Geometrical Optics

Consider the refraction at the face AB. Snell's law,

$\sin i/\sin r = \sin 60^\circ /\sin r = \sqrt 3 $,

gives $r = 30^\circ $.

The geometry in BCQP gives

$\angle BPQ = 30^\circ + 90^\circ = 120^\circ $

$\angle CQP = 360^\circ - (135^\circ + 60^\circ + 120^\circ ) = 45^\circ $.

Thus, the angle of incidence at Q is ${i_1} = 45^\circ $. The critical angle for prism to air refraction is given by $\sin {i_c} = 1/\sqrt 3 $. Since $\sin {i_1} = 1/\sqrt 2 > 1\sqrt 3 $, we get ${i_1} > {i_c}$ i.e., the angle of incidence is greater than the critical angle. Thus, the ray undergoes total internal reflection at Q. The laws of reflection gives ${r_1} = {i_1} = 45^\circ $. In triangle QRD, $\angle QRD = 60^\circ $ and hence the angle of incidence at R is ${i_2} = 30^\circ $.

Applying Snell's law at face AD, we get

$\sqrt 3 \times \sin 30^\circ = 1 \times \sin e$

or, $\sqrt 3 \times {1 \over 2} = \sin e$

$\sin e = {{\sqrt 3 } \over 2}$

or, $e = {\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) = 60^\circ $

From figure,

The angle between the incident ray and the emergent ray is 90$^\circ$.

Hence, option (c) is correct and option (d) is incorrect.

Note : Angle between incident and emergent rays is the same as the angle between the two faces = 90$^\circ$.

First refraction through lens

Here, u = $-$30 cm, f = +15 cm

Using ${1 \over v} - {1 \over u} = {1 \over f}$

${1 \over v} - {1 \over { - 30}} = {1 \over { + 15}}$ or ${1 \over v} = {1 \over {15}} - {1 \over {30}}$

v = +30 cm

$\therefore$ Image I₁ is real and formed at a distance of 30 cm on the right side of the lens.

Distance of image I₁ from the mirror is

30 cm $-$ 10 cm = 20 cm

The image I₁ acts as an virtual object for the mirror. The mirror forms an image I₂ at a distance of 20 cm in front of it.

The image I₂ acts as an object for the lens.

Second refraction through lens

Here, u = +10 cm, f = +15 cm

${1 \over v} - {1 \over { + 10}} = {1 \over { + 15}}$ or ${1 \over v} = {1 \over {15}} + {1 \over {10}}$ $\Rightarrow$ v = +6 cm

The final image I₃ is real and at a distance of 16 cm from the mirror.

If $\mu_1<\mu_2$ then the ray bends towards the normal after refraction at $\mu_1-\mu_2$ interface. Incident rays parallel to the optic axis bend towards the optic axis by the convex lens and away from the optic axis by the concave lens. If $\mu_1>\mu_2$ then the ray bends away from the normal. If $\mu_2=\mu_3$ then the ray goes straight without bending at the $\mu_2-\mu_3$ interface. If $\mu_2>\mu_3$ then the ray bends away from the normal.

Using mirror formula for the concave mirror, we have

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

Option (A) : When the value is (42, 56), we have

${1 \over { - 56}} + {1 \over { - 42}} = {{ - 98} \over {56 \times 42}} \Rightarrow f = {{ - 56 \times 42} \over {98}} = - 24$ cm

Option (B) : When the value is (48, 48), we have

${1 \over { - 48}} + {1 \over { - 48}} = {{ - 1} \over {24}} \Rightarrow f = - 24$ cm

Option (C) : When the value is (66, 33), we have

${1 \over { - 33}} + {1 \over { - 66}} = {{ - 1} \over {22}} \Rightarrow f = - 22$ cm

Option (D) : When the value is (78, 39), we have

${1 \over { - 39}} + {1 \over { - 78}} = {{ - 1} \over {26}} \Rightarrow f = - 26$ cm

Since options (A) and (B) give $f = - 24$ cm, which is the same as that given in the problem, we conclude that the options (C) and (D) as the correct options.

The velocity of the ball when it has fallen through 7.2 m is

$V = \sqrt {2g(7.2)} = 12$ m/s

For a general height $h$ above the water surface, we have

${{h'} \over h} = {4 \over 3} \Rightarrow h' = {4 \over 3}h$

Therefore, $V' = {4 \over 3}V = {4 \over 3} \times 12 = 16$ m/s

where $V'$ is the velocity of ball with respect to the fish.

From Snell's laws

${\mu _1}\sin \theta = {\mu _2}\sin {\theta _2}$

Case I : Region I

${n_0}\sin \theta = {{{n_0}} \over 2}\sin \alpha $ ...... (i)

Case II : Region II

${{{n_0}} \over 2}\sin \alpha = {{{n_0}} \over 6}\sin {\theta _c}$ ..... (ii) ($\therefore$ $\beta = {\theta _c}$)

Case II : Region III

${{{n_0}} \over 6}\sin {\theta _c} = {{{n_0}} \over 8}\sin 90^\circ $

$ \Rightarrow {{{n_0}} \over 6}\sin {\theta _c} = {{{n_0}} \over 8}$ ..... (iii)

From equation (i), (ii) and (iii), we have,

${n_0}\sin \theta = {{{n_0}} \over 8}$

$ \Rightarrow \sin \theta = {1 \over 8}$

$ \Rightarrow \theta = {\sin ^{ - 1}}\left( {{1 \over 8}} \right)$

(A) $\to$ P, Q, R, S

$v = {f \over {1 - {f \over u}}},m = {{ - v} \over u}$

For different values of u, we get different values of $v$(+ve, $-$ve or $\infty$). Like wise $m$ comes out to be positive, negative, greater than 1 etc.

(B) $\to$ Q

In case of a convex mirror we always get virtual, diminished image, between P and F.

(C) $\to$ P, Q, R, S

When object is at infinity, a real, inverted and diminished image is formed at F. When the object is brought closer to the lens. The image moves away from the lens and increase in size. (Image is between F and 2F). when the object is at 2F', the image is formed at 2F. The image is real, inverted and is of the same size. When the object is moved closer between 2F' and F', the image is real, inverted and magnified. When the object is at F' the image is real, inverted, highly magnified and is formed at infinity.

(D) $\to$ P, Q, R, S

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

Here ${R_1} < {R_2}$

$\therefore$ $f$ is positive.

Therefore, it behaves like a convex lens.

For minimum deviation, the ray in the prism is parallel to base of the prism. This condition does not depend on the colour (i.e., wavelength of incident radiation. So, both the cases, by geometry, $r=30^\circ$.

Since the wave fronts are parallel in each medium, the beam is also a parallel beam travelling in the direction perpendicular to the wave fronts. Since the beam would bend inside medium 2 when travelling from medium 1 , the second medium has the higher refractive index.

Trick : For plane wave fronts the beam of light is parallel.

Since point c and d are on same wave front.

Therefore, $\phi_d=\phi_c$

Similarly, $\phi_e=\phi_f$

$\therefore~\phi_d-\phi_f=\phi_c-\phi_f$

About forming an inverted image: A concave mirror will make an upside-down (inverted) image only when the object is placed farther from the mirror than its focal point. This means the distance $x$ from the object to the pole should be bigger than the focal length $f$ (so, $x > f$).

When the object and image overlap: If you place the object at $x = 2f$ (twice the focal length), the image will appear at the same spot as the object. At this exact spot, moving your eye side to side does not make the image and object swap positions. This means you cannot see “parallax.”

Understanding parallax with two points: Imagine you have two points in space, $A$ and $B$. $A$ is closer to your eye, $B$ is farther away. If your eye is at point $E$ on the line joining $A$ and $B$, the following image shows this:

How parallax works: If you move your eye in one direction (say, to the left), the point that is closer to you ($A$) will appear to move in the opposite direction (to the right) compared to the background ($B$). The point that is farther away ($B$) will seem to move in the same direction as your eye (to the left).

For example, if you move your eye from $E$ to $E_1$ (leftwards), the line from your eye to the far point $B$ changes from $E B$ to $E_1 B$. The near point $A$ looks like it shifts to the right compared to this line. The line from your eye to the near point $A$ changes from $E A$ to $E_1 A$, and the far point $B$ now seems to move to the left relative to $A$.

You can try this! Put two fingers in front of your eyes (one closer, one farther), and move your head. You’ll see the closer finger appears to shift more, in the opposite direction you move your eye.

If the two points ($A$ and $B$) are at the same spot (they “coincide”), you won’t see any parallax. There is no shifting effect between them.

What happens in the mirror experiment: In this experiment, when you move your eye to the left, the image of the pin seems to move to the right of the real object pin. This shows the image is between your eye and the object pin. That means the object is closer to the mirror’s pole $P$ than its image is. For this to happen, the object must be between the focal length $f$ and twice the focal length $2f$. In other words, $f < x < 2f$.

a reflected ray and refracted ray and the angle b/w them would be less than (180$^\circ$ - 2$\theta$).

Statement 1

The formula connecting $ u, v $ and $ f $ for a spherical mirror is valid only for mirrors whose sizes are very small compared to their radii of curvature.

✅ True.

This is due to the paraxial approximation—the mirror formula $\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$ holds only when the mirror aperture is small so that rays make small angles with the principal axis. This avoids spherical aberration.

Statement 2

Laws of reflection are strictly valid for plane surfaces, but not for large spherical surfaces.

❌ False.

The laws of reflection (angle of incidence = angle of reflection) are fundamental and apply to all reflecting surfaces—plane, spherical, or irregular. What changes in large spherical mirrors is that reflected rays do not all converge at a single focus due to geometry, but the law of reflection itself still holds locally at every point on the surface.

Conclusion

• Statement 1: True

• Statement 2: False

✅ Correct Option: C

Option C: Statement 1 is True, Statement 2 is False.

Focal length of the lens is given by

$ f=\frac{\mathrm{R}}{\mu-1}=15 $

After silvering the plane surface of the lens, it converts into the thin plano-convex mirror.

Focal length is now changed.

$\therefore$ For mirror, new focal length $f$ is given by

$ f^{\prime}=\frac{R}{2(\mu-1)}=\frac{15}{2}=7.5 $

Thus, for the mirror, equation becomes,

$ \begin{gathered} \frac{1}{v}+\frac{1}{u}=\frac{1}{f^{\prime}} \\ {[u=\text { object distance from mirror }=-20 \mathrm{~cm}} \\ v=\text { image distance from mirror }=?] \\ \frac{1}{v}+\frac{1}{-20}=\frac{1}{-7.5} \end{gathered} $

$ \begin{aligned} & \frac{1}{v}=-\frac{1}{7.5}+\frac{1}{20}=\frac{-20+7.5}{7.5 \times 20} \\ & \frac{1}{v}=\frac{-12.5}{150} \\ & v=\frac{-150}{12.5} \\ & v=-12 \mathrm{~cm} \end{aligned} $

$\therefore-\operatorname{sign}$ Indicates that it is to the left of the system.

$ \text { From, figure, } \quad r=f \tan \alpha $

$ \begin{aligned} & & &r \propto f \\ & & r^2 & =f^2 \tan ^2 \alpha \\ & \therefore & \pi r^2 & =f^2 \tan ^2 \alpha \quad \text { (area of image) } \\ & \therefore & r^2 & \propto f^2 \end{aligned} $

Given : $v-u$ graph of a convex lens.

from

$ \begin{aligned} & \frac{1}{f}=\frac{1}{v}-\frac{1}{u} \\ & \frac{1}{f}=\frac{1}{10}-\frac{1}{-10}=\frac{1}{10}+\frac{1}{10} \end{aligned} $

$ \begin{aligned} &\begin{gathered} \frac{1}{f}=\frac{2}{10}=\frac{1}{5} \\ f=5 \mathrm{~cm} \end{gathered}\\ &\text { On differentiation of the given formula of } f \text {. }\\ &\begin{aligned} -\frac{d f}{f^2} & =\frac{-d v}{v^2}-\frac{d u}{u^2} \\ \frac{\Delta f}{f^2} & =\frac{\Delta v}{v^2}+\frac{\Delta u}{u^2} \\ \frac{\Delta f}{5^2} & =\frac{0.1}{10^2}+\frac{0.1}{10^2}=\frac{0.2}{100} \\ \Delta f & =\frac{0.2 \times 25}{100}=0.05 \mathrm{~cm} \\ f & =5 \pm 0.05 \mathrm{~cm} \end{aligned} \end{aligned} $

(A) Intensity of light received by the lens

• The amount of light collected (intensity received) depends on the aperture (radius R) of the objective lens.

• Larger radius ⇒ more light collected ⇒ brighter image.

✅ (A) → (P).

(B) Angular magnification

• For a simple telescope (in normal adjustment),

  $ M = \frac{f_0}{f_e}. $

• So, magnification depends on focal lengths $f_0, f_e$.

✅ (B) → (R).

(C) Length of telescope

• Optical length of telescope in normal adjustment is

  $ L = f_0 + f_e. $

• Again depends on focal lengths $f_0, f_e$.

✅ (C) → (R).

(D) Sharpness of image

• Sharpness (freedom from blurring) is affected by optical aberrations:

• Spherical aberration (S): due to large aperture.

• Chromatic aberration / dispersion (Q): due to different focal lengths for different colors.

• Both dispersion and spherical aberration degrade sharpness.

✅ (D) → (P, Q, S) — P indirectly since aperture radius affects spherical aberration, but specifically aberrations (Q, S).

However, let’s check options carefully:

Comparing to given options:

| A | D → (P, Q) (missing S) |

| B | D → (P, Q, S) ✅ |

| C | Wrong A and C mappings |

| D | Wrong C, D mappings |

✅ Correct option is Option B

$ \boxed{[\mathrm{A}\to(\mathrm{P}); \; \mathrm{B}\to(\mathrm{R}); \; \mathrm{C}\to(\mathrm{R}); \; \mathrm{D}\to(\mathrm{P},\mathrm{Q},\mathrm{S})]} $

If $c_{1}$ be the critical angle for the interface AB in figure.

$\begin{array}{rlr} n_{2} \sin c_{1} & =n_{1} \sin 90^{\circ} & \left(n_{1}=\sqrt{2}\right. \\ 2 \sin c_{1} & =\sqrt{2} \times 1 & \left.n_{2}=2\right) \\ \sin c_{1} & =\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}=\sin 45^{\circ} \\ c_{1} & =45^{\circ} \\ n_{1} & =\sqrt{2} \\ n_{2} & =2 \\ \hline n_{3} & =\sqrt{3} \end{array}$

Since, AB and CD are parallel, the angle of incidence for both will be equal.

If $c_{2}$ be the critical angle for the interface CD.

$\begin{aligned} n_{2} \sin c_{2} & =n_{3} \sin 90^{\circ} \quad\left(n_{3}=\sqrt{3}\right) \\ 2 \sin c_{2} & =\sqrt{3} \times 1 \\ \sin c_{2} & =\frac{\sqrt{3}}{2}=\sin 60^{\circ} \\ c_{2} & =60^{\circ} \end{aligned}$

The minimum value of the angle of incidence must be equal to the greater of the values of $c_{1}$ and $c_{2}$.

$\therefore$ Minimum angle of incidence is $60^{\circ}$.

(A) At minimum deviation $r_{1}=r_{2}=30^{\circ}$

From Snell's law $\mu=\frac{\sin i_{1}}{\sin r_{1}}$

or $\quad \sqrt{3}=\frac{\sin i_{1}}{\sin 30^{\circ}}=\frac{\sin i_{1}}{\frac{1}{2}}$

$\therefore \quad \sin i_{1}=\sqrt{3} \times \frac{1}{2}=\frac{\sqrt{3}}{2}=\sin 60^{\circ}$

$\therefore \quad i_{1}=60^{\circ}$

(B) In the position shown below net derivation suffered by the ray of light should be minimum.

Rotate DCE about C for final emergent ray to have minimum deviation.

This is achieved when the second prism is rotated anti-clockwise by $60^{\circ}$ about C.

The resolving power (RP) of an optical instrument is inversely proportional to the wavelength (λ) of light used. So, if we denote the resolving powers corresponding to λ₁ and λ₂ as RP₁ and RP₂ respectively, we can express this relationship as :

RP $ \propto $ ${1 \over \lambda }$

Therefore, the ratio of the resolving powers corresponding to λ₁ and λ₂ is given by the inverse of the ratio of the wavelengths. In LaTeX notation, this relationship can be written as :

$\frac{RP_1}{RP_2} = \frac{\lambda_2}{\lambda_1}$

Substituting the given wavelengths into this equation, we get :

$\frac{RP_1}{RP_2} = \frac{5000 \, \text{Å}}{4000 \, \text{Å}} = \frac{5}{4}$

So, the correct answer is :

Option D : 5 : 4.

Optical fibers work on the principle of total internal reflection. When light is transmitted through the fiber, it is reflected off the inner walls of the fiber in such a way that it remains within the fiber, allowing it to carry the light signal over great distances with minimal loss.

So, the correct answer is :

Option A : total internal reflection.