, \mu_w=\frac{4}{3}\) Focal length \((f)\) of glass convex lens is given by

\(
\frac{1}{f}=\left(\mu_g-1\right)\left(\frac{2}{R}\right)
\)
or \(\frac{1}{f}=\left(\frac{3}{2}-1\right) \frac{2}{R}=\frac{1}{R}[latex] or [latex]f=R \dots(i)\)
Focal length \(\left(f^{\prime}\right)\) of water filled concave lens is given by
\(
\frac{1}{f^{\prime}}=\left(\mu_w-1\right)\left(-\frac{2}{R}\right) \text { or } \frac{1}{f^{\prime}}=\left(\frac{4}{3}-1\right)\left(-\frac{2}{R}\right)
\)
\(
\begin{aligned}
&=-\frac{2}{3 R}=-\frac{2}{3 f}\\
&\text { [Using eqn. (i)] }
\end{aligned}
\)
Equivalent focal length \(\left(f_{\text {eq }}\right)[latex] of lens system
[latex]
\frac{1}{f_{e q}}=\frac{1}{f}-\frac{2}{3 f}+\frac{1}{f}=\frac{3-2+3}{3 f}=\frac{4}{3 f}
\)
\(
\therefore \quad f_{e q}=\frac{3 f}{4}
\)

Hint

(c) Magnification in the mirror, \(m=-\frac{v}{u}\)
\(
m=-2 \Rightarrow v=2 u
\)
As \(v[latex] and [latex]u\) have same signs so the mirror is concave and image formed is real.
\(
\begin{aligned}
& m=-\frac{1}{2} \Rightarrow v=\frac{u}{2} \Rightarrow \text { Concave mirror and real image. } \\
& m=+2 \Rightarrow v=-2 u
\end{aligned}
\)
As \(v[latex] and [latex]u\) have different signs but magnification is 2 so the mirror is concave and image formed is virtual.
\(
m=+\frac{1}{2} \Rightarrow v=-\frac{u}{2}
\)
As \(v\) and \(u\) have different signs with magnification \(\left(\frac{1}{2}\right)\) so the mirror is convex and image formed is virtual.

Hint

(d) Given, \(i=45^{\circ}, A=60^{\circ}\)
Since the ray undergoes minimum deviation, therefore, angle of emergence from second face, \(e=i=45^{\circ}\)
\(
\begin{aligned}
\therefore \quad \delta_m & =i+e-A=45^{\circ}+45^{\circ}-60^{\circ}=30^{\circ} \\
\mu & =\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{60^{\circ}+30^{\circ}}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} \\
& =\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{1}{\sqrt{2}} \times \frac{2}{1}=\sqrt{2}
\end{aligned}
\)

Hint

(c)

Here \(\mu=1.5\)
\(l=\) length of the slab
\(x=\) position of air bubble
from one side
As per question, total apparent length of slab \(=5+3\)
or \(\frac{x}{\mu}+\frac{(l-x)}{\mu}=8[latex] or [latex]\frac{l}{\mu}=8\)
\(\therefore l=8 \mu=8 \times 1.5=12 \mathrm{~cm}\)

Hint

(b) Here, \(v=-400 \mathrm{~cm}=-4 \mathrm{~m}, u=\infty, f=\) ?

Using lens formula, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) or \(\frac{1}{-4}-\frac{1}{\infty}=\frac{1}{f}\) or \(f=-4 \mathrm{~m}\)
Lens should be concave.
Power of lens \(=\frac{1}{f}=\frac{1}{-4}=-0.25 \mathrm{D}\)

Hint

(b) Here \(f_o=40 \mathrm{~cm}, f_e=4 \mathrm{~cm}\)
Tube length \((l)=\) Distance between lenses
\(
=v_o+f_e
\)
For objective lens,
\(
u_o=-200 \mathrm{~cm}, v_o=?
\)
\(
\frac{1}{v_o}-\frac{1}{u_0}=\frac{1}{f_o} \text { or } \frac{1}{v_o}-\frac{1}{-200}=\frac{1}{40}
\)
\(
\text { or } \frac{1}{v_o}=\frac{1}{40}-\frac{1}{200}=\frac{4}{200} \quad \therefore v_0=50 \mathrm{~cm}
\)
\(
\therefore \quad l=50+4=54 \mathrm{~cm}
\)

Hint

(b)

\(
\begin{aligned}
&\text { For eye-piece lens, }\\
&\begin{aligned}
& m=\frac{f}{f+u}=\frac{h_1}{h_0} \Rightarrow \frac{f_e}{f_e-\left(f_o+f_e\right)}=\frac{I}{L} \\
& \Rightarrow \frac{f_o}{f_e}=-\frac{L}{I}=\text { Magnification of the telescope }
\end{aligned}
\end{aligned}
\)

Hint

(b) As beam of light is incident normally on the face \(A B\) of the right angled prism \(A B C\), so no deviation occurs at face \(A B\) and it passes straight and strikes the face \(A C\) at an angle of incidence, \(i=45^{\circ}\).
For total reflection to take place at face \(A C\),
\(
i>i_c \text { or } \sin i>\sin i_c
\)
where \(i_c\) is the critical angle.
But as here \(i=45^{\circ}\) and \(\sin i_c=\frac{i}{\mu}\)
\(\therefore \quad \sin 45^{\circ}>\frac{1}{\mu}\) or \(\frac{1}{\sqrt{2}}>\frac{1}{\mu}\) or \(\mu>\sqrt{2}=1.414\)

As \(\mu_{\text {red }}(=1.39)<\mu(=1.414)\) while \(\mu_{\text {green }}(=1.44)\) and \(\mu_{\text {blue }}(=1.47)>\mu(=1.414)\), so only red colour will be transmitted through face \(A C\) while green and blue colours will suffer total internal reflection.
So the prism will separate red colour from the green and blue colours as shown in the given figure.

Hint

(c)

Using lens maker’s formula,
\(
\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)
\)
\(
\begin{aligned}
& \Rightarrow \frac{1}{\mathrm{f}_1}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\infty}-\frac{1}{-20}\right) \\
& \Rightarrow \mathrm{f}_1=40 \mathrm{~cm} \\
& \Rightarrow \frac{1}{\mathrm{f}_2}=\left(\frac{1.7}{1}-1\right)\left(\frac{1}{-20}-\frac{1}{+20}\right) \\
& \Rightarrow \mathrm{f}_2=-\frac{100}{7} \mathrm{~cm}
\end{aligned}
\)
Similarly,
\(
\begin{aligned}
& \Rightarrow \frac{1}{\mathrm{f}_3}=\left(\frac{1.5}{1}-1\right)\left(\frac{1}{\infty}-\frac{1}{-20}\right) \\
& \Rightarrow \mathrm{f}_3=40 \mathrm{~cm}
\end{aligned}
\)
Now according to the condition
\(
\begin{aligned}
& \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}+\frac{1}{\mathrm{f}_3} \\
& \Rightarrow \frac{1}{\mathrm{f}_{\mathrm{cq}}}=\frac{1}{40}+\frac{1}{-100 / 7}+\frac{1}{40} \\
& \therefore \quad \mathrm{f}_{\mathrm{eq}}=-50 \mathrm{~cm}
\end{aligned}
\)
Therefore, the focal length of the combination is -50 cm .

Note: For a system of lenses net focal length \(F_{\text {net }}\)
\(
=\frac{1}{f_1}+\frac{1}{f_2}+\ldots+\frac{1}{f_n}
\)
Net power \(P_{\text {net }}=P_1+P_2+\ldots+P_n\)

Hint

(d)
\(
\text { As } \mu=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)} \text { or } \cot \frac{A}{2}=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}
\)
\(
\frac{\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)} \text { or } \sin \left(\frac{\pi}{2}-\frac{A}{2}\right)=\sin \left(\frac{A}{2}+\frac{\delta}{2}\right)
\)
\(
\begin{aligned}
& \frac{\pi}{2}-\frac{A}{2}=\frac{A}{2}+\frac{\delta}{2} \\
& \therefore \quad \delta=\pi-2 A=180^{\circ}-2 A
\end{aligned}
\)

Hint

(d) Magnifying power of a microscope,
\(
m=\left(\frac{L}{f_o}\right)\left(\frac{D}{f_e}\right)
\)
where \(f_o\) and \(f_e\) are the focal lengths of the objective and eyepiece respectively and \(L\) is the distance between their focal points and \(D\) is the least distance of distinct vision. If \(f_o\) increases, then \(m\) will decrease.
Magnifying power of a telescope, \(m=\frac{f_o}{f_e}\)
where \(f_o\) and \(f_e\) are the focal lengths of the objective and eyepiece respectively.
If \(f_o\) increases, then \(m\) will increase.

Hint

(b)

On reflection from the silvered surface, the incident ray will retrace its path, if it falls normally on the surface.
By geometry, \(r=A\)
Applying Snell’s law at surface \(P Q\),
\(
1 \sin i=\mu \sin r
\)
\(
\begin{aligned}
& 1 \sin i=\mu \sin r \\
& \mu=\frac{\sin i}{\sin r}=\frac{\sin 2 A}{\sin A}=2 \cos A
\end{aligned}
\)

Hint

(a) The combination of two lenses 1 and 2 is as shown in figure.

\(
\therefore \quad \frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}
\)
\(
\begin{aligned}
&\text { According to lens maker’s formula }\\
&\begin{gathered}
\frac{1}{f_1}=\left(\mu_1-1\right)\left(\frac{1}{\infty}-\frac{1}{-R}\right)=\frac{\left(\mu_1-1\right)}{R} \\
\frac{1}{f_2}=\left(\mu_2-1\right)\left(\frac{1}{-R}-\frac{1}{\infty}\right)
\end{gathered}
\end{aligned}
\)
\(
=\left(\mu_2-1\right)\left(-\frac{1}{R}\right)=-\frac{\left(\mu_2-1\right)}{R}
\)
\(
\therefore \quad \frac{1}{f}=\frac{\left(\mu_1-1\right)}{R}-\frac{\left(\mu_2-1\right)}{R}
\)
\(
\frac{1}{f}=\frac{\left(\mu_1-\mu_2\right)}{R} ; f=\frac{R}{\left(\mu_1-\mu_2\right)}
\)

Hint

(b) Converging power of cornea,
\(
P_c=+40 \mathrm{D}
\)
Least converging power of eye lens, \(P_e=+20 \mathrm{D}\)
Power of the eye-lens, \(P=P_c+P_e\)
\(
=40 \mathrm{D}+20 \mathrm{D}=60 \mathrm{D}
\)
Power of the eye lens
\(
\begin{aligned}
& P=\frac{1}{\text { Focal length of the eye lens }(f)} \\
& f=\frac{1}{P}=\frac{1}{60 \mathrm{D}}=\frac{1}{60} \mathrm{~m}=\frac{100}{60} \mathrm{~cm}=\frac{5}{3} \mathrm{~cm}
\end{aligned}
\)
Distance between the retina and cornea-eye lens = Focal length of the eye lens
\(
=\frac{5}{3} \mathrm{~cm}=1.67 \mathrm{~cm}
\)

Hint

(a) The reddish appearance of the sun at sunrise and sunset is due to the scattering of light by the atmosphere, where shorter wavelengths like blue are scattered more, leaving the longer wavelengths like red to reach our eyes.

Hint

(d) Different angles as shown in the figure.

\(
\begin{aligned}
& \theta+40^{\circ}=90^{\circ} \\
\therefore \quad & \theta=90^{\circ}-40^{\circ}=50^{\circ}
\end{aligned}
\)

Hint

(a) According to lens maker’s formula
\(
\frac{1}{f}=\left(\frac{\mu_g}{\mu_L}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
\)
where \(\mu_g\) is the refractive index of the material of the lens and \(\mu_L\) is the refractive index of the liquid in which lens is dipped.
As the biconvex lens dipped in a liquid acts as a plane sheet of glass, therefore
\(
\begin{gathered}
f=\infty \Rightarrow \frac{1}{f}=0 \\
\therefore \frac{\mu_g}{\mu_L}-1=0 \text { or } \mu_g=\mu_L
\end{gathered}
\)

This implies that the liquid must have refractive index equal to glass.
Note: A transparent solid is invisible in a liquid of same refractive index because of no refraction.

Hint

(c)

\(
\therefore \quad d=2 f_1+f_2
\)

Hint

(c) Magnifying power, \(m=\frac{f_0}{f_e}=9 \dots(i)\)
where \(f_o\) and \(f_e\) are the focal lengths of the objective and eyepiece respectively
Also, \(f_o+f_e=20 \mathrm{~cm} \dots(ii)\)
On solving (i) and (ii), we get
\(
f_o=18 \mathrm{~cm}, f_e=2 \mathrm{~cm}
\)

Hint

\(
\text { (b) As } \mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}
\)
\(
\begin{aligned}
& \mu=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin A}{\sin \left(\frac{A}{2}\right)}=\frac{2 \sin \left(\frac{A}{2}\right) \cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)} \\
& =2 \cos \left(\frac{A}{2}\right)
\end{aligned}
\)
As \(\delta=i+e-A\)
At minimum deviation, \(\delta=\delta_m, i=e\)
\(
\begin{aligned}
& \therefore \quad \delta_m=2 i-A \text { or } 2 i=\delta_m+A \\
& i=\frac{\delta_m+A}{2}=\frac{A+A}{2}=A \quad\left(\because \delta_m=A(\text { given })\right) \\
& \quad i_{\min }=0^{\circ} \Rightarrow A_{\min }=0^{\circ}
\end{aligned}
\)
Then, \(\mu_{\text {max }}=2 \cos 0^{\circ}=2\)
\(
\because \quad i_{\max }=\frac{\pi}{2} \Rightarrow A_{\max }=\frac{\pi}{2}
\)
Then, \(\mu_{\min }=2 \cos 45^{\circ}=2 \times \frac{1}{\sqrt{2}}=\sqrt{2}\) So refractive index lies between 2 and \(\sqrt{2}\).

Hint

(d)

Here, \(f=-10 \mathrm{~cm}\)
For end \(A, u_A=-20 \mathrm{~cm}\)
Image position of end \(A\),
\(
\begin{aligned}
& \frac{1}{v_A}+\frac{1}{u_A}=\frac{1}{f} \\
& \frac{1}{v_A}+\frac{1}{(-20)}=\frac{1}{(-10)} \text { or } \frac{1}{v_A}=\frac{1}{-10}+\frac{1}{20}=-\frac{1}{20} \\
& v_A=-20 \mathrm{~cm}
\end{aligned}
\)
For end \(B, u_B=-30 \mathrm{~cm}\)
Image position of end \(B\),
\(
\begin{aligned}
& \frac{1}{v_B}+\frac{1}{u_B}=\frac{1}{f} \\
& \frac{1}{v_B}+\frac{1}{(-30)}=\frac{1}{(-10)} \text { or } \frac{1}{v_B}=\frac{1}{-10}+\frac{1}{30}=-\frac{2}{30} \\
& v_B=-15 \mathrm{~cm}
\end{aligned}
\)
Length of the image
\(
=\left|v_A\right|-\left|v_B\right|=20 \mathrm{~cm}-15 \mathrm{~cm}=5 \mathrm{~cm}
\)

Hint

(c) Given, Radius of curvature: \(R=20 \mathrm{~cm}\)
Object distance: \(u=30 \mathrm{~cm}\)
Object height: \({h}_o=2 \mathrm{~cm}\)

Lens maker’s formula: \(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
For a biconvex lens with equal radii: \(\frac{1}{f}=\frac{2(n-1)}{R}\)
Thin lens equation: \(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)
Magnification: \(M=\frac{\boldsymbol{h}_{\boldsymbol{i}}}{\boldsymbol{h}_{\boldsymbol{o}}}=\frac{\boldsymbol{v}}{\boldsymbol{u}}\)
Refractive index of air is \(n=1\)
\(
\frac{1}{f}=(1.5-1)\left(\frac{1}{20}-\frac{1}{-20}\right)
\)
\(
f=20 \mathrm{~cm}
\)
Calculate the image distance
Apply the thin lens equation:
\(
\begin{aligned}
& \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\
& \frac{1}{20}=\frac{1}{v}+\frac{1}{30}
\end{aligned}
\)
\(
v=60 \mathrm{~cm}
\)
\(
\begin{aligned}
&\text { Apply the magnification formula: }\\
&\begin{aligned}
M & =-\frac{v}{u} \\
M & =-\frac{60}{30} \\
M & =-2
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&\text { Apply the magnification formula: }\\
&\begin{aligned}
M & =\frac{h_i}{h_o} \\
-2 & =\frac{h_i}{2} \\
h_i & =-4 \mathrm{~cm}
\end{aligned}
\end{aligned}
\)
Since \(v\) is positive, the image is real.
Since \(M\) is negative, the image is inverted.
The image height is 4 cm .
\(
\text { The image is real, inverted, and has a height of } 4 \mathrm{~cm} \text {. }
\)

Hint

(b) Difference between apparent and real depth of a pond is due to refraction. Other three are due to total internal reflection.

Total internal reflection occurs when light travels from a denser medium to a less dense medium at an angle greater than the critical angle, causing the light to reflect back into the denser medium.

Optical fiber: Light bounces off the walls of the fiber repeatedly due to total internal reflection, allowing light to travel through the fiber.

Mirage on hot summer days: Hot air near the ground acts as a less dense medium compared to the cooler air above it, causing light to bend and create a mirage.

Brilliance of diamond: The high refractive index of diamond causes light to undergo multiple internal reflections, creating its characteristic brilliance.

Apparent depth of a pond: This phenomenon is due to refraction, not total internal reflection. Light bends as it moves from water (denser) to air (less dense), making the pond appear shallower than it actually is.

Hint

(c) Angle of the first prism: \(A_1=15^{\circ}\)
Refractive index of the first prism: \(\mu_1=1.5\)
Refractive index of the second prism: \(\mu_2=1.75\)
Condition: dispersion without deviation

How to solve:
Set the sum of the deviations produced by the two prisms to zero and solve for the angle of the second prism.
Step 1: Calculate the deviation produced by the first prism.
The deviation \(\delta_1\) produced by the first prism is given by:
\(
\begin{aligned}
& \delta_1=\left(\mu_1-1\right) A_1 \\
& \delta_1=(1.5-1) \times 15^{\circ} \\
& \delta_1=0.5 \times 15^{\circ} \\
& \delta_1=7.5^{\circ}
\end{aligned}
\)

Step 2: Calculate the deviation produced by the second prism.
The deviation \(\delta_2\) produced by the second prism is given by:
\(
\begin{aligned}
& \delta_2=\left(\mu_2-1\right) A_2 \\
& \delta_2=(1.75-1) A_2 \\
& \delta_2=0.75 A_2
\end{aligned}
\)

Step 3: Apply the condition for dispersion without deviation.
For dispersion without deviation, the net deviation is zero:
\(
\begin{aligned}
& \delta_1-\delta_2=0 \\
& 7.5^{\circ}-0.75 A_2=0
\end{aligned}
\)

Step 4: Solve for the angle of the second prism \(A_2\).
\(
\begin{aligned}
& 0.75 A_2=7.5^{\circ} \\
& A_2=\frac{7.5^{\circ}}{0.75} \\
& A_2=10^{\circ}
\end{aligned}
\)
Solution: The angle of the second prism should be \(10^{\circ}\).

Hint

(d) The lens formula is \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\).
For a converging beam, the object distance \(u\) is positive.
For a diverging lens, the focal length \(f\) is negative.
Use the lens formula to find the focal length of the lens.

Step 1: Calculate the object distance \(u\).
The object distance is the distance where the rays would have converged without the lens.
\(
u=10 \mathrm{~cm}
\)

Step 2: Apply the lens formula.
Use the lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) with \(v=15 \mathrm{~cm}\) and \(u=10 \mathrm{~cm}\).
\(
\frac{1}{f}=\frac{1}{15}-\frac{1}{10}
\)
\(
f=-30 \mathrm{~cm}
\)
Solution: The focal length of the diverging lens is -30 cm .

Hint

(c) For total internal reflection, \(\sin i>\sin C\) where, \(i=\) angle of incidence, \(C=\) critical angle
But, \(\sin C=\frac{1}{\mu} \quad \therefore \quad \sin i>\frac{1}{\mu} \quad\) or \(\quad \mu>\frac{1}{\sin i}\)
\(
\mu>\frac{1}{\sin 45^{\circ}} \quad\left(i=45^{\circ}(\text { Given })\right)
\)

\(
\begin{aligned}
& \mu \geq \frac{1}{\sin C} \geq \sqrt{2} \geq 1.414 \\
& \Rightarrow \mu=1.50
\end{aligned}
\)

Hint

(c) Focal length of the lens remains same.
Intensity of image formed by lens is proportional to area exposed to incident light from object.
i.e., Intensity \(\propto\) area
or \(\frac{I_2}{I_1}=\frac{A_2}{A_1}\)
Initial area, \(A_1=\pi\left(\frac{d}{2}\right)^2=\frac{\pi d^2}{4}\)
\(
\begin{aligned}
&\text { After blocking, exposed area, }\\
&A_2=\frac{\pi d^2}{4}-\frac{\pi(d / 2)^2}{4}=\frac{\pi d^2}{4}-\frac{\pi d^2}{16}=\frac{3 \pi d^2}{16}
\end{aligned}
\)
\(
\therefore \quad \frac{I_2}{I_1}=\frac{A_2}{A_1}=\frac{\frac{3 \pi d^2}{16}}{\frac{\pi d^2}{4}}=\frac{3}{4}
\)
\(
I_2=\frac{3}{4} I_1=\frac{3}{4} I \left(\because \quad I_1=I\right)
\)
Hence, focal length of a lens \(=f\), intensity of the image \(=\frac{3 I}{4}\)

Hint

(c) Refractive index for medium \(M_1\) is
\(
\mu_1=\frac{c}{v_1}=\frac{3 \times 10^8}{1.5 \times 10^8}=2
\)
Refractive index for medium \(M_2\) is
\(
\mu_2=\frac{c}{v_2}=\frac{3 \times 10^8}{2.0 \times 10^8}=\frac{3}{2}
\)
For total internal reflection, \(\sin i \geq \sin C\)
where \(i=\) angle of incidence, \(C=\) critical angle
But \(\sin C=\frac{\mu_2}{\mu_1}\)
\(
\therefore \sin i \geq \frac{\mu_2}{\mu_1} \geq \frac{3 / 2}{2} \Rightarrow i \geq \sin ^{-1}\left(\frac{3}{4}\right)
\)

Hint

(a)

Angle of prism,
\(
A=r_1+r_2
\)
For minimum deviation
\(
r_1=r_2=r \quad \therefore A=2 r
\)
Given, \(A=60^{\circ}\)
\(
\text { Hence, } r=\frac{A}{2}=\frac{60^{\circ}}{2}=30^{\circ}
\)

Hint

(d) The focal length of the combination
\(
\frac{1}{f}=\frac{1}{f_1}+\frac{1}{f_2}
\)
\(
\begin{aligned}
&\therefore \text { Power of the combinations, }\\
&P=\frac{f_1+f_2}{f_1 f_2}\left(\because P=\frac{1}{f}\right)
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (c) } \frac{\text { size of image }}{\text { size of object }}=\left|\frac{v}{u}\right| \\
& \Rightarrow \text { size of the image }=\frac{1.39 \times 10^9 \times 10^{-1}}{1.5 \times 10^{11}}=0.92 \times 10^{-3} \mathrm{~m} \\
& \text { size of the image }=9.2 \times 10^{-4} \mathrm{~m}
\end{aligned}
\)

Hint

(d)

From pythagorus theorem, Hypotenuse comes out to be 5 cm .
Since, \(\frac{1}{\mu}=\frac{\sin i}{\sin 90^{\circ}}\)
\(
\mu=\frac{1}{\sin i}=\frac{5}{3}
\)
Speed, \(v=\frac{c}{\mu}=\frac{3 \times 10^8}{5 / 3}=1.8 \times 10^8 \mathrm{~m} / \mathrm{s}\)

Hint

(b) By using \(v=n \lambda\)
Here, \(n=2 \times 10^{14} \mathrm{~Hz}\)
\(
\begin{aligned}
& \lambda=5000 \AA=5000 \times 10^{-10} \mathrm{~m} \\
& v=2 \times 10^{14} \times 5000 \times 10^{-10}=10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)
Refractive index of the material,
\(
\mu=c / v=\frac{3 \times 10^8}{10^8}=3
\)

Hint

\(
\text { (a) Apparent depth } \begin{aligned}
& =\frac{\text { real depth }}{\mu}=\frac{3}{1.5} \\
& =2 \mathrm{~cm}
\end{aligned}
\)
As image appears to be raised by 1 cm , therefore, microscope must be moved upwards by 1 cm .

Hint

(d) Focal length of convex lens \(f_1=25 \mathrm{~cm}\)
Focal length of concave lens \(f_2=-25 \mathrm{~cm}\)
Power of combination in dioptres,
\(
P=P_1+P_2=\frac{100}{f_1}+\frac{100}{f_2}=\frac{100}{25}-\frac{100}{25}=0 .
\)

Hint

(c)

\(
\begin{aligned}
&\text { Using law of triangle, we get } \angle r=30^{\circ}\\
&\begin{aligned}
& \mu=\frac{\sin i}{\sin r} \\
& \Rightarrow \sqrt{2} \times \sin 30^{\circ}={sini} \\
& \Rightarrow {sini}=\frac{1}{\sqrt{2}} \\
& \Rightarrow {i}=45^{\circ}
\end{aligned}
\end{aligned}
\)

Hint

(b) Red and green rays emerge from two points, propagating in two different parallel directions.

Hint

(a) Since the lens is equiconvex, the radius of curvature of each half is same, say \(R\). We know from Lens maker’s formula
\(
\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
\)
(considering the lens to be placed in air).
Here \(R_1=R, R_2=-R\) by convention
\(
\therefore \frac{1}{f}=(\mu-1) \frac{2}{R} \Rightarrow(\mu-1) \frac{1}{R}=\frac{1}{2 f}
\)
If we cut the lens along \(X O X^{\prime}\) then the two halves of the lens will be having the same radii of curvature and so, focal length \(f^{\prime}=f\)
But when we cut it along \(Y O Y^{\prime}\) then, we will have
\(
\begin{aligned}
R_1 & =R \text { but } R_2=\infty \\
\therefore \quad \frac{1}{f^{\prime \prime}} & =(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=(\mu-1) \frac{1}{R}=\frac{1}{2 f} \\
\Rightarrow f^{\prime \prime} & =2 f
\end{aligned}
\)

Hint

(b) When refractive index of lens is equal to the refractive index of liquid, the lens behave like a plane surface with focal length infinity.

\(
\frac{1}{f}=\left(\mu-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
\)
\(
\begin{aligned}
&\text { Here } \mu=\frac{\mu_{\text {convexlens }}}{\mu_{\text {liquid }}}=1\\
&\therefore f=\infty
\end{aligned}
\)

Hint

(b) Using the lens formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
Given \(v=d\), for equal size image \(|v|=|u|=d\)
By sign convention \(u=-d\)
\(
\therefore \quad \frac{1}{f}=\frac{1}{d}+\frac{1}{d} \quad \text { or } f=\frac{d}{2}
\)

Hint

(c)

Applying Snell’s law of refraction at \(A\), we get
\(
\begin{aligned}
& \mu=\frac{\sin i}{\sin r}=\frac{\sin 45^{\circ}}{\sin r} \\
\therefore & \sin r=1 / \sqrt{2} \mu \\
\therefore & r=\sin ^{-1}\left(\frac{1}{\sqrt{2} \mu}\right) \dots(i)
\end{aligned}
\)
Applying the condition of total internal reflection at \(B\), we get
\(
i_c=\sin ^{-1}(1 / \mu) \dots(ii)
\)
where \(i_c\) is the critical angle.
Now, \(r+i_c=90^{\circ}=\pi / 2\)
\(
\therefore \quad \sin ^{-1} \frac{1}{\sqrt{2} \mu}=\frac{\pi}{2}-\sin ^{-1} \frac{1}{\mu}
\)
\(
\text { or, } \sin ^{-1} \frac{1}{\sqrt{2} \mu}=\cos ^{-1} \frac{1}{\mu}
\)
\(
\begin{aligned}
& \therefore \quad \frac{1}{\sqrt{2} \mu}=\frac{\sqrt{\mu^2-1}}{\mu} \text { or } \frac{1}{2}=\mu^2-1 \\
& \therefore \mu=\sqrt{3 / 2}
\end{aligned}
\)

Hint

(c) Angular limit of resolution of eye is the ratio of wavelength of light to diameter of eye lens.
Angular limit of resolution of eye
\(
=\frac{\text { Wavelength of light }}{\text { Diameter of eye lens }}
\)
ie, \(\theta=\frac{\lambda}{d} \ldots(i)\)
If \(y\) is the minimum resolution between two objects at distance \(D\) from eye, then
\(
\theta=\frac{y}{D} . . \text { (ii) }
\)
From Eqs. (i) and (ii), we have
\(
\frac{y}{D}=\frac{\lambda}{d}
\)
or \(y=\frac{\lambda D}{d} \ldots(iii)\)
Given, \(\lambda=5000 Å=5 \times 10^{-7} \mathrm{~m}\),
\(
D=50 \mathrm{~m}, d=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}
\)
Substituting in Eq. (iii), we get
\(
\begin{aligned}
& y=\frac{5 \times 10^{-7} \times 50}{2 \times 10^{-3}} \\
& =12.5 \times 10^{-3} \mathrm{~m} \\
& =1.25 \mathrm{~cm}
\end{aligned}
\)

Hint

(a) Optical fibres are based on total internal reflection.
Explanation:
Optical fibres work by guiding light through the core of the fibre by repeatedly reflecting it off the inner walls. This phenomenon is called total internal reflection, which occurs when light strikes the boundary between two mediums with different refractive indices at an angle greater than the critical angle. The light is then trapped within the core and travels along the fibre.

Why other options are incorrect:
Less scattering:
While some scattering of light does occur within an optical fibre, it’s not the fundamental principle behind its operation. Scattering is the random redirection of light as it interacts with particles in the medium. In a well-designed optical fibre, scattering is minimized to ensure efficient transmission of light.

Refraction:
Refraction is the bending of light as it passes from one medium to another. Refraction occurs when light enters the fibre initially, but it’s total internal reflection that keeps the light confined within the core.

Less absorption coefficient:
A low absorption coefficient means the light is not easily absorbed by the fibre material. This is important for signal transmission, but it’s not the defining principle of optical fibre function. Total internal reflection is the primary mechanism that dictates how light travels through the fibre.

Hint

(b) The velocity, intensity, and wavelength of light changes when it propagates from one medium to another. But the frequency of the light does not change. Hence \(\mathrm{n}=\mathrm{n}^{\prime}\).

Hint

\(
\begin{aligned}
&\text { (b)  } \theta \text { is the critical angle. }\\
&\begin{array}{ll}
\therefore & \theta=\sin ^{-1}(1 / \mu)=\sin ^{-1}(3 / 5) \\
\text { or, } & \sin \theta=3 / 5 \\
\therefore & \tan \theta=3 / 4=r / 4 \\
\text { or, } & r=3 \mathrm{~m} .
\end{array}
\end{aligned}
\)

Hint

(c) Total apparent depth,
\(
y=y_1+y_2=5+2=7 \mathrm{~cm} .
\)
If \(x\) is real depth \(=\) thickness of slab, then as
\(
\mu=\frac{\text { real depth }}{\text { apparent depth }}=\frac{x}{y}
\)
or, \(x=\mu y=1.5 \times 7=10.5 \mathrm{~cm}\).

Hint

(b) internal reflection and dispersion.

Explanation: A rainbow is formed when sunlight enters a raindrop, undergoes refraction (bending of light), then internal reflection inside the drop, and finally disperses into its constituent colors as it exits the droplet, creating the colorful spectrum we see.

Hint

(a) The silvered plano convex lens behaves as a concave mirro; whose focal length is given by
\(
\begin{aligned}
&\frac{1}{F}=\frac{2}{f_1}+\frac{1}{f_m}\\
&\text { If plane surface is silvered }\\
&\begin{aligned}
& f_m=\frac{R_2}{2}=\frac{\infty}{2}=\infty \\
& \therefore \frac{1}{f_1}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& =(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right)=\frac{\mu-1}{R} \\
& \therefore \frac{1}{F}=\frac{2(\mu-1)}{R}+\frac{1}{\infty}=\frac{2(\mu-1)}{R}
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
&F=\frac{R}{2(\mu-1)}\\
&\text { Here, } R=10 \mathrm{~cm}, \mu=1.5\\
&\therefore F=\frac{10}{2(1.5-1)}=10 \mathrm{~cm}
\end{aligned}
\)

Hint

(b) Concept: The mirror’s length is equal to the height of the man standing in front of it at a distance of \(d\). We trace different regions of the observer’s body using the law of reflection, which states that the angle of incidence equals the angle of reflection. The angle is the angle of incidence between the man’s legs and his line of vision. Because the spectator can see his legs, he can also see any part of his body between them. Minimum required length of the mirror is half the height of object
\(
=\frac{6 \text { feet }}{2}=3 \text { feet }
\)

Hint

(a)  First, calculate the image distance \(v\) using the lens formula. Then, calculate the magnification of the objective lens \({M}_o\). Finally, calculate the magnification of the eyepiece \(M_e\) using the total magnification formula.

Step 1: Calculate the image distance \(v\) using the lens formula.
The lens formula is:
\(\frac{1}{f_o}=\frac{1}{v}-\frac{1}{u}\)
Substitute the given values:
\(\frac{1}{\frac{1}{4}}=\frac{1}{v}-\frac{1}{-\frac{1}{3.8}}\)
\(4=\frac{1}{v}+3.8\)
\(
v=5 \mathrm{~cm}
\)
Step 2: Calculate the magnification of the objective lens \(M_o\).
The magnification of the objective lens is:
\(M_o=\frac{v}{u}\)
Substitute the values:
\(M_o=\frac{5}{\frac{1}{3.8}}\)
\({M}_o=5 \times 3.8\)
\(M_o=19\)

Step 3: Calculate the magnification of the eyepiece \(M_c\).
The total magnification is:
\(\quad M=M_o \times M_e\)
Substitute the values:
\(95=19 \times M_e\)
Solve for \({M}_e\) :
\(M_e=\frac{95}{19}\)
\(M_e=5\)

Solution: The magnification of the eyepiece is 5 .

Hint

(b) The refractive index \((\mu)\) of a prism of angle A , and minimum deviation \(\delta_m\) is given by
\(
\begin{aligned}
&\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin A / 2}\\
&\text { Given, } \mu=\cot \frac{A}{2}\\
&\begin{aligned}
& \therefore \cot \frac{A}{2}=\frac{\sin \frac{\left(A+\delta_m\right)}{2}}{\sin A / 2} \\
& \Rightarrow \frac{\cot A / 2}{\sin A / 2}=\frac{\sin \frac{\left(A+\delta_m\right)}{2}}{\sin A / 2} \\
& \Rightarrow \cos \frac{A}{2}=\sin \left(\frac{A+\delta_m}{2}\right) \\
& \therefore \sin \left(90^{\circ}-\frac{A}{2}\right)=\sin \left(\frac{A+\delta_m}{2}\right) \\
& \Rightarrow 90^{\circ}-\frac{A}{2}=\frac{A+\delta_m}{2} \\
& \Rightarrow 180^{\circ}-A=A+\delta_m \\
& \Rightarrow \delta_m=180^{\circ}-2 A=\pi-2 A
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (b) } R_1=60 \mathrm{~cm}, R_2=\infty, \mu=1.6 \\
& \frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \frac{1}{f}=(1.6-1)\left(\frac{1}{60}\right) \Rightarrow f=100 \mathrm{~cm} .
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (b) Frequency }(\mathrm{n})=100 \mathrm{~Hz} \\
& v=\mathrm{n} \lambda \\
& \lambda=\frac{3 \times 10^8}{100}
\end{aligned}\\
&\text { [where, velocity of light }(\mathrm{v})=3 \times 10^8 \mathrm{~m} / \mathrm{s} \text { ] }\\
&\lambda=3 \times 10^6 \mathrm{~m} .
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) Angle of minimum deviation }\\
&\begin{aligned}
& \mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)} \Rightarrow \sqrt{3}=\frac{\sin \left(\frac{60^{\circ}+\delta_m}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} \\
& \Rightarrow \sin \left(30^{\circ}+\frac{\delta_m}{2}\right)=\frac{\sqrt{3}}{2} \Rightarrow 30^{\circ}+\frac{\delta_m}{2}=60^{\circ} \\
& \Rightarrow \delta_m=60^{\circ} .
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
& \text { (a) } n>\frac{\sin r}{\sin i} \\
& \text { i.e., } n>\frac{\sin 90^{\circ}}{\sin 45^{\circ}} \Rightarrow n>\sqrt{2}
\end{aligned}
\)

Hint

(a) For lens, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(
u=-30, f=20, v=60 \mathrm{~cm}
\)
To have an upright image of the object, coincide with it, image should tend to form at centre of curvature of convex mirror. Therefore, the distance of convex mirror from the lens
\(
=60-10=50 \mathrm{~cm}
\)

Hint

(a) \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)

Since the refractive index of violet colour \(\left(\mu_v\right)\) is greater than the refractive index of red colour \(\left(\mu_r\right)\), therefore focal length of violet colour is less than the focal length of red colour or in other words, \(f_v<f_r\).

Hint

(d) Length of astronomical telescope \(\left(f_o+f_e\right)=44\) cm and ratio of focal length of the objective lens to that of the eye piece \(\frac{f_o}{f_e}=10\)
From the given ratio, we find that \(f_o=10 f_e\).
Therefore \(10 f_e+f_e=44\) or \(f_e=4 \mathrm{~cm}\) and focal length of the objective \(\left(f_o\right)\)
\(
=44-f_e=44-4=40 \mathrm{~cm}
\)

Hint

(a) For e.m. wave entering from air to glass slab \((\mu)\), frequency remains \(n\), wavelength, \(\lambda^{\prime}-\frac{\lambda}{\mu}\) and velocity, \(v^{\prime}=\frac{v}{\mu}\)
Note: When electromagnetic wave enters in other medium, frequency remains unchanged while wavelength and velocity become \(\frac{1}{\mu}\) times.

Hint

(a) According to Snell’s law,
\(
\mu=\frac{\sin i}{\sin r^{\prime}}=\frac{\sin i}{\sin \left(90^{\circ}-r\right)}=\frac{\sin i}{\cos r}
\)
From law of reflection, \(i=r\)
\(
\therefore \quad \mu=\frac{\sin r}{\cos r}=\tan r
\)
Critical angle \(=\sin ^{-1}(\mu)=\sin ^{-1}(\tan r)\).

Hint

(b) Focal length \(f_1=80 \mathrm{~cm}\) and \(f_2=-50 \mathrm{~cm}\) (Minus sign due to concave lens)
Power of the combination ( \(P\) )
\(
=P_1+P_2=\frac{100}{f_1}+\frac{100}{f_2}=\frac{100}{80}-\frac{100}{50}=-0.75 \mathrm{D}
\)

Hint

(b) For a convex lens, \(f_R>f_V\) or \(f_V<f_R\). For a concave lens, focal length is negative.
\(\therefore \quad\left|F_V\right|<\left|F_R\right|\) or \(F_V>F_R\) as the smaller negative value is bigger.

Hint

(d) Total thickness \(=t\); Refrative index \(=\mu\)
Speed of light in Glass plate \(=\frac{c}{\mu}\)
Time taken \(=\frac{t}{\left(\frac{c}{\mu}\right)}=\frac{\mu t}{c}\).
where, \(t=\) thickness of glass plate

Note:
\(
\begin{aligned}
&\text { Speed of light in medium }\\
&v=\frac{\text { Speed of light in vaccum }}{\mu \text { of medium }}
\end{aligned}
\)

Hint

(c) Thickness of glass plate \((t)=6 \mathrm{~cm}\);
Distance of the object \((u)=8 \mathrm{~cm}\).
And distance of the image \((v)=12 \mathrm{~cm}\).
Let \(x=\) Apparent position of the silvered surface in cm.
Since the image is formed due to reflection at the silvered face and by the property of mirror image Distance of object from the mirror \(=\) Distance of image from the mirror or, \(x+8=12+6-x \Rightarrow x=5 \mathrm{~cm}\).
Therefore, refractive index of glass
\(
=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{6}{5}=1.2 .
\)

Hint

(c) By displacement method, size of object
\(
(O)=\sqrt{I_1 \times I_2} .
\)
Therefore area of source of light \((A)=\sqrt{A_1 A_2}\)

Hint

\(
\begin{aligned}
&\text { (c) Time of exposure } t \propto(f \text { – number })^2\\
&\therefore \frac{t}{\left(\frac{1}{200}\right)}=\left(\frac{5.6}{2.8}\right)^2=4 \text { or } t=0.02 \mathrm{~s}
\end{aligned}
\)

Hint

(c) If two or more lenses are combined together in such a way that this combination produces images of different colours at the same point and of the same size, then this property is called ‘achromatism’. Concave and convex type of lenses are used for this combination.

Hint

(b) Angle between two mirrors \((\theta)=60^{\circ}\). Number of images formed by the inclined mirror
\(
(n)=\frac{360^{\circ}}{\theta}-1=\frac{360^{\circ}}{60^{\circ}}-1=6-1=5
\)
When two plane mirrors are inclined to each other at an angle \(\theta\) then number of images ( \(n\) ) formed \(n=\frac{360}{\theta}\) when object is placed asymmetrically, and \(n=\frac{360}{\theta}-1\) when object is placed symmetrically.

Hint

(a) According to Rayleigh, the amount of scattering is inversely proportional to the fourth power of the wavelength.

Hint

(a) In order to cut off all the light coming out of water surface, angle C should be equal to critical angle.

\(
\begin{aligned}
& \text { i.e. } \sin C=\frac{1}{\mu}=\frac{1}{5 / 3}=\frac{3}{5} \\
& \therefore \quad \tan C=3 / 4 \\
& \text { Now, } \tan C=\frac{r}{h} \\
& r=h \tan C=4 \times \frac{3}{4}=3 \mathrm{~m} \\
& \text { Diameter of disc }=2 r=6 \mathrm{~m}
\end{aligned}
\)

Hint

(a) Magnifying power of telescope,
\(
M=f_o / f_e
\)
To produce largest magnifications \(f_o>f_e\) and \(f_o\) and \(f_e\) both should be positive (convex lens).
Therefore \(f_e=+15 \mathrm{~cm}\).

Hint

\(
\text { (a) When the angle of prism is small, } \delta=(\mu-1) \mathrm{A}
\)

Explanation:

The refractive index of a prism is given by,
\(
\mu=\frac{\sin \left(\frac{A+\delta}{2}\right)}{\sin \left(\frac{A}{2}\right)}
\)
For a very thin prism, we can assume that \(\sin \theta \approx \theta\)
\(
\therefore \mu=\frac{\frac{A+\delta}{2}}{\frac{A}{2}}
\)
or, \(\mu=\frac{A+\delta}{A}\)
or, \(\delta=\mu A-A\)
or, \(\delta=(\mu-1) A\) is the required angle of deviation.

Hint

(d) For red light, focal length of lens is maximum because \(\mathrm{f} \propto \lambda\) and \(\lambda\) is maximum for red light.

Hint

\(
\begin{aligned}
& \text { (c) } v_g=\frac{c}{\mu}=\frac{3 \times 10^8}{\frac{3}{2}}=2 \times 10^8 \mathrm{~m} / \mathrm{s} \\
& t=\frac{x}{v_g}=\frac{4 \times 10^{-3}}{2 \times 10^8}=2 \times 10^{-11} \mathrm{~s}
\end{aligned}
\)

Hint

(c) From \(\mu=\frac{\mathrm{c}}{\mathrm{v}}=\frac{\mathrm{n} \lambda_{\mathrm{v}}}{\mathrm{n} \lambda_{\mathrm{m}}}, \lambda_{\mathrm{m}}=\frac{\lambda_{\mathrm{v}}}{\mu}\)
Here, \(\mathrm{c}=\) velocity of light in medium and \(\mathrm{v}=\) velocity of light in vacuum;
\(\mu=\) refractive index of the medium.
Hence, wavelength in medium \(\left(\lambda_{\mathrm{m}}\right)<\lambda_{\mathrm{a}}\)
( \(\because \mu>1\), given)
So, the required wavelength decreases.

Hint

\(
\text { (a) } \lambda_g=\frac{\lambda_a}{\mu}=\frac{5460}{1.5}=3640 Å
\)

Hint

(d) Ray optics is valid when characteristic dimensions are much larger than the wavelength of light.

Explanation: Ray optics is a model that simplifies light behavior by treating it as rays traveling in straight lines. This approximation works well when the objects interacting with the light are significantly larger than the wavelength of the light, allowing us to ignore diffraction effects. If the dimensions are of the same size or smaller as the wavelength, then diffraction effects become significant, and ray optics is no longer accurate.

Why other options are incorrect:
(a) of the same order as the wavelength of light:

If the dimensions are of the same order as the wavelength of light, diffraction effects become prominent, making ray optics invalid.
(b) much smaller than the wavelength of light:

When dimensions are much smaller than the wavelength of light, we are in the realm of wave optics, where diffraction dominates. Ray optics is not applicable in this case.
(c) of the order of one millimetre:

While one millimeter can be considered large compared to the wavelength of visible light (around 0.4-0.7 micrometers), ray optics is typically valid for dimensions much larger than just a few millimeters. For example, a millimetre is 1000 times the wavelength of visible light. To be truly considered valid in ray optics, the dimensions would need to be many orders of magnitude larger than the wavelength.

Hint

(c) Lens maker’s formula: \(\frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\), where \(f\) is the focal length, \(n\) is the refractive index of the lens material relative to the surrounding medium, and \(R_1\) and \(\boldsymbol{R}_{\mathbf{2}}\) are the radii of curvature of the lens surfaces.

How to Solve?
Use the lens maker’s formula to find the focal length of the lens in air and in the liquid, then divide the two equations to eliminate the radii of curvature.
Step 1: Find the ratio of focal lengths in air and liquid
Write the lens maker’s formula for the lens in air:
\(\frac{1}{f_a}=\left(n_g-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
Write the lens maker’s formula for the lens in the liquid:
\(\frac{1}{f_l}=\left(\frac{n_g}{n_l}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
Divide the first equation by the second equation:
\(\frac{f_l}{f_a}=\frac{n_g-1}{\frac{n_g}{n_l}-1}\)

Step 2: Calculate the focal length in the liquid
Substitute the given values into the equation:
\(\frac{f_l}{2}=\frac{1.5-1}{\frac{1.5}{1.25}-1}\)
\(
\frac{f_l}{2}=2.5
\)
\(
f_l=5 \mathrm{~cm}
\)
The focal length of the lens when immersed in the liquid is 5 cm.

Hint

(d) For series combination of lens
\(
\begin{aligned}
& p_{\mathrm{eff}}=p_1+p_2+p_3+p_4=4 p \\
& m_{\mathrm{eff}}=m_1 \times m_2 \times m_3 \times m_4=m^4
\end{aligned}
\)

Hint

(c)

\(
\begin{aligned}
& m=\frac{L}{f_o} \times \frac{D}{f_e} \\
& =\frac{40}{2} \times \frac{25}{4} \\
& m=125
\end{aligned}
\)