NEET Practice Q & A

Hint

(b) When white light is passed through a prism, violet light shows the maximum deviation. This is because violet light has the shortest wavelength and therefore bends (refracts) the most when passing through the prism.

Hint

(b) Only a convex mirror consistently forms erect and diminished images.

Explanation:
The object height (man’s height) is \(h_o=6 \mathrm{~m}\).
The image height is \(h_i=2 \mathrm{~m}\).
The image is erect.
Magnification \(m=\frac{h_i}{h_o}\).
Convex mirrors always form erect and diminished images.
Plane mirrors form erect images of the same size.
Concave mirrors form erect and magnified images when the object is between the pole and focal point.
Magnification \(m=\frac{h_i}{h_o}=\frac{1}{3}\).
\(
\text { The image is erect and diminished }(m<1) \text {. }
\)

Hint

(c) The field of view in convex mirror is more as compared to plane or concave mirror. This is why it is used as rear view mirror in vehicles.

Explanation: A convex mirror has the largest field of view because its outward curvature causes light rays to diverge, allowing it to capture a wider area of the surrounding environment compared to a plane or concave mirror.

Hint

(a) Sky appears blue due to scattering of light. In absence of atmosphere, no scattering will occur. Therefore sky will be seen black.

Hint

(d) An air bubble in water shines due to total internal reflection at its outer surface. Here, light is propagating from denser medium (water) to rarer medium (air) and if \(i>\theta_c\), then total internal reflection takes place at the surface of the bubble. Hence, air bubble appears to be shinning in water.

Hint

(c) atmospheric particles scatter red light more than the blue light.

Explanation: The blue color of the sky is due to the scattering of light by atmospheric particles, where shorter wavelengths like blue are scattered more than longer wavelengths like red. If atmospheric particles were to scatter red light more than blue light, the sky would appear red instead.

Hint

(b) A passenger in an aeroplane may see a primary and a secondary rainbow as concentric circles.

Explanation: From an airplane, the rainbow appears as a full circle because there’s no obstruction from the ground like on the Earth’s surface, allowing you to see both the primary and secondary rainbows concentrically.

Hint

(b) The focal length of a lens is inversely proportional to the refractive index of the lens and the refractive index of the lens is inversely proportional to the square of wavelength. Therefore, the focal length is directly dependent on wavelength; it increases when the wavelength is increased.

Hint

(b) If a transparent plastic bag filled with air is completely immersed in water, it behaves as a convergent lens.

Explanation:
When the bag is filled with air, it acts as a concave lens because air has a lower refractive index than the plastic. However, when submerged in water, the refractive index of the surrounding water is closer to that of the plastic, causing the light to bend differently and converge. This makes the bag behave like a convergent lens.

Hint

(c) When light reflects at the interface from air to glass, it is moving from a medium with a lower refractive index (air) to a medium with a higher refractive index (glass). In this case, the reflected wave undergoes a phase change of \(\pi\) radians (or 180 degrees). This phenomenon is similar to a wave on a string being inverted when it hits a fixed point. The phase change is caused by the difference in the refractive indices of the two media.

Note: When a light wave travelling from a medium of refractive index \(\mu_1[latex] towards a medium of refractive index [latex]\mu_2\) undergo a \(180^{\circ}\) phase change on reflection when \(\mu_2>\mu_1\). No phase change occurs in the reflected light if \(\mu_2<\mu_1\).

Hint

(b) \(|m|=\frac{f_o}{f_e}\)
When \(f_e\) is halved, so \(m^{\prime}=\frac{2 \times f_o}{f_e}=2 m\)

Hint

(c) The radius of the glass sphere is \(R=6 \mathrm{~cm}\).
The refractive index of the glass is \(\mu_1=1.5\).
The object is placed at the center of the sphere.
Rays originating from the center of a sphere strike the surface normally.
Rays striking a surface normally pass undeviated.
Since the object is at the center of the sphere, all light rays from the object travel along the radius.
These rays strike the surface of the sphere at a \(90^{\circ}\) angle (normally).
When light rays strike a surface normally, they do not refract or deviate. Therefore, the rays continue in a straight line without bending.
Since the rays pass undeviated, they appear to originate from the same point as the object.
Thus, the virtual image is formed at the center of the sphere.
The distance of the image from the surface is equal to the radius of the sphere.
Distance \(=R=6 \mathrm{~cm}\).
The distance of the virtual image from the surface of the sphere is 6 cm.

Hint

(d) The image is thrice the original size, so the magnification is \(\pm 3\). The object distances are \(u_1=8 \mathrm{~cm}\) and \(u_2=16 \mathrm{~cm}\).
The magnification formula for a lens is \(m=\frac{f}{f+u}\).
For the object at \(u_1=8 \mathrm{~cm}\), the image is virtual and magnified, so \(m=+3\). Using the magnification formula: \(3=\frac{f}{f-8} \dots(i)\).
For the object at \(u_2=16 \mathrm{~cm}\), the image is real and magnified, so \(m=-3\). Using the magnification formula: \(-3=\frac{f}{f-16} \dots(ii)\).
Solving Eqs. (i) and (ii), we get
\(
f=12 \mathrm{~cm}
\)

Hint

\(
\begin{aligned}
& \text { (b) As, } \frac{c}{v}=\frac{\sin i}{\sin r} \\
& \Rightarrow \quad v=\frac{c \times \sin r}{\sin i}=\frac{3 \times 10^8 \times \sin 30^{\circ}}{\sin 45^{\circ}}=2.12 \times 10^8 \mathrm{~m} / \mathrm{s}
\end{aligned}
\)

Hint

(b) The resolving power of telescope \(=\frac{a}{1.22 \lambda}\)
where, \(a=\) aperture of telescope.
\(\therefore\) Resolving power \(\propto\) Aperture of telescope
If aperture of telescope is decreased, then the resolving power will decrease.

Hint

(c) The refracting angle \(A\) of a prism is given by \(A=r_1+r_2\), where \(r_1\) and \(r_2\) are the angles of refraction at the first and second surfaces, respectively.
For the light ray to emerge from the second surface, the angle of incidence at this surface, which is \(r_2\), must be less than or equal to the critical angle \(\boldsymbol{C}\).
So, \(r_2 \leq C\).
Consider the maximum possible value for \(r_1\).
The maximum value for \(r_1\) is also \(C\), as if \(r_1>C\), total internal reflection would occur at the first surface if the ray were reversed.
Thus, \(r_1 \leq C\).
Determine the maximum refracting angle.
The maximum refracting angle \(A_{\text {max }}\) occurs when both \(r_1[latex] and [latex]r_2\) are at their maximum possible values, which is \(C\).
\(
A_{\max }=C+C=2 C
\)
Solution: The maximum refracting angle of a prism is \(2 C\).

Hint

(c) If refractive index of lens \(n_1\) is equal to refractive index of liquid \(n_2\), then lens behaves as a plane glass plate and becames invisible in the medium.

Hint

(b) Molecules of a medium after absorbing incoming light radiations, emit them in all directions. This phenomenon is called scattering. According to scientist Rayleigh, the intensity of scattered light \(\propto \frac{1}{\lambda^4} \propto f^4\)
where, \(f\) is the frequency.

Hint

(d) The phenomenon not associated with total internal reflection is Dispersion of light.

Explanation: Total internal reflection occurs when light traveling in a denser medium strikes a boundary with a less dense medium at an angle greater than the critical angle, causing the light to be reflected back into the denser medium. Dispersion of light, on the other hand, is the separation of white light into its constituent colors when it passes through a prism or other medium, which is not related to the angle of incidence or total internal reflection.

Hint

(d) Given, wavelength in vacuum \(=5890 Å\)
In a medium of refractive index, the wavelength of ligh decreases to \(\frac{\lambda}{\mu}=\frac{5890}{1.5} \cong 3927 Å\).

Hint

\(
\begin{aligned}
&\text { (d) Resultant focal length of combination, }\\
&F_{\text {comb }}=\frac{f_1 f_2}{f_1+f_2}=\frac{f(-f)}{f+(-f)}=\infty
\end{aligned}
\)

Hint

(a) When a ray of light passes from a denser medium to a rarer medium, then it bends away from the normal at the interface of the two media.
The angle of incidence is measured with respect to the normal at the refractive boundary. It is given by

\(
C=\sin ^{-1}\left(\frac{n_2}{n_1}\right)
\)
where, \(C\) is critical angle, \(n_2\) is the refractive index of rarer medium and \(n_1\) of the denser medium.
\(
\begin{aligned}
\text { Given, } & & n_2 & =1.33, n_1=1.50 \\
& & C & =\sin ^{-1}\left(\frac{1.33}{1.50}\right) \\
\Rightarrow & & C & =\sin ^{-1}\left(\frac{8}{9}\right)
\end{aligned}
\)

Hint

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

Hint

(c) Real depth \(=1 \mathrm{~m}\)
Apparent depth \(=1-0.1=0.9 \mathrm{~m}\)
Refractive index, \(\mu=\frac{\text { Real depth }}{\text { Apparent depth }}=\frac{1}{0.9}=\frac{10}{9}\)

Hint

(d) Under minimum deviation, \(i=e=30^{\circ}\), so angle between emergent ray and second refracting surface is \(90^{\circ}-30^{\circ}=60^{\circ}\).

Hint

(b) In any medium other than air or vacuum, the velocities of different colours of light are different.Therefore, both red and green colours are refracted at different angles of refractions. Hence, after emerging from glass slab through opposite parallel face, they appear at two different points and move in two different parallel directions.

Hint

(a) When object is placed between centre of curvature and focus (i.e., \(f<u<2 f\) ), then image will be between \(2 f\) and \(\infty\), real, inverted, large in size and \(m>-1\)

Explanation: 

For a concave mirror, the focal length \(f\) is positive. The center of curvature \(\boldsymbol{C}\) is at \(2 f\).
The focal length is \(f=10 \mathrm{~cm}\).
The center of curvature is \(C=2 f=2 \times 10 \mathrm{~cm}=20 \mathrm{~cm}\).
The object distance is \(u=15 \mathrm{~cm}\).
Since \(10 \mathrm{~cm}<15 \mathrm{~cm}<20 \mathrm{~cm}\), the object is placed between the focal point \(F\) and the center of curvature \(\boldsymbol{C}\).
When an object is placed between \(F\) and \(C\) of a concave mirror, the image formed is real, inverted, and magnified.

Hint

(a) Let \(O\) be the size of the object.
Magnification for the magnified image: \(m_1=\frac{x_1}{O}\).
Magnification for the diminished image: \(m_2=\frac{x_2}{O}\).
The product of magnifications is \(1: m_1 m_2=1\).
Substitute the expressions for \(m_1\) and \(m_2: \frac{x_1}{O} \times \frac{x_2}{O}=1\).
Simplify the equation: \(\frac{x_1 x_2}{O^2}=1\).
Rearrange to find \(O: O^2=x_1 x_2\).
Take the square root: \(O=\sqrt{x_1 x_2}\).
The size of the object is \(\sqrt{x_1 x_2}\).

Hint

(d) Here, angle of prism, \(A=60^{\circ}\)
For minimum deviation, \(A=2 r\) or \(r=\frac{A}{2}=\frac{60^{\circ}}{2}=30^{\circ}\) for both colours.

Hint

(c) For real image, object should be placed at distance \(2 f\) for minimum distance between object and its real image.

Hint

(c) The magnifying power of the telescope in normal adjustment,
\(
m=\frac{f_o}{f_e}=\frac{100}{5}=20
\)

Hint

\(
\begin{aligned}
&\text { (b) We have, } n_a v_a=n_g v_g\\
&\begin{aligned}
\frac{n_g}{n_a} & =\frac{v_a}{v_g} \\
\frac{3}{2} & =\frac{3 \times 10^8}{v_g} \\
v_g & =2 \times 10^8 \\
\text { Time } & =\frac{\text { Distance }}{\text { Speed }} \\
t & =\frac{4 \times 10^{-3}}{2 \times 10^8} \\
t & =2 \times 10^{-11} \mathrm{~s}
\end{aligned}
\end{aligned}
\)

Hint

(b) In going from one medium to another, frequency remains same but wavelength and velocity decrease with the increase in refractive index i.e., wavelength becomes \(\lambda^{\prime}=\lambda / \mu\) and velocity becomes,
\(
c=\frac{v}{\mu}
\)

Hint

(a) For combination of one convex and one concave lens
\(
\begin{array}{rlrl}
\frac{1}{F} & =\frac{1}{f_1}-\frac{1}{f_2} \\
\Rightarrow & F & =\frac{f_1 f_2}{f_2-f_1}
\end{array}
\)
If \(f_1>f_2\), then \(F\) will be negative.

Hint

\(
\begin{aligned}
&\text { (d) We have, }\\
&C=\sin ^{-1}\left(\frac{v_1}{v_2}\right)=\sin ^{-1}\left(\frac{1.8 \times 10^8}{2.4 \times 10^8}\right)=\sin ^{-1}\left(\frac{3}{4}\right)
\end{aligned}
\)

Hint

(a) We have, \(f_y=\sqrt{f_R f_b}\)
\(
f_y=\sqrt{100.5 \times 99.5}=99.99
\)
Also, \(\omega=\frac{f_R-f_b}{f_y}=\frac{100.5-99.5}{99.99}\)
Dispersive power, \(\omega=\frac{1}{99.99}=0.01\)

Hint

(b) The number of images formed depends upon the angle between the mirrors. If two mirrors make an angle \(\theta\) with each other, then the number of images formed is
\(
n=\frac{360^{\circ}}{\theta}-1
\)
When mirrors are kept mutually perpendicular to each other, then \(\theta=90^{\circ}\).
\(
n=\frac{360}{90}-1=3
\)

Hint

(c) As, \(n=\frac{360^{\circ}}{60^{\circ}}=6\)
Since, \(n\) is an even integer, so total number of images will be \((n-1)\) or 5.

Hint

(c) The magnifying power \((m)\) of a telescope, \(m=-\frac{f_o}{f_e}\).
Now, if the focal length of the eyepiece \(\left(f_e\right)\) is doubled, then the new magnification would be equal to \(m^{\prime}=-\frac{f_o}{2 f_e}=\frac{m}{2}\).

Hint

(a) The minimum angular separation between two objects, so that they are just resolved is called resolving limit. For eye, it is \(1^{\prime}=\left[\frac{1}{60}\right]^{\circ}\).
Reciprocal of resolving limit is called resolving power,
\(
\begin{aligned}
(\mathrm{RP}) & =\frac{1}{\left(\frac{1}{60}\right)^{\circ}}=\frac{1}{\frac{1}{60} \times \frac{\pi}{180}} \\
& =\frac{10800}{\pi}=3.436 \times 10^3 \mathrm{rad}
\end{aligned}
\)

Hint

(b) Resolving power of microscope \(=\frac{2 \mu \sin \theta}{\lambda}\)
Since, \(\lambda_R>\lambda_B\), hence resolving power increases.

Hint

\(
\begin{aligned}
&\text { (a) Resolving limit of telescope }\\
&=\frac{1.22 \lambda}{a}=\frac{1.22 \times\left(5000 \times 10^{-10}\right)}{0.1}=6.1 \times 10^{-6} \mathrm{rad}
\end{aligned}
\)

Hint

(b) In an astronomical telescope, \(L=f_o+f_e\)
When tube length is decreased, then the image formed by the objective lens will lie between principal focus and optical centre of eye lens instead of lying at the focus of eye lens. Therefore, a virtual image will be formed at a finite distance from the eye lens.

Hint

(d) In this case, \(|m|=\frac{f_o}{f_e}=5 \dots(i)\) and length of telescope \(=f_o+f_e=36 \dots(ii)\)
On solving Eqs. (i) and (ii), we get
\(
f_e=6 \mathrm{~cm}, f_o=30 \mathrm{~cm}
\)

Hint

(b) Magnification, \(|M|=\frac{f_o}{f_e}=10\) or \(f_o=10 f_e\)
Given,
\(
\begin{aligned}
f_e+f_o & =1.1 \mathrm{~m} \\
f_e+10 f_e & =1.1 \times 100 \mathrm{~cm} \\
11 f_e & =110 \\
f_e & =10
\end{aligned}
\)
\(
\begin{aligned}
&\text { Magnification at least distance of distinct vision, }\\
&\begin{aligned}
M_D & =\frac{f_o}{f_e}\left(1+\frac{f_e}{D}\right)=10\left(1+\frac{10}{25}\right) \\
& =10\left(\frac{35}{25}\right)=14
\end{aligned}
\end{aligned}
\)

Hint

(d) Given, \(m=+\frac{1}{2}\), \(f=2 \mathrm{~m}=200 \mathrm{~cm}, u=\) ?
\(\therefore\) Magnification, \(m=\frac{f}{(f-u)}\)
\(
\begin{aligned}
+\frac{1}{2} & =\frac{200}{(200-u)} \\
200-u & =400 \\
u & =-200 \mathrm{~cm} \\
u & =-2 \mathrm{~m}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\begin{array}{ll}
\text { (c) We have, }  \frac{1}{\sin C}=\frac{\lambda_1}{\lambda_2} \\
\Rightarrow  \frac{1}{\sin C}=\frac{6000}{3000} \\
\Rightarrow  \frac{1}{\sin C}=\frac{6}{3}=2 \\
\Rightarrow  \sin C=\frac{1}{2}
\end{array}\\
&\Rightarrow \text { Critical angle between two medium, } C=\sin ^{-1}\left(\frac{1}{2}\right)
\end{aligned}
\)

Hint

(c) For concave mirror,

Using the relation, \(\frac{1}{v}-\frac{1}{7.5}=-\frac{1}{5}\)
\(
\therefore \quad v=-15 \mathrm{~cm}
\)
Therefore position of final Image will be at the pole of the convex mirror.

Hint

(d) Length of the telescope when final image is formed at least distance of distinct vision,
\(
\begin{aligned}
L & =f_o+u_e=f_o+\frac{f_e D}{f_e+D} \\
& =50+\frac{5 \times 25}{(5+25)}=\frac{325}{6} \mathrm{~cm}
\end{aligned}
\)

Hint

(c) We have, \(\frac{I}{O}=\frac{f}{(f-u)}\)
\(
\begin{array}{ll}
\Rightarrow & \frac{I}{5}=\frac{-10}{-10-(-100)} \\
\Rightarrow & I=-0.55
\end{array}
\)
Size of the image, \(I=0.55 \mathrm{~cm}\)

Hint

(a) We have, the effective power given by
\(
P_{\mathrm{eff}}=P_1+P_2
\)
Also, we know \(P=\frac{1}{f}\)
Hence, \(P_{\text {eff }}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{0.20}+\frac{1}{0.25}=5+4=9 \mathrm{D}\)

Hint

(c) Hypermetropia can be removed by using a convex lens. Focal length of used lens, \(f=+d=+40 \mathrm{~cm}\)
\(=+(\) defected near point \()\)
\(
\therefore \text { Power of lens }=\frac{100}{f(\mathrm{~cm})}=\frac{100}{40}=2.5 \mathrm{D}
\)

Hint

(c) We have, \(\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)\)
\(
\frac{1}{30}=(\mu-1)\left(\frac{1}{10}\right)
\)
The refractive index of the material of the lens,
\(
\mu=\frac{4}{3}=1.33
\)

Hint

(c) Refractive index of the material of prism,
\(
\mu=\frac{\sin \left(\frac{A+\delta_m}{2}\right)}{\sin \left(\frac{A}{2}\right)}
\)
\(
\begin{aligned}
& \sqrt{3}=\frac{\sin \left(\frac{60^{\circ}+\delta_m}{2}\right)}{\sin \left(\frac{60^{\circ}}{2}\right)} \\
& \frac{\sqrt{3}}{2}=\sin \left(30^{\circ}+\frac{\delta_m}{2}\right)
\end{aligned}
\)
\(
\therefore \text { Minimum deviation of the prism, } \delta_m=60^{\circ}
\)

Hint

(a) As, \(A+\delta=i+e\)
\(
\Rightarrow 30^{\circ}+30^{\circ}=60^{\circ}+e
\)
Angle of emergence, \(e=60^{\circ}-60^{\circ}=0\)
\(
r_1=A=30^{\circ}
\)
Refractive index, \(\mu=\frac{\sin i}{\sin r}=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\sqrt{3}=1.732\)

Hint

(b) The divergence angle of the emergent beam will be equal to the divergence angle of the incident beam, which is \(\alpha\). This is because the glass slab causes a lateral displacement of the rays but does not change their angular separation.

Hint

(b) We have, \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(
\because \quad \frac{1}{30}-\frac{1}{(-u)}=\frac{1}{f} \text { or } \frac{1}{30}+\frac{1}{u}=\frac{1}{f} \dots(i)
\)
Similarly, \(\frac{1}{120}+\frac{1}{u-90}=\frac{1}{f} \dots(ii)\)
On solving Eqs. (i) and (ii), we get
The focal length of the lens, \(f=24 \mathrm{~cm}\)

Hint

\(
\begin{aligned}
& \text { (a) Given, } f_o+f_e=105 \\
& \qquad \begin{aligned}
f_o+\frac{f_o}{20} & =105 \\
\Rightarrow \quad f_o & =100 \mathrm{~cm}
\end{aligned}
\end{aligned}
\)

Hint

(b) We have, \(\frac{1}{f}=(1.5-1)\left(\frac{1}{10}+\frac{1}{10}\right)\)
\(\therefore\) Focal length, \(f=10 \mathrm{~cm}\)
Radius of curvature, \(R=2 f=20 \mathrm{~cm}\)

Hint

(d) The focal length of the lens,
\(
\begin{array}{rlrl}
\frac{1}{f} & =(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
\frac{1}{f} & =(1.5-1)\left(\frac{1}{20}-\frac{1}{30}\right) \\
\Rightarrow & \frac{1}{f} =0.5\left(\frac{30-20}{600}\right) \\
\text { or } & \frac{1}{f} =\frac{1}{2} \times \frac{10}{600}=\frac{1}{120}
\end{array}
\)
or focal length, \(f=120 \mathrm{~cm}\)

Hint

(b) Power of the lens \(=\frac{1}{\text { Focal length }}\)
Focal length of combination of convex and concave lenses is given by
\(
\frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2}
\)
where, \(f_1\) and \(f_2\) be the focal lengths of convex and concave lenses, respectively.
Now,
\(
\begin{aligned}
& \frac{1}{F}=\frac{1}{0.4}+\frac{1}{(-0.25)} \\
& \frac{1}{F}=-1.5
\end{aligned}
\)
Power,
\(
P=-1.5 \mathrm{D} \left(\because P=\frac{1}{F}\right)
\)

Hint

(a) Power of a biconvex lens is 10 D and radius of curvature of each surface \(=10 \mathrm{~cm}\).
\(
\begin{array}{ll}
\Rightarrow & P=\frac{1}{f}=(\mu-1)\left[\frac{1}{R_1}-\frac{1}{R_2}\right] \\
\Rightarrow & 10=(\mu-1)\left[\frac{100}{10}-\left(-\frac{100}{10}\right)\right] \\
\Rightarrow & 10=(\mu-1)[20] \\
\Rightarrow & (\mu-1)=\frac{1}{2}
\end{array}
\)
\(\Rightarrow\) Refractive index of the material, \(\mu=\frac{3}{2}\)

Hint

\(
\begin{aligned}
&\text { (d) Focal length of the combination, }\\
&\begin{aligned}
& \quad \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2} \\
& \Rightarrow \quad \frac{1}{80}=\frac{1}{20}+\frac{1}{f_2} \text { or } f_2=-\frac{80}{3} \mathrm{~cm} \\
& \therefore \text { Power of second lens, } P^{\prime}=\frac{100}{f_2}=\frac{100}{(-80 / 3)}=-3.75 \mathrm{D}
\end{aligned}
\end{aligned}
\)

Hint

(c) The maximum deviation can be, when it is reflected back from boundary.
\(
i=\theta_C=45^{\circ}
\)
\(\therefore\) Deviation, \(\delta=180^{\circ}-2 i=90^{\circ}\)

Hint

(a) When plano-convex lens is silvered at curved surface, then it becomes concave mirror of focal length,
\(
f=\frac{R}{2 \mu}=\frac{30}{2 \times 1.5}=10 \mathrm{~cm}
\)
If we have to find real image of same size of the object, then object must be placed at centre of curvature \(u=2 f=20 \mathrm{~cm}\).

Hint

\(
\begin{aligned}
&\text { (c) From the formula, } \sin C=\frac{1}{{ }_1 \mu_2} \Rightarrow \sin C={ }_2 \mu_1\\
&\begin{aligned}
\therefore \quad \frac{\mu_1}{\mu_2} & =\frac{v_2}{v_1} \Rightarrow \sin C=\frac{10 x / t_2}{x / t_1} \\
\sin C & =\frac{10 t_1}{t_2} \\
C & =\sin ^{-1}\left(\frac{10 t_1}{t_2}\right)
\end{aligned}
\end{aligned}
\)

Hint

(b) From geometry of the figure, Angle, \(\theta=90^{\circ}-22^{\circ}=68^{\circ}\)

Hint

\(
\begin{aligned}
&\begin{array}{ll}
\text { (c) Magnifying power, } m_{\infty}= & \frac{\left(L_{\infty}-f_o-f_e\right) \cdot D}{f_o f_e} \\
\Rightarrow & 45=\frac{\left(L_{\infty}-1-5\right) \times 25}{1 \times 5}
\end{array}\\
&\Rightarrow \text { The length of the tube, } L_{\infty}=15 \mathrm{~cm}
\end{aligned}
\)

Hint

\(
\text { (a) } u \text { changes from } 0 \text { to }-\infty \text {, then } v \text { will change from } 0 \text { to }+f \text {. }
\)

Hint

(a) The focal length of the concave lens is \(f=20 \mathrm{~cm}\).
The lens is placed in contact with a plane mirror.
The power of a plane mirror is considered infinite, \(\boldsymbol{P}_{\boldsymbol{m}}=\infty\).
The equivalent focal length \(F\) of a lens-mirror combination is given by \(\frac{1}{F}=\frac{2}{f_l}+\frac{1}{f_m}\).
For a concave lens, the focal length is negative: \(f_l=-20 \mathrm{~cm}\).
The focal length of a plane mirror is infinite: \(f_m=\infty\).
The formula for the equivalent focal length \(F\) is: \(\frac{1}{F}=\frac{2}{f_I}+\frac{1}{f_m}\).
Substitute the values: \(\frac{1}{F}=\frac{2}{-20 \mathrm{~cm}}+\frac{1}{\infty}\).
Simplify: \(\frac{1}{F}=-\frac{1}{10 \mathrm{~cm}}+0\).
Solve for \(F: F=-10 \mathrm{~cm}\).
A negative focal length indicates a convex mirror.
The magnitude of the focal length is 10 cm.
The combination acts as a convex mirror of focal length 10 cm.

Hint

(c) We can represent this diagrammatically as

\(
\text { Thus, we see that } C=\sin ^{-1}\left(\frac{1}{\mu}\right)=\sin ^{-1}\left(\frac{3}{4}\right)
\)

Hint

(d) The focal length of the plano-convex lens is 20 cm with the plane surface being silvered.
So, the net power of the lens is given by \(P=\frac{1}{f} \times 2=\frac{2}{f}\)
Thus, the net focal length, \(\frac{f}{2}=\frac{20}{2}=10 \mathrm{~cm}\)

Hint

\(
\begin{aligned}
&\text { (d) Given, } u=m f\\
&\begin{aligned}
& \therefore \frac{1}{v}-\frac{1}{(-m f)}=-\frac{1}{f} \Rightarrow \frac{1}{v}=-\frac{1}{f}\left(1+\frac{1}{m}\right) \\
& \Rightarrow \quad \frac{1}{v}=\frac{m+1}{-m f} \Rightarrow-\frac{v}{u}=\frac{1}{1+m}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) According to lens Maker’s formula, }\\
&\begin{aligned}
& \frac{1}{f} =(\mu-1)\left(\frac{1}{R}-\frac{1}{(-R)}\right) \\
\Rightarrow \frac{1}{f} =(\mu-1)\left(\frac{2}{R}\right) \dots(i)
\end{aligned}
\end{aligned}
\)
On cutting,
\(
\begin{aligned}
& \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\
& \frac{1}{f^{\prime}}=(\mu-1)\left(\frac{1}{R}\right) \dots(ii)
\end{aligned}
\)
From Eqs. (i) and (ii), we get \(f^{\prime}=2 f\).

Hint

(b) As optical paths are equal, hence
\(
\begin{aligned}
n_g x_g & =n_w x_w \\
n_w & =n_g \frac{x_g}{x_w}
\end{aligned}
\)
Refractive index of the water,
\(
n_w=1.53 \times \frac{4.0}{4.5}=1.36
\)

Hint

\(
\begin{aligned}
&\text { (b) For total internal reflection, } i>C\\
&\begin{array}{ll}
\Rightarrow & \sin i>\sin C \Rightarrow \sin 45^{\circ}>\frac{1}{\mu} \\
\Rightarrow & \mu>\sqrt{2} \Rightarrow \mu>1.414
\end{array}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) From Snell’s law, }\\
&\begin{aligned}
\mu \sin i & =\text { constant } \\
\mu_4 \sin x & =\mu \sin i \\
\sin x & =\frac{\mu}{\mu_4} \sin i
\end{aligned}
\end{aligned}
\)

Hint

(b) As, incident ray and reflected ray are parallel to each other it means, deviation, \(\quad \delta=180^{\circ}\)
\(
\begin{aligned}
&\begin{aligned}
\because & \delta =360^{\circ}-2 \theta \\
\therefore & 180^{\circ} =360^{\circ}-2 \theta
\end{aligned}\\
&\text { Angle between two plane mirrors, }\\
&\theta=90^{\circ}
\end{aligned}
\)

Hint

\(
\text { (d) According to the question, } \frac{h}{\mu}=\frac{21}{2}=10.5
\)

\(
\text { Height, } h=10.5 \times \frac{4}{3} \Rightarrow h=14 \mathrm{~cm}
\)

Hint

\(
\text { (b) Given, } A C=60 \mathrm{~m} \text { and } \angle B C B^{\prime}=90^{\circ}
\)

\(
\begin{aligned}
&\text { From the figure, }\\
&\begin{aligned}
& \therefore \text { Angle, } \angle B C A=\frac{\angle B C B^{\prime}}{2}=45^{\circ} \\
& \quad \text { We have, } \tan 45^{\circ}=\frac{h}{A C} \\
& \Rightarrow A C=h \Rightarrow h=60 \mathrm{~m}
\end{aligned}
\end{aligned}
\)

Hint

(d) At second surface, there is no refraction, so
\(
\begin{array}{ll}
& r_2=0 \\
\therefore & r_1=A=30^{\circ}
\end{array}
\)
From Snell’s law,
\(
\begin{array}{ll}
& n=\frac{\sin i_1}{\sin r_1} \\
\Rightarrow & n=\frac{\sin i_1}{\sin 30^{\circ}} \\
\Rightarrow & \sin i_1=\sqrt{2} \times \frac{1}{2}=\frac{1}{\sqrt{2}} \\
\Rightarrow & \sin i_1=\frac{1}{\sqrt{2}} \Rightarrow i_1=45^{\circ}
\end{array}
\)

Hint

\(
\text { (c) Given, } P_1+P_2=9 \dots(i)
\)
\(
\begin{aligned}
&\text { We have, } \quad P=P_1+P_2-d P_1 P_2\\
&\Rightarrow \quad \frac{27}{5}=9-\frac{20}{100} \times P_1 P_2
\end{aligned}
\)
\(
\begin{gathered}
\text { Again }\left(P_2-P_1\right)^2=\left(P_1+P_2\right)^2-4 P_1 P_2 \\
P_2-P_1=3 \dots(ii)
\end{gathered}
\)
\(
\begin{aligned}
&\text { On solving Eqs. (i) and (ii), we get }\\
&P_1=3 \text { and } P_2=6
\end{aligned}
\)

Hint

\(
\text { (c) From the figure in } \triangle A B C, 45^{\circ}+90^{\circ}+2 \alpha=180^{\circ}
\)

\(
\begin{gathered}
\alpha=22.5^{\circ} \\
\angle O B A=90^{\circ}-\alpha
\end{gathered}
\)
\(
\begin{aligned}
& \text { In } \triangle O A B, \theta+45^{\circ}+90^{\circ}-\alpha=180^{\circ} \\
& \Rightarrow \quad \theta=67.5^{\circ}
\end{aligned}
\)

Hint

(a)
\(
\begin{aligned}
\text { Power of system } & =\frac{1}{f_1}+\frac{1}{f_2}-\frac{d}{f_1 f_2} \\
& =\frac{1}{1}+\left(\frac{1}{-0.25}\right)-\frac{0.75}{(1)(-0.25)} \\
& =1-4+3=-3+3=0
\end{aligned}
\)
Since, power of the system is zero, therefore the incident parallel beam of light will remain parallel after emerging from the system.

Hint

(c) Angle of incidence, \(i=\theta\)
Angle of equilateral prism,
\(
\begin{aligned}
& A=60^{\circ} \\
& \mu=\sqrt{3}
\end{aligned}
\)
In the case of minimum deviation, the refracted ray inside the prism is parallel to the base.
\(
\begin{array}{ll}
\text { Hence, } & r=\frac{A}{2}=\frac{60}{2}=30^{\circ} \\
\therefore & \mu=\frac{\sin i}{\sin r} \\
\Rightarrow & \mu=\frac{\sin \theta}{\sin r} \\
\Rightarrow & \sqrt{3}=\frac{\sin \theta}{\sin 30^{\circ}} \\
\Rightarrow & \sin \theta=\sqrt{3} \sin 30^{\circ} \\
\Rightarrow & \sin \theta=\sqrt{3} \times \frac{1}{2} \\
\Rightarrow & \sin \theta=\sin 60^{\circ} \\
\Rightarrow & \theta=60^{\circ}
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (c) For dispersion without deviation, }\\
&\begin{aligned}
& \frac{A}{A^{\prime}}=\frac{\left(\mu^{\prime}-1\right)}{(\mu-1)} \\
& \frac{6^{\circ}}{A^{\prime}}=\frac{0.75}{0.50} \\
& A^{\prime}=\frac{0.50 \times 6^{\circ}}{0.75} \\
& A^{\prime}=4^{\circ}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Here, } n=1.5 \text {, as per sign convention followed }\\
&\begin{aligned}
& R_1 & =+20 \mathrm{~cm} \text { and } R_2=-20 \mathrm{~cm} \\
\therefore \quad & \frac{1}{f} =(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)
\end{aligned}
\end{aligned}
\)
\(
\begin{aligned}
& =(1.5-1)\left[\frac{1}{(+20)}-\frac{1}{(-20)}\right]=0.5 \times \frac{2}{20}=\frac{1}{20} \\
& f=+20 \mathrm{~cm}
\end{aligned}
\)
Incident rays travelling parallel to the axis of lens will converge at its second principal focus, hence
\(
L=+20 \mathrm{~cm}
\)

Hint

\(
\begin{aligned}
&\text { (d) By Rayleigh’s law of scattering, }\\
&\begin{aligned}
& I \propto \frac{1}{\lambda^4} \text { or } \frac{I_2}{I_1}=\left(\frac{\lambda_1}{\lambda_2}\right)^4 \\
& \frac{I_2}{A}=\left(\frac{440}{660}\right)^4 \text { or } I_2 \cong \frac{A}{6}
\end{aligned}
\end{aligned}
\)

Hint

(c) Light should fall normally on the silvered face, i.e.
\(
\begin{array}{ll}
& r_2=0 \\
\therefore & r_1=A=30^{\circ}
\end{array}
\)
Now, refractive index, \(\mu=\frac{\sin i_1}{\sin r_1}\) or \(\sqrt{2}=\frac{\sin i_1}{\sin 30^{\circ}}\)
Angle of incidence, \(i_1=45^{\circ}\)

Hint

(c) For objective, \(\frac{1}{v_o}-\frac{1}{-200}=\frac{1}{50}\)
\(
\therefore \quad v_o=\frac{200}{3} \mathrm{~cm}
\)
For eyepiece, \(\frac{1}{-25}-\frac{1}{\left(-u_e\right)}=\frac{1}{5}\)
\(
\therefore \quad u_e=\frac{25}{6}
\)
Therefore, separation between objective and eye-piece,
\(
L=v_o+u_e=\frac{200}{3}+\frac{25}{6}=\frac{425}{6}=71 \mathrm{~cm}
\)

Hint

(d) Critical angle, \(\theta_C=\sin ^{-1}\left(\frac{1}{1.5}\right)=41.8^{\circ}\)
Incident angle, \(i_1=30^{\circ}\)
\(
\begin{gathered}
\therefore \quad \sin r_1=\frac{\sin i_1}{\mu}=\frac{\sin \left(30^{\circ}\right)}{1.5} \\
r_1=19.5^{\circ}
\end{gathered}
\)
or angle of emergence, \(r_2=A-r_1=90^{\circ}-19.5^{\circ}=70.5^{\circ}\)
Since, \(\quad r_2>\theta_C\)
\(\therefore\) The ray will not emerge from the prism.

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (c) We have, } \delta=i_1+i_2-A \\
& 45^{\circ}=\left(i_1+i_2\right)-60^{\circ} \\
& \therefore \quad i_1+i_2=105^{\circ} \dots(i) \\
& \text { and } \quad i_1-i_2=20^{\circ} \dots(ii)
\end{aligned}\\
&\text { On solving Eqs. (i) and (ii), we get }\\
&i_1=62.5^{\circ} \text { and } i_2=42.5^{\circ}
\end{aligned}
\)

Hint

(d) For terrestrial telescope, magnification,
\(
\begin{array}{ll}
& M=\frac{f_o}{f_e}=\frac{80}{f_e}=20 \\
\Rightarrow & f_e=4 \mathrm{~cm}
\end{array}
\)
Hence, length of terrestrial telescope
\(
\begin{aligned}
& =f_o+f_e+4 f \\
& =80+4+4 \times 20 \\
& =164 \mathrm{~cm}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (c) We assume that, final image is formed at infinity and }\\
&\begin{aligned}
& M_{\infty}  =\frac{\left(L_{\infty}-f_o-f_e\right) \cdot D}{f_o f_e}=\frac{L D}{f_o f_e} \\
\text { or } & 400  =\frac{20 \times 25}{0.5 \times f_e} \\
\therefore & f_e  =2.5 \mathrm{~cm}
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (c) We have, } \frac{1}{f}=\frac{2 \mu}{R_2}-\frac{2(\mu-1)}{R_1}
\)

\(
\begin{gathered}
\frac{1}{-10}=\frac{2 \times 1.5}{\infty}-\frac{2(1.5-1)}{R} \\
R=10 \mathrm{~cm}
\end{gathered}
\)

Hint

(c) The separation between the images formed by extreme wavelengths of visible range is called the longitudinal chromatic aberration, given by
\(
f_1-f_2=\omega \times f
\)
where, \(\omega\) is dispersive power.
Given,
\(
\begin{aligned}
\omega & =0.02, f=20 \mathrm{~cm} \\
f_1-f_2 & =0.02 \times 20=0.40
\end{aligned}
\)

Hint

(a) Here, \(\quad x=u+v \dots(i)\)
Magnification, \(\quad m=\frac{f}{f-u}=\frac{f-v}{f}\)
Image is real, magnification is negative
\(
\begin{aligned}
-m & =\frac{f}{f-u} \\
u & =\frac{(m+1)}{m} f \\
-m & =\frac{f-v}{f}
\end{aligned}
\)
\(
\begin{aligned}
&\Rightarrow \quad v=(m+1) f\\
&\text { Putting the value of } v \text { in Eq. (i), we get }\\
&\begin{aligned}
& x =\frac{(m+1)}{m} f+(m+1) f \\
\Rightarrow & f =\frac{m x}{(m+1)^2}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) Linear magnification, }\\
&\begin{aligned}
& \qquad \begin{aligned}
m=\frac{I}{O}= & \frac{f}{u-f}=\frac{-10}{-25+10} \\
m & =\frac{10}{15}=\frac{2}{3} \\
\text { Areal magnification } & =m^2=\frac{4}{9} \\
\text { Area of image } & =\frac{4}{9} \times \text { Area of object } \\
& =\frac{4}{9} \times(3)^2=4 \mathrm{~cm}^2
\end{aligned}
\end{aligned}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\begin{aligned}
& \text { (d) As, } n=\frac{\text { Real depth }}{\text { Apparent depth }} \\
& \Rightarrow \quad n=\frac{6}{4}=\frac{3}{2} \\
& \text { Also, } \quad \frac{n_1}{u}-\frac{n_2}{v}=\frac{n_1-n_2}{R} \\
& \Rightarrow \frac{1.5}{6}-\frac{4}{17}=\frac{1.5-1}{R}
\end{aligned}\\
&\text { Radius of curvature, } R=34 \mathrm{~cm}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) We have, } \frac{1}{v}+\frac{1}{u}=\frac{1}{f}\\
&\begin{aligned}
\Rightarrow \quad \frac{1}{v}-\frac{1}{20} & =\frac{1}{-15} \\
v & =-60 \mathrm{~m} \\
\text { Image speed } & =\left(\frac{v^2}{u^2}\right)(\text { Object speed })=\left(\frac{60}{20}\right)^2(5)=45 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}
\)

Hint

(a) From condition of achromatism,
\(
\begin{array}{rlrl}
& & \frac{\omega_1}{f_1}+\frac{\omega_2}{f_2} & =0 \\
\Rightarrow & \frac{\omega_1}{\omega_2} & =-\frac{f_1}{f_2} \\
\Rightarrow & \frac{5}{3} & =\frac{-(-15)}{f_2}
\end{array}
\)
Focal length, \(f_2=9 \mathrm{~cm}\) (convex)

Hint

\(
\begin{aligned}
&\text { (b) Angle } i \text { at both faces will be } 45^{\circ} \text {. For TIR to take place, }\\
&\begin{array}{ll}
& i>\theta_C \text { or } \sin i>\sin \theta_C \\
\therefore & \frac{1}{\sqrt{2}}>\frac{1}{\mu} \text { or } \mu>\sqrt{2}
\end{array}
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (a) As, per question, }\\
&\because \frac{1}{-(f+1)}-\frac{1}{-(f-1)}=\frac{1}{f}
\end{aligned}
\)

\(
\text { On solving, we get } f=(\sqrt{2}+1) \mathrm{cm}
\)

Hint

\(
\text { (c) We have, } \frac{\mu_2}{v}=\frac{\mu_1}{u}-\frac{\mu_2-\mu_1}{R}
\)
\(
\begin{aligned}
& \frac{1}{v}=\frac{1.5}{-6}-\frac{1-1.5}{-10} \\
& v=-5 \mathrm{~cm}
\end{aligned}
\)

Hint

(c) Shift from the slab, \(S=\left(1-\frac{1}{\mu}\right) t=2 \mathrm{~cm}\)
Let \(x\) be the distance between object and mirror, then
\(
x-2=40 \text { or } x=42 \mathrm{~cm}
\)

Hint

(d) Lens should be convex. Images should be magnified, if object at 16 cm length should lie between \(f\) and \(2 f\). Because corresponding to \(6 \mathrm{~cm}(<f)\), length image should be virtual.
From the given options, if \(f=11 \mathrm{~cm}\).
Then, \(6 \mathrm{~cm},<f\) and \(11 \mathrm{~cm}<16 \mathrm{~cm}<22 \mathrm{~cm}\) or \(f<16 \mathrm{~cm}<2 f\)

Hint

(b) Using mirror formula, \(\frac{1}{v}+\frac{1}{(-u)}=\frac{1}{-f}\)
Multiplying with \(u, \frac{u}{v}=\frac{-u}{f}+1=\frac{f-u}{f}\)
\(\therefore\) Magnification, \(m=-\frac{v}{u}=\left(\frac{f}{f-u}\right)\)
Now, \(\quad|d v|=m^2|d u|\)
Size of image \(=\left(\frac{f}{f-u}\right)^2 \cdot b=\left(\frac{f}{u-f}\right)^2 \cdot b\)

Hint

\(
\text { (d) We have, } 90^{\circ}-\theta+2 \theta=90^{\circ}
\)

\(
\begin{aligned}
&\therefore \quad \theta=0^{\circ}\\
&\text { Angle of reflected ray with horizontal }=90^{\circ}-\theta=90^{\circ}
\end{aligned}
\)

Hint

(d) When total internal reflection just takes place from lateral surface, \(i=C\), i.e. \(60^{\circ}=C\)
\(
\Rightarrow \quad \sin 60^{\circ}=\sin C=\frac{1}{\mu} \Rightarrow \mu=\frac{2}{\sqrt{3}}
\)
Time taken by light to traverse some distance in a medium,
\(
t=\frac{\mu x}{c}=\frac{\frac{2}{\sqrt{3}} \times 10^3}{3 \times 10^8}=3.85 \mu \mathrm{~s}
\)

Hint

(c) When the lens in air, we have
\(
P_a=\frac{1}{f_a}=\frac{\mu_g-\mu_a}{\mu_a}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
\)
When the lens is in liquid, we have
\(
P_l=\frac{1}{f_l}=\frac{\mu_g-\mu_l}{\mu_l}\left[\frac{1}{R_1}-\frac{1}{R_2}\right]
\)
Given,
\(
P_a=5, P_l=-1, \mu_a=1 \text { and } \mu_g=1.5
\)
On solving, we get \(\mu_l=\frac{5}{3}\)

Hint

(b) We have, \(\frac{\mu_2}{v_1}-\frac{\mu_1}{\infty}=\frac{\mu_2-\mu_1}{R} \dots(i)\)
and \(\frac{\mu_3}{f}-\frac{\mu_2}{v_1}=\frac{\mu_3-\mu_2}{-R} \dots(ii)\)
From Eqs. (i) and (ii), we get
\(
\frac{\mu_3}{f}=\frac{\mu_2-\mu_1-\mu_3+\mu_2}{R}=\frac{2 \mu_2-\left(\mu_1+\mu_3\right)}{R}
\)
Lens will become diverging, if \(f\) is negative i.e.
\(
2 \mu_2<\left(\mu_1+\mu_3\right)
\)

Hint

\(
\text { (b) For end } A \text { of the rod, } u=2 f-\frac{f}{3}=\frac{5 f}{3}
\)

\(
\begin{aligned}
&\begin{array}{ll}
\text { By using, } & \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \Rightarrow \frac{1}{-f}=\frac{1}{v}+\frac{1}{(-5 f / 3)} \\
\therefore & v=-\frac{5}{2} f
\end{array}\\
&\text { So, length of image }=\frac{5}{2} f-2 f=\frac{f}{2}
\end{aligned}
\)

Hint

\(
\text { (c) At } P, u=v \text { which is possible only when } u=2 f
\)

\(
\text { If we take another point } Q \text { just above } P \text {, then } v>2 f \text {. }
\)

Hint

\(
\begin{aligned}
&\text { (a) For real image, }\\
&\begin{aligned}
u & =-x \\
v & =+m x \\
\therefore \quad \frac{1}{m x}-\frac{1}{-x} & =\frac{1}{f} \dots(i)
\end{aligned}
\end{aligned}
\)
For virtual image, \(\quad u=-y\)
Then, \(\quad v=-m y\)
\(
\therefore \quad \frac{1}{-m y}-\frac{1}{(-y)}=\frac{1}{f} \dots(ii)
\)
On solving Eqs. (i) and (ii), we get
\(
f=\frac{x+y}{2}
\)

Hint

(a) Biconvex lens is cut perpendicularly to the principal axis, it will become a plano-convex lens.
Focal length of biconvex lens,
\(
\begin{aligned}
& \frac{1}{f}=(n-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right) \\
& \frac{1}{f}=(n-1) \frac{2}{R} \quad\left(\because R_1=R, R_2=-R\right)
\end{aligned}
\)
\(
f=\frac{R}{2}(n-1) \dots(i)
\)
\(
\begin{aligned}
&\text { For plano-convex lens, }\\
&\begin{aligned}
& \frac{1}{f_1}=(n-1)\left(\frac{1}{R}-\frac{1}{\infty}\right) \\
& f_1=\frac{R}{(n-1)} \dots(ii)
\end{aligned}
\end{aligned}
\)
On comparing Eqs. (i) and (ii),we see the focal length become double.
As power of lens, \(P \propto \frac{1}{\text { focal length }}\).
Hence, power will become half.
New power \(=\frac{4}{2}=2 \mathrm{D}\)

Hint

\(
\text { (c) We have, refractive index, } \mu=\frac{\sin i}{\sin r} \Rightarrow \sqrt{3}=\frac{\sin 60^{\circ}}{\sin r}
\)

\(
\sin r=\frac{1}{2}
\)
Refracted angle, \(r=30^{\circ}\)
The angle between the reflected and refracted rays,
\(
\theta=180^{\circ}-i-r=90^{\circ}
\)

Hint

\(
\begin{aligned}
&\text { (d) Equation of straight line } A B \text {, }\\
&\frac{1}{m}=\tan \theta u-\frac{b a}{c}
\end{aligned}
\)

From, \(\frac{1}{v}-\frac{1}{-u}=\frac{1}{+f}\)
\(
v=\frac{u f}{u-f} \text { or } \frac{v}{u}=m=\frac{f}{u-f} \Rightarrow \frac{1}{m}=\left(\frac{1}{f}\right) u-1
\)
Comparing this with equation of straight line, \(1 / f\) is the slope of line, which is \(b / c\). So, the focal length of the lens used is \(\frac{c}{b}\).

Hint

\(
\begin{aligned}
&\text { (a) We have, } \frac{1}{3 x}-\frac{1}{(-x)}=\frac{1}{+30}\\
&\therefore \quad x=40 \mathrm{~cm} \text { and } 3 x=120 \mathrm{~cm}
\end{aligned}
\)

To decrease the magnification, object should be moved towards \(2 F_1\).
Hence, image will move towards \(2 F_2\). Let displacement is \(y\), then
\(
2(40+6)=(120-y) \quad\left(\because m=\frac{v}{u}\right)
\)
\(
y=28 \mathrm{~cm}
\)

Hint

\(
\begin{aligned}
&\text { (c) Due to glass slab, increase in path }\\
&\begin{array}{rlrl}
& & =(\mu-1) t=(1.5-1) 3 \mathrm{~cm} \\
& & =1.5 \mathrm{~cm} \\
\therefore \quad u^{\prime} =-(21+1.5) \mathrm{cm} \\
\text { or } & u^{\prime} =-22.5 \mathrm{~cm}
\end{array}
\end{aligned}
\)
\(
\text { Now, } \quad \frac{1}{-22.5}+\frac{1}{v}=\frac{1}{-5}
\)
\(
\begin{aligned}
& \frac{1}{v}=\frac{1}{22.5}-\frac{1}{5} \text { or } \frac{1}{v}=\frac{5-22.5}{22.5 \times 5} \\
& v=\frac{22.5 \times 5}{17.5} \text { or } v=-6.43 \mathrm{~cm}
\end{aligned}
\)
\(
\text { The position of the final image }=-(6.43-1.5) \mathrm{cm}=-4.93 \mathrm{~cm}
\)

Hint

(a) From the relation,
\(
\begin{aligned}
\frac{1}{v}+\frac{1}{u} & =\frac{1}{f} \text { or } \frac{1}{v}-\frac{1}{u}=\frac{1}{-f} \\
\frac{1}{v} & =\frac{1}{u}-\frac{1}{f} \text { or } v=\left(\frac{u f}{f-u}\right)
\end{aligned}
\)
Since, \(u>f, v\) is negative or
\(
|v|=\left(\frac{u f}{u-f}\right)>f
\)
The end which is at infinity will have its image at focus.
\(\therefore\) Length of image,
\(
L=|v|-f=\frac{f^2}{u-f}
\)

Hint

(c) The reflected ray will rotate by angle \(2 \theta\). For TIR to take place at water-air boundary,
\(
\begin{array}{rlrl}
& \sin 2 \theta & >\sin \theta_C \text { or } \sin 2 \theta>\frac{1}{\mu} \\
\therefore & \theta & >\frac{1}{2} \sin ^{-1}\left(\frac{3}{4}\right)
\end{array}
\)

Hint

\(
\begin{aligned}
&\text { (a) Velocity of image }\\
&\begin{aligned}
v_r & =\sqrt{v^2+v^2-2 v \cdot v \cdot \cos 2 \theta} \\
& =2 v \sin (\theta)
\end{aligned}
\end{aligned}
\)

Hint

\(
\text { (b) We know that, } \quad \frac{1}{\mu}=\frac{\sin i}{\sin r^{\prime}} \dots(i)
\)
\(
\begin{aligned}
&\text { But }\\
&\begin{aligned}
r+r^{\prime} & =90^{\circ} \text { or } r^{\prime}=90^{\circ}-r \\
\sin r^{\prime} & =\sin \left(90^{\circ}-r\right) \\
& =\cos r
\end{aligned}
\end{aligned}
\)
\(
\sin r^{\prime}=\cos i \quad(\text { since }, i=r) \dots(ii)
\)
From Eqs. (i) and (ii), we get
\(
\begin{array}{ll}
\therefore & \frac{1}{\mu}=\frac{\sin i}{\cos i} \text { or } \frac{1}{\mu}=\tan i \\
\text { or } & \sin i_C=\tan i=\tan r
\end{array}
\)
The critical angle for pair of media is,
\(
i_C=\sin ^{-1}(\tan r)
\)

Hint

\(
\begin{aligned}
& \text { (a) } A Q=A R \\
& \Rightarrow \angle A Q R=\angle A R Q=60^{\circ} \\
& \therefore \quad r_1=r_2=30^{\circ} \\
& \quad \sin i_1=\mu \sin r_1
\end{aligned}
\)

\(
\begin{aligned}
& \Rightarrow \quad \sin i_1=\sqrt{3} \sin 30^{\circ} \\
& \quad=\sqrt{3} \times \frac{1}{2}=\frac{\sqrt{3}}{2} \\
& \Rightarrow \quad i_1=60^{\circ} \\
& \text { Similarly, } i_2=60^{\circ} \\
& \text { In prism, deviation, } \delta=i_1+i_2-A \\
& \quad=60^{\circ}+60^{\circ}-60^{\circ}=60^{\circ}
\end{aligned}
\)

Hint

\(
\text { (a) From the Snell’s law, } \mu_1 \sin i_1=\mu_2 \sin i_2
\)

\(
\therefore \quad \frac{\mu_1 \alpha}{\sqrt{a^2+b^2}}=\frac{\mu_2 \alpha}{\sqrt{\alpha^2+\beta^2}}\left(\because \sin \theta=\frac{\text { Height }}{\text { Hypotenuse }}\right)
\)
\(
\left(\begin{array}{rl}
\because r_A & =a \hat{\mathbf{i}}+b \hat{\mathbf{j}} \text { and } \\
r_B & =\alpha \hat{\mathbf{i}}+\beta \hat{\mathbf{j}} \text { are unit vectors }
\end{array}\right)
\)
\(
\begin{aligned}
\sqrt{a^2+b^2} & =\sqrt{\alpha^2+\beta^2}=1 \\
\mu_1 \alpha & =\mu_2 \alpha
\end{aligned}
\)

Hint

\(
\begin{aligned}
&\text { (b) Here, angle of incidence, } i=45^{\circ}\\
&\frac{\text { Lateral shift }(d)}{\text { Thickness of glass slab }(t)}=\frac{1}{\sqrt{3}}
\end{aligned}
\)

\(
\text { Lateral shift, } d=\frac{t \sin \delta}{\cos r}=\frac{t \sin (i-r)}{\cos r} \Rightarrow \frac{d}{t}=\frac{\sin (i-r)}{\cos r}
\)
\(
\begin{aligned}
\frac{d}{t} & =\frac{\sin i \cos r-\cos i \sin r}{\cos r} \\
\frac{d}{t} & =\frac{\sin 45^{\circ} \cos r-\cos 45^{\circ} \sin r}{\cos r}=\frac{\cos r-\sin r}{\sqrt{2} \cos r} \\
\frac{d}{t} & =\frac{1}{\sqrt{2}}(1-\tan r) \text { or } \frac{1}{\sqrt{3}}=\frac{1}{\sqrt{2}}(1-\tan r) \\
\tan r & =1-\frac{\sqrt{2}}{\sqrt{3}}
\end{aligned}
\)
\(
\text { Angle of refraction, } r=\tan ^{-1}\left(1-\frac{\sqrt{2}}{\sqrt{3}}\right)
\)

Hint

\(
\text { (b) Fish is observer and bird is object. }
\)

\(
\begin{aligned}
&\text { Apparent distance between } F \text { and } B \text { at some instant will be }\\
&\begin{array}{rlrl}
y  =(x+\mu h) \\
\left(-\frac{d y}{d t}\right) & =\left(-\frac{d x}{d t}\right)+(\mu)\left(-\frac{d h}{d t}\right) \\
9 & =3+\frac{4}{3}\left(-\frac{d h}{d t}\right) \\
\therefore \quad-\frac{d h}{d t} & =4.5 \mathrm{~m} / \mathrm{s}
\end{array}
\end{aligned}
\)

Hint

\(
\text { (d) Refractive index, } \mu=\frac{\sin 60^{\circ}}{\sin 30^{\circ}}=\frac{\sqrt{3} / 2}{1 / 2}=\sqrt{3}
\)

Hint

(b) In first case, when sun is at infinity, i.e.

\(
\begin{aligned}
u & =\infty \\
\frac{1}{f} & =\frac{1}{-32}+\frac{1}{(-\infty)} \\
f & =-32 \mathrm{~cm}
\end{aligned}
\)
When water is filled in that tank upto a height of 20 cm, the image formed by the mirror will act as virtual object \(O\) for water surface.
\(
B I=B O \times \frac{3}{4}=12 \times \frac{3}{4}=9 \mathrm{~cm}
\)