Geometrical Optics

We are tasked with finding the relative refractive index given the relation between the critical angle and the Brewster’s angle.

Step 1: Recall the definitions

• Critical angle (from denser medium $n_1$ to rarer medium $n_2$, with $n_1 > n_2$):

$ \sin \theta_c = \frac{n_2}{n_1} = \frac{1}{\mu} $

where $\mu = \frac{n_1}{n_2}$ is the relative refractive index $> 1$.

• Brewster’s angle:

$ \tan \theta_B = \frac{n_2}{n_1} = \frac{1}{\mu} $

Step 2: Given condition

$ \frac{\sin \theta_c}{\sin \theta_B} = \eta = 1.28 $

So:

$ \frac{1/\mu}{\sin \theta_B} = 1.28 $

Step 3: Relation between $\theta_B$ and $\mu$

Since $\tan \theta_B = \tfrac{1}{\mu}$,

$ \sin \theta_B = \frac{\tan \theta_B}{\sqrt{1+\tan^2 \theta_B}} = \frac{\tfrac{1}{\mu}}{\sqrt{1+\tfrac{1}{\mu^2}}} = \frac{1}{\sqrt{\mu^2+1}} $

Step 4: Substitute into ratio

$ \frac{\sin \theta_c}{\sin \theta_B} = \frac{\tfrac{1}{\mu}}{\tfrac{1}{\sqrt{\mu^2+1}}} = \frac{\sqrt{\mu^2+1}}{\mu} $

Given this equals 1.28:

$ \frac{\sqrt{\mu^2+1}}{\mu} = 1.28 $

Step 5: Solve for $\mu$

$ \sqrt{\mu^2+1} = 1.28 \mu $

Square both sides:

$ \mu^2 + 1 = 1.6384 \, \mu^2 $

$ 1 = (1.6384 - 1)\mu^2 $

$ 1 = 0.6384 \mu^2 $

$ \mu^2 = \frac{1}{0.6384} \approx 1.565 $

$ \mu \approx 1.25 $

Step 6: Find the relative index of the two media in the question’s terms

Here we defined $\mu = \frac{n_1}{n_2} \approx 1.25$.

But the relative refractive index of the two media usually means $\frac{n_2}{n_1}$ (rarer/denser), since that's how it appears in options (<1).

$ \frac{n_2}{n_1} = \frac{1}{\mu} \approx \frac{1}{1.25} = 0.8 $

✅ Final Answer:

$ \boxed{0.8} $

Correct Option: C

We use the Lens Maker's Formula for each planoconvex lens :

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

For left Plano-convex lens $\left(\mathrm{L}_1\right)$ :

The refractive index of the lens is $\mu_1=1.5$

Radii of curvatures are $\mathrm{R}_1=15 \mathrm{~cm}$ (curved surface), $\mathrm{R}_2=\infty$ (plane surface)

$ \frac{1}{f_1}=(1.5-1)\left(\frac{1}{15}-\frac{1}{\infty}\right)=0.5 \times \frac{1}{15}=\frac{1}{30} $

$\Rightarrow $ $\mathrm{f}_1=+30 \mathrm{~cm}$

For right Plano-convex lens $\left(\mathrm{L}_2\right)$ :

The refractive index of lens $\mu_2=1.2$

Radii of curvatures are $\mathrm{R}_1=\infty$, (Plane surface) and $\mathrm{R}_2=-12 \mathrm{~cm}$ ( (curved surface)

$ \frac{1}{f_2}=(1.2-1)\left(\frac{1}{\infty}-\frac{1}{-12}\right)=0.2 \times \frac{1}{12}=\frac{1}{60} $

$\Rightarrow $ $\mathrm{f}_1=+60 \mathrm{~cm}$

Since the lenses are thin and in contact, the total power is the sum of individual powers :

$ \frac{1}{F}=\frac{1}{f_1}+\frac{1}{f_2} $

$\Rightarrow $ $\frac{1}{F}=\frac{1}{30}+\frac{1}{60}=\frac{2+1}{60}=\frac{3}{60}=\frac{1}{20}$

$\Rightarrow $ $\mathrm{F}=+20 \mathrm{~cm}$

Given object distance $\mathrm{u}=-30 \mathrm{~cm}$

Using the Lens Formula: $\frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{~F}}$

$ \frac{1}{v}-\frac{1}{-30}=\frac{1}{20} $

$\Rightarrow \frac{1}{v}+\frac{1}{30}=\frac{1}{20}$

$\Rightarrow $ $\frac{1}{v}=\frac{1}{20}-\frac{1}{30}=\frac{3-2}{60}=\frac{1}{60}$

$ \mathrm{v}=+60 \mathrm{~cm} $

Magnification for a lens is given by : $\mathrm{m}=\mathrm{v} / \mathrm{u}$

$ m=\frac{60}{-30}=-2 $

The negative sign indicates that the image is real and inverted, and its size is twice the size of the object. Hence, the correct option is (A).

The refractive index $n$ of a prism with refracting angle $A$ and angle of minimum deviation $\delta_m$ is given by the formula:

$ \mathrm{n}=\frac{\sin \left(\frac{\mathrm{A}+\delta_{\mathrm{m}}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)} $

According to question the angle of minimum deviation is equal to the refracting angle:

$ \delta_{\mathrm{m}}=\mathrm{A} $

Putting in the formula,

$ \mathrm{n}=\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}=\frac{\sin (A)}{\sin \left(\frac{A}{2}\right)} $

$\Rightarrow $ $\mathrm{n}=\frac{2 \sin \left(\frac{\mathrm{~A}}{2}\right) \cos \left(\frac{\mathrm{A}}{2}\right)}{\sin \left(\frac{\mathrm{A}}{2}\right)}=2 \cos \left(\frac{\mathrm{~A}}{2}\right)$

The value of the refractive index depends on the possible values for the refracting angle $A$.

As A approaches $0^{\circ}, \cos \left(\frac{\mathrm{A}}{2}\right) \rightarrow \cos \left(0^{\circ}\right)=1$.

Therefore, n approaches $2 \times 1=2$.

For minimum deviation to occur, the internal angle of incidence $\mathrm{r}=\frac{\mathrm{A}}{2}$ must be less than the critical angle C to let the light emerge from prism.

$ \sin \left(\frac{A}{2}\right)<\sin (C)=\frac{1}{n} $

Using $\mathrm{n}=2 \cos \left(\frac{\mathrm{~A}}{2}\right)$,

$ 2 \sin \left(\frac{\mathrm{~A}}{2}\right) \cos \left(\frac{\mathrm{A}}{2}\right)<1 \Rightarrow \sin (\mathrm{~A})<1 $

This condition is satisfied for all $\mathrm{A}<90^{\circ}$.

If $\mathrm{A}=90^{\circ}$, then $\mathrm{n}=2 \cos \left(45^{\circ}\right)=2 \times \frac{1}{\sqrt{2}}=\sqrt{2}$

For a practical transparent prism where $0<\mathrm{A}<90^{\circ}$, the refractive index must fall in the range:

$ \sqrt{2}<\mathrm{n}<2 $

Hence, the correct option is (B).

If the focal length of a lens is $f$ (in meters) then its power is given as $P=\frac{1}{f} \ldots$ (i)

The Lens Maker's formula is given as $\frac{1}{f}=(\mu-1)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$, where $\mu$ is the refractive index of lens with respect to the medium, $R_1$ and $R_2$ are the radii of curvatures (measured using sign convention).

Using the Lens Maker's Formula and power formula :

$ \mathrm{P}=\frac{1}{\mathrm{f}}=(\mu-1)\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right) $

Here, $\mu_1=$ refractive index of biconvex lens, $R_1 \& R_2$ are the radii of curvatures of front and back curvature of biconvex lens.

Then power of biconvex lens is,

$ P_{\text {convex }}=(1.5-1)\left(\frac{1}{R_1}-\left(\frac{1}{-R_2}\right)\right)=0.5\left(\frac{1}{R_1}+\frac{1}{R_2}\right) $

For the Plano-Concave Lens

Refractive Index $\mu_2=1.7$

The curvature of the plano-concave lens matches the back surface of the biconvex lens. Thus, its curved surface has radius $R_2$.

For a plano-concave lens, one surface is flat $R_3=\infty$ and the other is concave (radius $R_4=R_2$ ).

So, the power of Plano-concave lens is,

$ P_{\text {concave }}=(1.7-1)\left(\frac{1}{R_3}-\frac{1}{R_4}\right)=0.7\left(\frac{1}{\infty}-\frac{1}{R_2}\right)=-\frac{0.7}{R_2} $

The magnitudes of power are the same

$ \left|P_{\text {convex }}\right|=\left|P_{\text {concave }}\right| $

$\Rightarrow $ $0.5\left(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}\right)=\frac{0.7}{\mathrm{R}_2}$

$\Rightarrow $ $\frac{5}{\mathrm{R}_1}=\frac{7}{\mathrm{R}_2}-\frac{5}{\mathrm{R}_2}$

$\Rightarrow $ $\frac{5}{\mathrm{R}_1}=\frac{2}{\mathrm{R}_2}$

$\Rightarrow $ $\frac{\mathrm{R}_1}{\mathrm{R}_2}=\frac{5}{2}$

Here, $R_1 $ & $ R_2$ are the radii of curvatures of front and back curvatures whose ratio is $5: 2$.

Hence, the correct option is (B).

Since the image is real, the mirror must be a concave mirror.

For a real image formed by a concave mirror, the magnification is negative.

$ \mathrm{m}=-3 $

We know that $m=-\frac{v}{u}$, where $m$ is the magnification of the image, $v$ is the image distance and $u$ is the object distance.

$\Rightarrow $ $ -3=-\frac{v}{u} \Rightarrow v=3 u $

The distance between the object and the image is 40 cm . In a concave mirror, for a real magnified image, both the object ( $u$ ) and the image ( $v$ ) are on the same side of the mirror.

The distance between them is $|\mathrm{v}|-|\mathrm{u}|$.

$ |\mathrm{v}|-|\mathrm{u}|=40 \mathrm{~cm} $

Substituting $\mathrm{v}=3 \mathrm{u}$,

$ 3|\mathrm{u}|-|\mathrm{u}|=40 $

$\Rightarrow $ $ 2|\mathrm{u}|=40 \Rightarrow|\mathrm{u}|=20 \mathrm{~cm} $

Using sign convention, the object distance is negative; $\mathrm{u}=-20 \mathrm{~cm}$.

Therefore, the image distance is :

$ \mathrm{v}=3 \times(-20)=-60 \mathrm{~cm} $

Using the mirror formula :

$ \frac{1}{f}=\frac{1}{v}+\frac{1}{u} $

Substituting the values :

$ \frac{1}{f}=\frac{1}{-60}+\frac{1}{-20} $

$\Rightarrow $ $ \frac{1}{f}=\frac{-1-3}{60} \Rightarrow f=\frac{60}{-4}=-15 \mathrm{~cm} $

Therefore, the focal length of the concave mirror is -15 cm . The negative sign indicates it is a concave mirror.

Hence, the correct option is (D)

The power $(\mathrm{P})$ of the lens is given as +5 D .

The relationship between power and focal length (f) in meters is $\mathrm{P}=\frac{1}{\mathrm{f}}$.

$ \mathrm{f}=\frac{1}{\mathrm{P}}=\frac{1}{5} \mathrm{~m}=20 \mathrm{~cm} $

For a convex lens forming a real image, the lens formula is:

$ \frac{1}{f}=\frac{1}{v}-\frac{1}{u} $

Using sign convention for a real object ( $u$ is negative) and a real image ( $v$ is positive):

$ \frac{1}{20}=\frac{1}{v}-\frac{1}{-u} \Rightarrow \frac{1}{20}=\frac{1}{v}+\frac{1}{u} $

The theoretical image distance $\left(\mathrm{v}_{\text {th }}\right)$ for each person can be determined for given object distance $(\mathrm{u})$ using $\mathrm{v}_{\text {th }}=\frac{\mathrm{uf}}{\mathrm{u}-\mathrm{f}}$ (where u is the magnitude).

For person $\mathrm{P}_1$ the object distance

Most readings have slight experimental errors (within 2-4 cm). However, for Person $\mathrm{P}_3$, the recorded value of 37 cm is significantly different from the theoretical value of 46.6 cm (an error of 9.6 cm ). This makes $\mathrm{P}_3$ 's reading distinctly incorrect compared to the others.

Therefore, the data recorded by person $P_3$ is mathematically inconsistent with the properties of a lens with 20 cm focal length.

Hence, the correct option is (C).

Light is traveling from the prism material (denser medium) to the coating material (rarer medium).

Refractive index of Prism $\mathrm{n}_1=\mathrm{n}$

Refractive index of Coating $\mathrm{n}_2=\frac{\mathrm{n}}{2}$

According to question at the exit surface, the light exactly meets the condition of critical angle.

The formula for the critical angle is :

$ \sin \theta_{\mathrm{C}}=\frac{\mathrm{n}_{\text {rare }}}{\mathrm{n}_{\text {denser }}} $

$\Rightarrow $ $\sin \theta_C=\frac{n / 2}{n}=\frac{1}{2}$

$ \theta_C=\sin ^{-1}(1 / 2)=30^{\circ} $

So, the angle of incidence at the second face $\left(\mathrm{r}_2\right)$ is equal to the critical angle :

$ r_2=30^{\circ} $

Applying the condition for minimum deviation, the light ray passes symmetrically through the prism.

$ \mathrm{r}_1=\mathrm{r}_2=30^{\circ} $

The light ray inside the prism is parallel to the base (for an equilateral prism).

For any prism, the relationship between the Prism Angle A and the internal angles is:

$ \begin{gathered} \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 \\ \mathrm{~A}=30^{\circ}+30^{\circ} \\ \mathrm{A}=60^{\circ} \end{gathered} $

Hence, the correct option is (A).

Light hits the surface at an angle of incidence $i$.

Part of the light bounces back. According to the Law of Reflection, the angle of reflection is equal to the angle of incidence (i).

Part of the light enters the second medium. It bends at an angle of refraction $r$.

The reflected and refracted rays are perpendicular to each other. This means the angle between the reflected ray and the refracted ray is $90^{\circ}$.

The sum of angles on a straight line is $180^{\circ}$.

$ \mathrm{i}+90^{\circ}+\mathrm{r}=180^{\circ} \Rightarrow \mathrm{i}+\mathrm{r}=90^{\circ} $

$ \Rightarrow \mathrm{r}=90^{\circ}-\mathrm{i} $

Now using the fundamental law of refraction, Snell's Law :

$ \mu_1 \sin (\mathrm{i})=\mathrm{n}_2 \sin (\mathrm{r}) \Rightarrow \mu_1 \sin (\mathrm{i})=\mathrm{n}_2 \sin \left(90^{\circ}-\mathrm{i}\right) $

$\Rightarrow \mu_1 \sin (\mathrm{i})=\mathrm{n}_2 \cos (\mathrm{i})$

$\Rightarrow $ $ \tan (\mathrm{i})=\frac{\mu_2}{\mu_1} $

The refractive index of first medium is $\mu_1=2$ and the refractive index of second medium is $\mu_2=2 \sqrt{3}$

$ \tan (i)=\frac{2 \sqrt{3}}{2}=\sqrt{3} \Rightarrow i=60^{\circ} $

Therefore, the angle of incidence is $60^{\circ}$.

Hence, the correct option is (B).

A compound microscope consists of two lenses objective lens (lens near the object) and the eyepiece (lens near the eye.). To achieve high magnification, objective lens must have a small focal length compared to eyepiece.

So,

• 1. Focal length of Objective $=f_0=2 \mathrm{~cm}$

• 2. Focal length of Eyepiece $=\mathrm{f}_{\mathrm{e}}=5 \mathrm{~cm}$

During normal adjustment in optical instruments the final image is formed at infinity. This allows the observer to view the image with a relaxed eye.

The total magnification ( M ) of a microscope is the product of the magnification of the objective ( $\mathrm{m}_{\mathrm{o}}$ ) and the magnification of the eyepiece ( $\mathrm{m}_{\mathrm{e}}$ ).

$ \mathrm{M}=\mathrm{m}_{\mathrm{o}} \times \mathrm{m}_{\mathrm{e}} $

The eyepiece acts like a simple magnifying glass. For an image formed at infinity, the angular magnification is the ratio of the distinct vision distance (D) to the focal length ( $\mathrm{f}_{\mathrm{e}}$ ).

$ m_e=\frac{D}{f_e} $

Where Least Distance of Distinct Vision $\mathrm{D}=25 \mathrm{~cm}$.

The objective forms a real, inverted image.

The tube length $(\mathrm{L})$ is defined as the distance between the second focal point of the objective and the first focal point of the eyepiece.

$ m_o \approx \frac{L}{f_o} $

So, the total magnification of the microscope is,

$ \mathrm{M}=\left(\frac{\mathrm{L}}{\mathrm{f}_{\mathrm{o}}}\right) \times\left(\frac{\mathrm{D}}{\mathrm{f}_{\mathrm{e}}}\right) $

Putting the values,

$ M=\left(\frac{10 \mathrm{~cm}}{2 \mathrm{~cm}}\right) \times\left(\frac{25}{5}\right)=25=5^2 $

$\Rightarrow $ $ 5^2=5^k \Rightarrow k=2 $

Therefore, the value of k is 2.

Hence, the correct option is (C).

When two thin prisms are combined to produce dispersion without deviation, they are arranged such that their refracting angles are inverted relative to each other. The net deviation produced by the combination is zero.

$\delta_{\text {net }}=\delta_1-\delta_2=0$

$\Rightarrow $ $ \delta_1=\delta_2 $

The deviation $(\delta)$ produced by a thin prism is given by the formula :

$ \delta=(\mu-1) \mathrm{A} $

where :

• $\mu$ is the refractive index of the material.

• A is the angle of the prism.

Since the deviations must be equal and opposite :

$ \left(\mu_1-1\right) \mathrm{A}_1=\left(\mu_2-1\right) \mathrm{A}_2 $

The angle of prism for first prism is $\mathrm{A}_1=5^{\circ}$ and its refractive index is $\mu_1=1.72$

The refractive index of the second prism is $\mu_2=1.9$

$ (1.72-1) \times 5^{\circ}=(1.9-1) \times \mathrm{A}_2 $

$\Rightarrow $ $ 3.6=0.9 \times \mathrm{A}_2 \Rightarrow \mathrm{~A}_2=\frac{3.6}{0.9}=4^{\circ} $

Therefore, the angle of the second prism is $4^{\circ}$. Hence, the correct option is (B).

For total internal reflection to occur, light must travel from a denser medium (lower speed) to a rarer medium (higher speed).

The speed of light in medium A (Denser) is $\mathrm{v}_{\mathrm{A}}=2.4 \times 10^8 \mathrm{~m} / \mathrm{s}$

The speed of light in medium $B$ (rarer) is $v_B=2.7 \times 10^8 \mathrm{~m} / \mathrm{s}$

Since $\mathrm{v}_{\mathrm{A}}<\mathrm{v}_{\mathrm{B}}$, medium A is indeed optically denser than medium B , satisfying the condition for TIR.

The critical angle is determined by Snell's Law at the boundary where the angle of refraction is $90^{\circ}$ :

$ n_A \sin \left(\theta_c\right)=n_B \sin \left(90^{\circ}\right) \Rightarrow \sin \left(\theta_c\right)=\frac{n_B}{n_A} $

Since the refractive index n is inversely proportional to the speed of light $\left(\mathrm{n}=\frac{\mathrm{c}}{\mathrm{v}}\right)$,

$ \frac{n_B}{n_A}=\frac{c / v_B}{c / v_A}=\frac{v_A}{v_B} $

So,

$ \sin \left(\theta_c\right)=\frac{v_A}{v_B} $

$\Rightarrow $ $\sin \left(\theta_C\right)=\frac{2.4 \times 10^8}{2.7 \times 10^8}=\frac{8}{9}$

$ \tan \left(\theta_c\right)=\frac{8}{\sqrt{9^2-8^2}}=\frac{8}{17} \Rightarrow \theta_c=\tan ^{-1}\left(\frac{8}{\sqrt{17}}\right) $

Therefore, the critical angle is $\tan ^{-1}\left(\frac{8}{\sqrt{17}}\right)$.

Hence, the correct option is (D).

The parallax method is an experimental technique used to locate the exact position of an image. It involves using an object pin and an image pin. The position of the image is found by placing the image pin such that there is no relative shift (no parallax) between the image pin and the actual image of the object pin when you move your eye side to side.

This technique requires the image to be a real image that is formed in the actual space in front of the mirror. So, to use the parallax method, the concave mirror is required because it produces a real image.

The image formation cases for a concave mirror based on the object's position :

• Between the Pole (P) and Focus (F) :

The mirror forms a virtual, erect, and magnified image behind the mirror. A virtual image cannot be located using a physical image pin in standard parallax methods because the light rays do not actually converge at the image location.

At the Focus (F) : The image is formed at infinity.

Between F and C : The image so formed is real and inverted in front of mirror.

At C : The mirror forms a real and inverted image in front of the mirror.

While placing the object between F and C does form a real image, stating it ONLY works here is incorrect because placing it beyond C also works. Hence, option (A) is incorrect.

Placing the object between P and F forms a virtual image, making the standard parallax method impossible. Hence, option (B) is incorrect.

Placing the object at any point beyond the focus (F) the nature of image formed is real (whether it is formed beyond C , at C , or between C and F ). This allows the image pin to be used to remove parallax. Hence,

While placing it beyond C forms a real image, restricting it to ONLY beyond C is incorrect because placing it between F and C also works. Hence, the option (D) is incorrect.

Therefore, the correct option is (C).

The focal length of convex lens $=\mathrm{f}_1=+5 \mathrm{~cm}$

The focal length of concave lens $=\mathrm{f}_2=-4 \mathrm{~cm}$

Initial object distance: $u=-10 \mathrm{~cm}$ with respect to convex lens.

Using lens formula for first lens,

$\frac{1}{v_1}-\frac{1}{u}=\frac{1}{f_1} \Rightarrow \frac{1}{v_1}-\frac{1}{-10}=\frac{1}{5}$

$\Rightarrow \frac{1}{v_1}=\frac{1}{5}-\frac{1}{10}=\frac{1}{10}$

$ \mathrm{v}_1=10 \mathrm{~cm} $

So, the image by first lens is formed at a distance of 10 cm from the first lens. The positive sign shows that the first image is formed right of first lens.

Case 1 : when lenses are combined together without any gap ( $\mathrm{d}=0$ )

When an object is placed at distance $u$, the first lens forms an image at distance $v_1$.

Using lens formula:

$ \frac{1}{\mathrm{v}_1}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}_1} \ldots \text { (i) } $

This image acts as a virtual object for the second lens. Since there is no gap, the object distance for the second lens is simply $\mathrm{v}_1$. It forms the final image at distance v .

$ \frac{1}{v}-\frac{1}{v_1}=\frac{1}{f_2} \ldots (ii)$

From both the equations,

$\left(\frac{1}{v_1}-\frac{1}{u}\right)+\left(\frac{1}{v}-\frac{1}{v_1}\right)=\frac{1}{f_1}+\frac{1}{f_2}$

$\Rightarrow $ $\frac{1}{v}-\frac{1}{u}=\frac{1}{f_1}+\frac{1}{f_2}=\frac{1}{f_{e q}}$

$\Rightarrow $ $\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2}$

$\Rightarrow $ $\frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{1}{5}+\frac{1}{-4}=\frac{4-5}{20}=-\frac{1}{20}$

$\Rightarrow $ $ \mathrm{f}_{\mathrm{eq}}=-20 \mathrm{~cm} $

Using the lens formula for equivalent lens :

$ \frac{1}{v}-\frac{1}{u}=\frac{1}{f_{e q}} $

$\Rightarrow $ $\frac{1}{v}-\frac{1}{-10}=-\frac{1}{20}$

$\Rightarrow $$\frac{1}{v}=-\frac{1}{20}-\frac{1}{10}$

$\Rightarrow $ $\frac{1}{v}=\frac{-1-2}{20}=-\frac{3}{20}$

$\Rightarrow $ $ \mathrm{v}=-\frac{20}{3} \mathrm{~cm} $

The magnification $\mathrm{m}_1$ for the combined system is the ratio of image distance to object distance:

$ m_1=\frac{v}{u}=\frac{-20 / 3}{-10}=\frac{2}{3} $

Case 2 : When lenses are separated by a gap of d = 1 cm

Refraction through the first lens (Convex, $\mathrm{f}_1=+5 \mathrm{~cm}$ )

The object is still at $\mathrm{u}_1=-10 \mathrm{~cm}$.

$ \frac{1}{v_1}-\frac{1}{u_1}=\frac{1}{f_1} $

$\Rightarrow $ $\frac{1}{v_1}-\frac{1}{-10}=\frac{1}{5}$

$\Rightarrow $ $\frac{1}{v_1}+\frac{1}{10}=\frac{1}{5}$

$\Rightarrow $ $\frac{1}{v_1}=\frac{1}{5}-\frac{1}{10}=\frac{2-1}{10}=\frac{1}{10}$

$\Rightarrow $ $ \mathrm{v}_1=+10 \mathrm{~cm} $

The lateral magnification of the first lens is :

$ \mathrm{m}_{\mathrm{L} 1}=\frac{\mathrm{v}_1}{\mathrm{u}_1}=\frac{10}{-10}=-1 $

Refraction through the second lens (Concave, $\mathrm{f}_2=-4 \mathrm{~cm}$ )

The image formed by the first lens acts as the object for the second lens. Since the first image is 10 cm to the right of $L_1$, and $L_2$ is placed 1 cm to the right of $L_1$, the distance of this image from $L_2$ is:

$ \mathrm{u}_2=\mathrm{v}_1-\mathrm{d}=10-1=+9 \mathrm{~cm} $

It is a virtual object for the second lens, hence the positive sign.

Now, applying the lens formula for $\mathrm{L}_2$ :

$ \frac{1}{\mathrm{v}_2}-\frac{1}{\mathrm{u}_2}=\frac{1}{\mathrm{f}_2} $

$\Rightarrow $ $\frac{1}{v_2}-\frac{1}{9}=\frac{1}{-4}$

$\Rightarrow $ $\frac{1}{v_2}=-\frac{1}{4}+\frac{1}{9}=\frac{-9+4}{36}=-\frac{5}{36}$

$\Rightarrow $ $ \mathrm{v}_2=-\frac{36}{5} \mathrm{~cm} $

The lateral magnification of the second lens is:

$ \mathrm{m}_{\mathrm{L} 2}=\frac{\mathrm{v}_2}{\mathrm{u}_2}=\frac{-36 / 5}{9}=-\frac{4}{5} $

So, the total magnification in second case is

$ \mathrm{m}_2=\mathrm{m}_{\mathrm{L} 1} \times \mathrm{m}_{\mathrm{L} 2}=(-1) \times\left(-\frac{4}{5}\right)=\frac{4}{5} $

So, the required ratio of magnification is $\left|\frac{\mathrm{m}_1}{\mathrm{~m}_2}\right|=\frac{2 / 3}{4 / 5}=\frac{5}{6}$.

The given prism is an equilateral, that means the angle of the prism (A) is $60^{\circ}$.

Refractive index of the prism is $\mu=\sqrt{2}$

The ray grazes the second surface. This means the angle of emergence (e) is $90^{\circ}$.

The angle of incidence at the first surface of prism is i .

Let the angle of refraction at first surface is $\mathrm{r}_1$.

The ray travels straight through the prism and hits the second face AC.

Let the angle of incidence at the second surface be $\mathrm{r}_2$.

The ray bends away from the normal as it exits into the air. Let this angle be e (angle of emergence).

The sum of angles $r_1$ and $r_2$ is equal to the angle of prism.

$ \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 $

Using Snell's law at the emergent surface (Face AC),

$ \sqrt{2} \cdot \sin \left(r_2\right)=1 \cdot \sin \left(90^{\circ}\right) $

$\Rightarrow $ $\sqrt{2} \cdot \sin \left(r_2\right)=1$

$\Rightarrow $ $ \sin \left(r_2\right)=\frac{1}{\sqrt{2}} \Rightarrow r_2=45^{\circ} $

Using $\mathrm{A}=\mathrm{r}_1+\mathrm{r}_2$

$ 60^{\circ}=\mathrm{r}_1+45^{\circ} $

$\Rightarrow $ $r_1=60^{\circ}-45^{\circ}$

$\Rightarrow $ $ r_1=15^{\circ} $

So, the angle of refraction at the incident surface is $15^{\circ}$.

Hence, the correct option is (D).

The incident ray is parallel to the base where the base angle is $45^{\circ}$.

If the ray is parallel to the base, the angle of incidence i with the normal to the first surface is also $45^{\circ}$

Let the refractive index of the prism be $\mu=\sqrt{2}$ and air be $\mu_{\text {air }}=1$.

Applying Snell's Law at the first surface :

$ \mu_{\mathrm{air}} \sin (\mathrm{i})=\mu \sin \left(\mathrm{r}_1\right) $

$\Rightarrow $ $1 \times \sin \left(45^{\circ}\right)=\sqrt{2} \times \sin \left(\mathrm{r}_1\right)$

$\Rightarrow $ $\frac{1}{\sqrt{2}}=\sqrt{2} \sin \left(\mathrm{r}_1\right)$

$\Rightarrow $ $\sin \left(r_1\right)=\frac{1}{2}$

$\Rightarrow $ $ r_1=30^{\circ} $

It is given that the emergent ray grazes along the second surface. This means the angle of emergence e is $90^{\circ}$.

This occurs when the angle of incidence inside the prism at the second surface $\left(\mathrm{r}_2\right)$ is equal to the critical angle (C).

Applying Snell's Law at the second surface :

$ \mu \sin \left(\mathrm{r}_2\right)=\mu_{\mathrm{air}} \sin \left(90^{\circ}\right) $

$\Rightarrow $ $\sqrt{2} \times \sin \left(\mathrm{r}_2\right)=1 \times 1$

$\Rightarrow $ $\sin \left(r_2\right)=\frac{1}{\sqrt{2}}$

$\Rightarrow $ $ \mathrm{r}_2=45^{\circ} $

For any prism, the relationship between the apex angle A and the internal refracted angles is :

$ \mathrm{A}=\mathrm{r}_1+\mathrm{r}_2 $

$\Rightarrow $ $ \mathrm{A}=30^{\circ}+45^{\circ}=75^{\circ} $

Sum of angles in a triangle is $180^{\circ}$ :

$ 45^{\circ}+\theta+\mathrm{A}=180^{\circ} $

$\Rightarrow $ $45^{\circ}+\theta+75^{\circ}=180^{\circ}$

$\Rightarrow $ $120^{\circ}+\theta=180^{\circ}$

$\Rightarrow $ $ \theta=60^{\circ} $

So, the angle $\theta$ of prism is $60^{\circ}$.

Hence, the correct option is (C).

Equivalent focal length

$\begin{aligned} & \frac{1}{\mathrm{f}}=\frac{1}{\mathrm{f}_1}+\frac{1}{\mathrm{f}_2} \\ & =\frac{1}{30}+\frac{1}{-20}=\frac{2-3}{60}=-\frac{1}{60} \\ & \mathrm{f}=-60 \mathrm{~cm} \end{aligned}$

Lens formula

$\begin{aligned} & \frac{1}{\mathrm{v}}-\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{-20}=\frac{1}{-60} \\ & \mathrm{v}=-15 \mathrm{~cm} \end{aligned}$

$\begin{aligned} \frac{1}{\mathrm{f}} & =\left(\frac{1.3-1}{1}\right)\left(\frac{1}{\infty}-\frac{1}{-30}\right) \\ & =\left(\frac{1.5-1}{1}\right)\left(\frac{1}{-30}-\frac{1}{-30}\right) \\ & =\frac{0.3}{30}+\frac{0.5}{60}=\frac{1}{100}+\frac{1}{120} \\ & =\frac{6+5}{600}=\frac{11}{600} \\ \mathrm{f} & =\frac{600}{11} \mathrm{~cm} \end{aligned}$

Refractive index has no relation with mass density because both have different meaning. Hence reason is incorrect.

So (A) is correct but (R) is not correct.

Magnification for a mirror is

$m=-\frac{v}{u}$

Given $m=\frac{1}{4}$, so

$-\frac{v}{u}=\frac{1}{4}\;\Rightarrow\; v=-\frac{u}{4}$

For a real object, $u<0$. Since $m$ is positive, the image is virtual, so $v>0$.

Let $u=-x$ (where $x>0$). Then

$v=-\frac{-x}{4}=\frac{x}{4}$

Distance between object and image (object in front, image behind) is

$x+v=40$

$x+\frac{x}{4}=40$

$\frac{5x}{4}=40 \Rightarrow x=32\ \text{cm}$

So,

$u=-32\ \text{cm},\quad v=+8\ \text{cm}$

Now use mirror formula:

$\frac{1}{f}=\frac{1}{v}+\frac{1}{u}=\frac{1}{8}-\frac{1}{32}=\frac{4-1}{32}=\frac{3}{32}$

$f=\frac{32}{3}\approx 10.7\ \text{cm}$

Correct option: C) $10.7\ \text{cm}$

$\begin{aligned} & \mathrm{r}+\theta_{\mathrm{C}}=90^{\circ} \\ & \mu_1 \sin \theta=\mu_2 \operatorname{sinr} \\ & \sin \theta=\frac{\mu_2}{\mu_1} \sin \left(90-\theta_{\mathrm{C}}\right) \\ & \sin \theta=\frac{\mu_2}{\mu_1} \cos \theta_{\mathrm{C}} \\ & \sin \theta_{\mathrm{C}}=\frac{\mu_1}{\mu_2} \end{aligned}$

$\begin{aligned} & \sin \theta=\frac{\mu_2}{\mu_1} \sqrt{1-\frac{\mu_1^2}{\mu_2^2}} \\ & \sin \theta=\sqrt{\frac{\mu_2^2-\mu_1^2}{\mu_1^2}}=\sqrt{\frac{\frac{25}{16}-1}{1}} \\ & \sin \theta=\frac{3}{4} \\ & \theta=\sin ^{-1}\left(\frac{3}{4}\right) \end{aligned}$

To find the power of a combination of two thin convex lenses, we use the formula for the equivalent focal length $ f_{\text{eq}} $ of the lens system when the lenses are placed coaxially with a certain distance $ d $ between them:

$ \frac{1}{f_{\text{eq}}} = \frac{1}{f_1} + \frac{1}{f_2} - \frac{d}{f_1 f_2} $

Given:

$ f_1 = 30 \, \text{cm} $

$ f_2 = 10 \, \text{cm} $

$ d = 10 \, \text{cm} $ (distance between the two lenses)

First, convert the focal lengths from centimeters to meters:

$ f_1 = 0.3 \, \text{m} $

$ f_2 = 0.1 \, \text{m} $

Plug in the values into the equation:

$ \frac{1}{f_{\text{eq}}} = \frac{1}{0.3} + \frac{1}{0.1} - \frac{0.1}{(0.3)(0.1)} $

Calculate each term:

$ \frac{1}{0.3} = \frac{10}{3} \approx 3.33 $

$ \frac{1}{0.1} = 10 $

$ \frac{0.1}{(0.3)(0.1)} = \frac{0.1}{0.03} = \frac{10}{3} \approx 3.33 $

Substitute these calculated values back into the formula:

$ \frac{1}{f_{\text{eq}}} = 3.33 + 10 - 3.33 $

$ \frac{1}{f_{\text{eq}}} = 10 $

Thus, the power of the lens combination is:

$ \text{Power} = \frac{1}{f_{\text{eq}}} = 10 \, \text{D} $

Therefore, the power of the lens system is 10 diopters (D).

To find the focal length of a lens with a refractive index of 1.6 when placed in water, given that its focal length in air is 12 cm and the refractive index of water is 1.28, we can use the lens maker's formula:

$ \frac{1}{f} = \left(\frac{\mu_L}{\mu_m} - 1\right) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

When the lens is in air, the refractive index of the medium, $\mu_m$, is 1. Therefore, for the lens in air:

$ \frac{1}{12} = (1.6 - 1) \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Simplifying gives:

$ \frac{1}{12} = \frac{6}{10} \left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

Solving for $\left(\frac{1}{R_1} - \frac{1}{R_2}\right)$:

$ \frac{1}{R_1} - \frac{1}{R_2} = \frac{10}{72} $

Now, when the lens is placed in water, $\mu_m = 1.28$. Substituting into the equation gives:

$ \frac{1}{f} = \left(\frac{1.6}{1.28} - 1\right) \left(\frac{10}{72}\right) $

Calculating each part:

$ \frac{1}{f} = \frac{32}{128} \times \frac{10}{72} $

$ \frac{1}{f} = \frac{1}{4} \times \frac{10}{72} $

Hence, solving for $f$:

$ f = 28.8 \, \text{cm} = 288 \, \text{mm} $

Thus, the focal length of the lens in water is 288 mm.

$\begin{aligned} &\text { For Convex mirror }\\ &\begin{aligned} & \frac{1}{\mathrm{v}}+\frac{1}{\mathrm{u}}=\frac{1}{\mathrm{f}} \\ & \frac{1}{\mathrm{v}}-\frac{1}{30}=\frac{1}{30} \\ & \frac{1}{\mathrm{v}}=\frac{2}{30}=\frac{1}{15} \Rightarrow \mathrm{v}=15 \mathrm{~cm} \end{aligned} \end{aligned}$

Image formed by convex mirror is at 45 cm from object so plane mirror should be placed midway at 22.5 cm from object so that both of their images may coinside,

Therefore distance between both mirrors

$=30-22.5=7.5 \mathrm{~cm}$

Correct Answer : Option 2

To solve this problem, start by using the magnification formula for spherical mirrors:

$ m = \frac{f}{f - u} $

Given that the initial magnification $ m = \frac{1}{2} $, and the object distance $ u = -40 $ cm (the negative sign indicates distance measured against the incoming light), we can write:

$ \frac{1}{2} = \frac{f}{f - (-40)} $

This simplifies to:

$ \frac{1}{2} = \frac{f}{f + 40} $

Cross-multiplying gives:

$ f + 40 = 2f $

This leads to:

$ f = 40 \text{ cm} $

To find the new object distance for a magnification of $ m = \frac{1}{3} $, we use the same formula:

$ m = \frac{f}{f - u} $

Substitute the known values:

$ \frac{1}{3} = \frac{40}{40 - u} $

Cross-multiplying gives:

$ 40 - u = 120 $

Solving for $ u $ results in:

$ u = -80 \text{ cm} $

Thus, the object needs to be placed 80 cm from the mirror, opposite to the initial placement of 40 cm. To achieve this repositioning, the object needs to be moved 40 cm further from its initial 40 cm distance, totaling a move of 40 cm away from the mirror.

When light is refracted through a prism and the angle of deviation is at its minimum, the following occurs:

The refracted ray inside the prism is parallel to the base of the prism.

The angle of incidence and the angle of emergence are equal.

There are two sets of angles of incidence that result in the same angle of deviation, except when the deviation is at its minimum.

These points are illustrated by the equation:

$ \delta = \mathrm{I} + \mathrm{e} - \mathrm{A} $

For the minimum angle of deviation ($\delta_{\text{min}}$), the condition $\mathrm{I} = \mathrm{e}$ is met, and the refracted ray becomes parallel to the base. Thus, statements A, C, and D are correct.

To determine the refractive index ($\mu$) of a thin convex lens, we use the lens maker's formula:

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

Given:

Focal length ($f$) = 12 cm

Radii of curvature ($R_1$ and $R_2$) = 10 cm and -15 cm, respectively

Substituting these values into the formula, we have:

$ \frac{1}{12} = (\mu - 1) \left( \frac{1}{10} - \frac{1}{-15} \right) $

Simplifying the equation:

$ \frac{1}{12} = (\mu - 1) \left( \frac{3 + 2}{30} \right) $

$ \frac{1}{12} = (\mu - 1) \times \frac{5}{30} $

$ \frac{1}{12} = (\mu - 1) \times \frac{1}{6} $

Solving for $\mu$:

$ \mu - 1 = \frac{1}{12} \times 6 $

$ \mu - 1 = \frac{1}{2} $

$ \mu = \frac{3}{2} $

Thus, the refractive index of the lens material is $\mu = 1.5$.

$\begin{aligned} & \text { Using } m=\frac{f}{u-f} \\ & m_1=\frac{6}{18-6}=\frac{1}{2} \end{aligned}$

$\mathrm{m}_2=\frac{6}{18+6}=\frac{1}{4} \quad \therefore \frac{\mathrm{~m}_2}{\mathrm{~m}_1}=\frac{1}{2}$

To replace a bi-convex lens with another convex lens that has different radii of curvature on each side (i.e., $ R_1 \neq R_2 $), while maintaining the same lens power, the radii must satisfy a specific relationship.

For the current bi-convex lens, both radii of curvature are given as $ \frac{1}{6} \, \text{cm} $. The lens formula for power is tied to the radii through:

$ \frac{1}{f_1} = (\mu - 1) \left( \frac{2}{R} \right) $

When this lens is replaced by a lens with different radii ($ R_1 $ and $ R_2 $), the condition that has to be met is:

$ \frac{1}{f_2} = (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) $

For power equivalence, the expressions for $ \frac{1}{f_1} $ and $ \frac{1}{f_2} $ must be equal:

$ (\mu - 1) \left( \frac{1}{R_1} + \frac{1}{R_2} \right) = (\mu - 1) \left( \frac{2}{R} \right) $

This simplifies to:

$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{R} $

Given $ R = \frac{1}{6} \, \text{cm} $, it follows that:

$ \frac{1}{R_1} + \frac{1}{R_2} = \frac{2}{\left(\frac{1}{6}\right)} $

Thus, simplifying the equation gives:

$ \frac{1}{R_1} + \frac{1}{R_2} = 12 $

Therefore, any valid pair of radii $ R_1 $ and $ R_2 $ must satisfy this equation.

To determine the image distance when a spherical surface separates two media with refractive indices of 1 and 1.5:

Use the lens maker's formula for a spherical surface:

$ \frac{\mu_2}{v} - \frac{\mu_1}{u} = \frac{\mu_2 - \mu_1}{R} $

Where:

$\mu_1$ = 1 (refractive index of the first medium)

$\mu_2$ = 1.5 (refractive index of the second medium)

$u$ = -0.2 m (object distance, negative as per convention)

$R$ = 0.4 m (radius of curvature)

Substituting the known values:

$ \frac{1.5}{v} - \frac{1}{-0.2} = \frac{1.5 - 1}{0.4} $

Simplify the equation:

$ \frac{1.5}{v} = \frac{0.5}{0.4} - \frac{1}{0.2} $

Calculate:

$ \frac{1.5}{v} = -\frac{1.5}{0.4} $

Solve for $v$:

$ v = -0.4 \, \text{m} $

The negative sign indicates that the image is 0.4 meters to the left of the spherical surface.

Location of image of A :-

$\begin{aligned} & \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \Rightarrow \frac{1}{v}-\frac{1}{-30}=\frac{1}{20} \Rightarrow \frac{1}{v}=\frac{1}{60} \Rightarrow v=60 \mathrm{~cm} \\ & \therefore m=2 \end{aligned}$

Since size of object is small wrt the location hence

$\begin{aligned} & d v=m^2 d u \Rightarrow d v=4 \times 1=4 \mathrm{~cm} \\ & h_i=m h_0 \Rightarrow h_i(d y)=2 \times 2=4 \mathrm{~cm} \end{aligned}$

$\therefore$ Angle made with principle axis $=-45^{\circ}$

Using lens formula,

For Ist case : ${1 \over f} = \left( {{a^{{\mu _g}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

where, ${R_1},{R_2}$ = radii of refracting surfaces

$f$ = focal length

${a^{{\mu _g}}} = {{^{{\mu _g}}} \over {^{{\mu _a}}}} = {{1.5} \over 1} = 1.5$

So, ${1 \over {24}} = (1.5 - 1)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ (As f = 24 cm given)

$ \Rightarrow {1 \over {24}} = 0.5\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ \Rightarrow {1 \over {12}} = \left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ \Rightarrow {1 \over {{R_1}}} - {1 \over {{R_2}}} = {1 \over {12}}$ .... (i)

For IInd case:

${1 \over f} = \left( {w{\mu _g} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ \Rightarrow {1 \over f} = \left( {{{1.5} \over {1.33}} - 1} \right)\left( {{1 \over {12}}} \right)$ ..... from (i)

$ \Rightarrow {1 \over f} = {{17} \over {133}} \times {1 \over {12}} \Rightarrow f = {{133 \times 12} \over {17}}$

$ \Rightarrow f = 93.88$ cm

$\approx$ 94 cm

Hence, option 1 is correct.

Using lens maker formula,

${{{\mu _2}} \over v} - {{{\mu _1}} \over 4} = {{{\mu _2} - {\mu _1}} \over R}$

For the image formed by the surface on the right side.

${\mu _1} = 1,\,{\mu _2} = 1.5,\,u = {{ - R} \over 2},{R_1} = - R$

so, ${{1.5} \over v} - {1 \over { - R/2}} = {{1.5 - 1} \over { - R}}$

$ \Rightarrow {{1.5} \over v} + {2 \over R} = {{ - 1} \over {2R}}$

$ \Rightarrow {{1.5} \over v} = {{ - 5} \over {2R}} \Rightarrow v = {{ - 3R} \over 5}$

Here '$-$' sign tells that image is formed left to B.

So, the distance of image from $P, = {{R - 3R} \over \Delta } = {{2R} \over 5}$

Now, for the surface on the left side,

${\mu _1} = 1,{\mu _2} = 1.5,\mu = {{3R} \over 2},{R_2} = R$

So, ${{1.5} \over v} - {1 \over {{{3R} \over 2}}} = {{1.5 - 1} \over R}$

$ \Rightarrow {{1.5} \over v} - {2 \over {3R}} = {1 \over {2R}}$

$ \Rightarrow {{1.5} \over v} = {1 \over R}\left( {{1 \over 2} + {2 \over 3}} \right)$

$ \Rightarrow {{1.5} \over v} = {7 \over {6R}}$

$ \Rightarrow v = {{9R} \over 7}$

So distance from $P = {{9R} \over 7} - R = {{2R} \over 7}$

So the distance between two images,

$ = {{2R} \over 5} - {{2R} \over 7} = 0.4R - 0.2857R$

$ \approx 0.4R - 0.286R = 0.114R$.

Here,

we will use lens maker formula,

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

So, the focal length of L$_1$ remains the same as the original lens because the curvature of the surfaces does not change

So, ${\left( f \right)_{{L_1}}} = f$ .... (1)

Now, for L$_3$, L$_3$ is plano-convex lens

${R_1} = R,\,{R_2} = \infty $

So, ${1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over \infty }} \right)$

$ \Rightarrow {1 \over {{{\left( f \right)}_{{L_3}}}}} = \left( {\mu - 1} \right){1 \over R}$ ..... (2)

For whole double convex lens,

${1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} - {1 \over { - R}}} \right)$ (As ${R_1} = R$, ${R_2} = - R$)

$ \Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{1 \over R} + {1 \over R}} \right)$

$ \Rightarrow {1 \over f} = \left( {\mu - 1} \right)\left( {{2 \over R}} \right)$ .... (3)

From (2) and (3),

${1 \over {{{\left( f \right)}_{{L_3}}}}} = {1 \over {2f}} \Rightarrow {\left( f \right)_{{L_3}}} = 2f$

Hence, ${{{{\left( f \right)}_{{L_1}}}} \over {{{\left( f \right)}_{{L_3}}}}} = {f \over {2f}} \quad\text{....(from (1)})\quad= 1:2$

$(u+f)(v-f)=f^2$

$\begin{aligned} & \mathrm{m}=-3=-\frac{\mathrm{v}}{\mathrm{u}} \text { and } \mathrm{v}-\mathrm{u}=20 \mathrm{~cm} \\ & \mathrm{f}=\frac{\mathrm{vu}}{\mathrm{v}+\mathrm{u}}=\frac{(-30)(-10)}{-30-10} \\ & \therefore \mathrm{R}=+15 \end{aligned}$

$\begin{aligned} & \frac{\mathrm{n}_2}{\mathrm{v}}-\frac{\mathrm{n}_1}{\mathrm{u}} \frac{\mathrm{n}_2-\mathrm{n}_1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}-\frac{1.3}{-13}=\frac{0.1}{\mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \frac{1.4}{\mathrm{v}}=\frac{1-\mathrm{R}}{10 \mathrm{R}} \\ & \mathrm{~m}=\frac{\mathrm{v} / \mathrm{n}_2}{\mathrm{u} / \mathrm{n}_1} \\ & -2 \times \frac{(-13)}{1.3}=\frac{10 \mathrm{R}}{1-\mathrm{R}} \\ & \mathrm{R}=\frac{2}{3} \mathrm{~cm} \end{aligned}$

We know, for thin prism,

$\mu = {{A + \delta } \over A}$

where, $\mu$ = refractive index

A = Angle of prism

$\delta$ = Angle of deviation

We can write, $\mu A = A + \delta $

$ \Rightarrow \delta = (\mu - 1)A$

given, ${\delta _{net}} = 0$

$ \Rightarrow ({\mu _1} - 1){A_1} - ({\mu _2} - 1){A_2} = 0$

$ \Rightarrow (1.54 - 1)4 - (1.72 - 1){A_2} = 0$

$ \Rightarrow {A_2} = {{54 \times 4} \over {72}} \Rightarrow {A_2} = 3^\circ $

from $\Delta OCB,\,\tan \theta = {{CB} \over {OC}} = {R \over R} = 1$

$ \Rightarrow \theta = 45^\circ $

We know, $\sin c = {1 \over \mu }$

for minimum $\mu$, sinc maximum $\Rightarrow$ C maximum

Here, ${c_{\max }} = \theta = 45^\circ $

So, $\mu = {1 \over {\sin 45^\circ }} = {1 \over {{1 \over {\sqrt 2 }}}}$

$ \Rightarrow \mu = \sqrt 2 $

$\begin{aligned} &\mathrm{H}=18 \mathrm{~km}\\ &\text { Size of camera film }=2 \mathrm{~cm} \times 2 \mathrm{~cm}\\ &\begin{aligned} & \mathrm{A}_{\text {image }}=400 \mathrm{~km}^2 \\ & \mathrm{x}=20 \times 10^3 \mathrm{~m}=2 \times 10^4 \mathrm{~m} \\ & \mathrm{y}=2 \times 10^{-2} \mathrm{~m} \\ & \frac{\mathrm{x}}{\mathrm{y}}=10^6=\frac{18 \mathrm{Km}}{\mathrm{f}} \\ & \mathrm{f}=18 \times 10^{-3} \mathrm{~m}=18 \mathrm{~mm} \\ & \mathrm{f}=1.8 \mathrm{~cm} \end{aligned} \end{aligned}$

$ P_1 = 2.5\,D,\quad f_1 = \frac{1}{P_1} = \frac{1}{2.5} = 0.4\,m $

After an increase in optical power by 0.1 D:

$ P_2 = 2.5\,D + 0.1\,D = 2.6\,D,\quad f_2 = \frac{1}{P_2} = \frac{1}{2.6} \approx 0.3846\,m $

The change in focal length is:

$ \Delta f = f_1 - f_2 = 0.4\,m - 0.3846\,m \approx 0.0154\,m $

The relative decrease in focal length is then calculated as:

$ \text{Relative decrease} = \frac{\Delta f}{f_1} = \frac{0.0154}{0.4} \approx 0.0385 \approx 0.04 $

Thus, the relative decrease in focal length is approximately 0.04.

$ \frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R} $

For a thin plano‑convex lens made of glass (refractive index $n_2 = 1.5$) immersed in a liquid (refractive index $n_1 = 1.2$), when the plane side is silver‐coated, a ray entering the curved surface from the liquid is refracted into the glass, reflects off the coated plane (thereby reversing its direction) and finally refracts back into the liquid at the same curved surface. In the paraxial approximation the net effect is equivalent to a mirror in the liquid with an effective focal length $f$.

A ray that is incident parallel to the optical axis (at height $h$) will, after this combined refraction–reflection–refraction process, emerge with an angular deviation corresponding to a mirror having a power given by:

$ \frac{h}{f} = \frac{2(n_2 - n_1)}{n_1R}\, h\,. $

This gives the relation:

$ \frac{1}{f} = \frac{2(n_2 - n_1)}{n_1 R}\,. $

Given that the effective focal length is $f = 0.2\,\text{m}$, substitute the values:

$ \frac{1}{0.2} = \frac{2(1.5 - 1.2)}{1.2\,R}\,. $

Simplify the numerator:

$ \frac{1}{0.2} = \frac{2(0.3)}{1.2\,R} = \frac{0.6}{1.2\,R}\,. $

Since

$ \frac{0.6}{1.2} = 0.5, $

the equation becomes:

$ 5 = \frac{0.5}{R}\,. $

Solving for $R$:

$ R = \frac{0.5}{5} = 0.10\,\text{m}\,. $

Thus, the radius of curvature of the curved surface of the lens is $0.10\,\text{m}$.

$ \frac{1}{f} = \left(\frac{n}{n'} - 1\right)\left(\frac{1}{R_1} - \frac{1}{R_2}\right) $

For a thin plano-convex lens, one surface is flat, meaning $R_2 = \infty$ and $\frac{1}{R_2} = 0.$ The formula reduces to:

$ \frac{1}{f} = \left(\frac{n}{n'} - 1\right)\frac{1}{R_1} $

Thus, the focal length is:

$ f = \frac{R_1}{\left(\frac{n}{n'} - 1\right)} $

For the First Lens (in Air)

Refractive index of lens, $n = 1.5$

Surrounding medium (air), $n' = 1$

Radius of curvature, $R_1 = 2 \text{ cm}$

Substitute into the formula:

$ f_1 = \frac{2}{\left(\frac{1.5}{1} - 1\right)} = \frac{2}{(1.5 - 1)} = \frac{2}{0.5} = 4 \text{ cm} $

For the Second Lens (in Liquid)

Refractive index of lens, $n = 1.5$

Surrounding medium (liquid), $n' = 1.2$

Radius of curvature, $R_1 = 3 \text{ cm}$

Substitute into the formula:

$ f_2 = \frac{3}{\left(\frac{1.5}{1.2} - 1\right)} $

First, calculate the term inside the parentheses:

$ \frac{1.5}{1.2} = 1.25 \quad \Rightarrow \quad 1.25 - 1 = 0.25 $

Now, compute $f_2$:

$ f_2 = \frac{3}{0.25} = 12 \text{ cm} $

Ratio of the Focal Lengths

$ \frac{f_1}{f_2} = \frac{4}{12} = \frac{1}{3} $

Thus, the ratio of $f_1$ to $f_2$ is:

$ 1 : 3 $

When a concave mirror is dipped in a liquid, its focal length is still determined by the geometry of the mirror. For a spherical concave mirror, the focal length is given by

$ f = \frac{R}{2} $

where $ R $ is the radius of curvature of the mirror. This derivation comes solely from geometrical considerations and the law of reflection, which states that the angle of incidence equals the angle of reflection. Since reflection is independent of the medium in which the mirror is immersed, the focal length remains unchanged.

Thus, even when the mirror is in a liquid of refractive index $ \mu $, its focal length is still

$ f. $

The correct answer is Option B: $ f $.

$ \textbf{Step 1: Symmetry at Minimum Deviation} $

In a prism, when the deviation is minimum, the path of light is symmetric. This means that the angle of incidence ($i$) is equal to the angle of emergence ($i'$), and the light inside the prism makes equal angles with the prism faces. If the prism angle is $A$, then the refracted angle at each interface is:

$ r = \frac{A}{2} $

$ \textbf{Step 2: Relating Deviation to the Angles} $

The formula for the deviation ($D$) in a prism is given by:

$ D = i + i' - A $

At minimum deviation, $i = i'$, so:

$ D_{\text{min}} = 2i - A $

The special condition given is that the minimum deviation is equal to the prism angle:

$ D_{\text{min}} = A $

Thus:

$ A = 2i - A \quad \Longrightarrow \quad 2i = 2A \quad \Longrightarrow \quad i = A $

$ \textbf{Step 3: Applying Snell's Law} $

At the first surface, Snell's law gives:

$ \sin i = \mu \sin r $

Substitute the values $i = A$ and $r = \frac{A}{2}$:

$ \sin A = \mu \sin\frac{A}{2} $

Given that the refractive index $\mu = \sqrt{3}$, we have:

$ \sin A = \sqrt{3} \sin\frac{A}{2} $

$ \textbf{Step 4: Solving the Equation} $

Recall the double-angle formula for sine:

$ \sin A = 2 \sin\frac{A}{2} \cos\frac{A}{2} $

Substitute this into the previous equation:

$ 2 \sin\frac{A}{2} \cos\frac{A}{2} = \sqrt{3} \sin\frac{A}{2} $

Assuming $\sin\frac{A}{2} \neq 0$, we can divide both sides by $\sin\frac{A}{2}$:

$ 2 \cos\frac{A}{2} = \sqrt{3} $

Solve for $\cos\frac{A}{2}$:

$ \cos\frac{A}{2} = \frac{\sqrt{3}}{2} $

Since:

$ \cos 30^\circ = \frac{\sqrt{3}}{2} $

It follows:

$ \frac{A}{2} = 30^\circ \quad \Longrightarrow \quad A = 60^\circ $

$ \textbf{Final Answer:} $

The angle of the prism is $60^\circ$.

We can derive the lateral shift (d) by considering the geometry of a ray entering a parallel-sided slab. Here’s a step‐by‐step explanation:

The slab has a thickness $h$ (measured perpendicular to the surfaces). When the ray enters the glass from air at angle $i$ (with respect to the normal), it refracts inside and makes an angle $r$.

Inside the slab, the ray travels a distance

$L = \frac{h}{\cos r}$

because the vertical component of the path must cover a distance $h$.

The ray inside the slab is shifted laterally (parallel to the surface). Its lateral displacement over the distance $L$ is

$L \sin r = \frac{h \sin r}{\cos r}.$

However, the emergent ray is parallel to the incident ray but not collinear. By constructing the appropriate geometry (extending the emergent ray backwards until it meets the incident ray's extension at the top interface), you find that the lateral shift between the incident ray and the emergent ray is given by

$d = \frac{h \sin(i - r)}{\cos r}.$

This expression correctly accounts for both the refraction inside the slab and the geometry of the displacement.

Thus, the correct answer is:

Option D: $\frac{h \sin (i-r)}{\cos r}.$

We wish to find the distance from the point where the ray joining the object (O) and its image (I) meets the spherical surface (P) to the object. The given condition is that

  $PO = PI,$

meaning that P is the midpoint of OI.

Let’s work through the steps:

Setup the coordinate system and geometry:

 • Choose the vertex of the spherical surface (the “pole”) as the origin (x = 0).

 • Let the optical axis be the x‑axis.

 • Since the object is in air, place O at x = –u (with u > 0).

 • Since the real image is formed inside the glass, place I at x = v (with v > 0).

 • The spherical surface has radius of curvature $R$ and its center of curvature is in the glass. Thus, we take the center of the sphere as $C=(R,0)$.

 • The vertex (the point on the sphere closest to the object) is then located at x = 0.

Position of P:

 • The ray joining O and I is along the x‑axis. Its midpoint is

  $x_P = \frac{-u + v}{2}.$

 • However, since the physical part of the spherical surface is the convex cap met first by the incident light, the ray in fact meets the surface at the vertex, which is at x = 0.

 • Equating the midpoint to 0, we have

  $\frac{-u + v}{2}=0 \quad \Longrightarrow \quad v = u.$

 • Thus, the object and its image are equidistant from the vertex, and

  $PO = |0 - (-u)| = u.$

Relate u and R using the refraction at a spherical surface:

 • The formula for refraction at a spherical surface (using the appropriate sign–convention) is

  $\frac{n_1}{u} + \frac{n_2}{v} = \frac{n_2 - n_1}{R},$

  where for our problem

   – $n_1 = 1$ (air)

   – $n_2 = 1.5$ (glass).

 • With $v = u$, substitute in:

  $\frac{1}{u} + \frac{1.5}{u} = \frac{0.5}{R} \quad \Longrightarrow \quad \frac{2.5}{u} = \frac{0.5}{R}.$

 • Solving for u:

  $u = \frac{2.5\,R}{0.5} = 5R.$

Conclusion:

 • Since $PO = u$, we have

  $PO = 5R.$

Thus, the distance $PO$ equals $5R$, which is Option A.

$ \begin{aligned} & \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2}{\mathrm{f}_{\mathrm{L}}}-\frac{1}{\mathrm{f}_{\mathrm{m}}} \\\\ & \mathrm{f}_{\mathrm{m}}=-\frac{\left|\mathrm{R}_2\right|}{2} \\\\ & \frac{1}{\mathrm{f}_{\mathrm{L}}}=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}\right) \end{aligned} $

Using lens maker formula, ${1 \over f} = \left( {{{{\mu _2}} \over {{\mu _1}}} - 1} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

Note : here, we will take air as medium for all three lenses so ${\mu _1} = 1$

${P_1} = {1 \over {{f_1}}} = \left( {{4 \over 3} - 1} \right)\left( {{1 \over \infty } - {1 \over { - {R_1}}}} \right)$

${P_1} = {1 \over 3}\left( {{1 \over {{R_1}}}} \right)$ ..... (1)

${P_2} = {1 \over {{f_2}}} = \left( {{3 \over 2} - 1} \right)\left( {{1 \over { - {R_1}}} - {1 \over { - {R_2}}}} \right)$

$ \Rightarrow {P_2} = {{ - 1} \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$ .... (2)

${P_3} = {1 \over {{f_3}}} = \left( {{4 \over 3} - 1} \right)\left( {{1 \over { - {R_2}}} - {1 \over \infty }} \right)$

$ \Rightarrow {P_3} = {{ - 1} \over 3}{1 \over {{R_2}}}$ ...... (3)

So, ${P_{eq}} = {1 \over {3{R_1}}} - {1 \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) - {1 \over {3{R_2}}}$

$ = {1 \over 3}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right) - {1 \over 2}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

$ = \left( {{1 \over 3} - {1 \over 2}} \right)\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

${P_{eq}} = {{ - 1} \over 6}\left( {{1 \over {{R_1}}} - {1 \over {{R_2}}}} \right)$

We know, for a combination of a lens and a mirror,

${P_{comb}} = 2{P_l} + {P_m}$ ...... (1)

Also using lens maker formula,

${1 \over {{f_l}}} = (\mu - 1)\left( {{1 \over R} - {1 \over { - R}}} \right) = (\mu - 1){2 \over R}$

So using eq. (i),

$ - {1 \over {{f_{comb}}}} = 2\left( {{1 \over {{f_l}}}} \right) - {1 \over {{f_m}}}$

$ \Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} - {1 \over {\left( { - {R \over 2}} \right)}}$

$ \Rightarrow - {1 \over {{f_{comb}}}} = {{4(\mu - 1)} \over R} + {2 \over R}$

$ \Rightarrow - {1 \over {{f_{comb}}}} = {2 \over R}(2\mu - 2 + 1) = {2 \over R}(2\mu - 1)$

$ \Rightarrow {f_{comb}} = {{ - R} \over {2(2\mu - 1)}}$

$ \Rightarrow {R_{comb}} = 2{f_{comb}} = {{ - 2R} \over { - 2(2\mu - 1)}} = {{ - R} \over {2\mu - 1}}$

Hence, distance $ = {R \over {2\mu - 1}}$