Geometrical Optics

In given system of lens and mirror, position of object O in front of lens is at a distance 2f. i.e. u = 2f = 40 cm

So, image (I₁) formed is real, inverted and at a distance, v = 2f = 2 $\times$ 20 = 40 cm, (behind lens) magnification, ${m_1} = {v \over u} = {{40} \over {40}} = 1$

Thus, size of image is same as that of that of object. This image (I₁) acts like a real object for mirror.

As object distance for mirror is

u = C = 2f = $-$ 20 cm

where, C = centre of curvature.

So, image (I₂) formed by mirror is at 2f.

$\therefore$ For mirror v = 2f = 2($-$ 10) = $-$ 20 cm

Magnification, ${m_2} = - {v \over u} = - {{( - 20)} \over {( - 20)}} = - 1$

Thus, image size is same as that of object.

The image I₂ formed by the mirror will act like an object for lens.

As the object is at 2f distance from lens, so image (I₃) will be formed at a distance 2f or 40 cm. Thus, magnification,

${m_3} = {v \over u} = {{40} \over {40}} = 1$

So, final magnification, $m = {m_1} \times {m_2} \times {m_3} = - 1$

Hence, final image (I₃) is real, inverted of same size as that of object and coinciding with object.

Given : ${i_1} = 60^\circ $; $A = 30^\circ $

Then angle of deviation is given by

$\delta = 1 - A = 60^\circ - 30^\circ = 30^\circ $

We know that,

${i_1} + {i_2} = A + \delta \Rightarrow 60^\circ + {i_2} = 30^\circ + 30^\circ \Rightarrow {i_2} = 0$

It means emergent ray leaves face AC normally.

For a convergent doublet of separated lens, we have

${1 \over f} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {d \over {{f_1}{f_2}}}$ ...... (1)

where d is separation between two lens, f₁ and f₂ are focal lengths of component lenses, f is resultant focal length. Therefore, Eq. (1) becomes

${1 \over {10}} = {1 \over {{f_1}}} + {1 \over {{f_2}}} - {2 \over {{f_1}{f_2}}} \Rightarrow {1 \over {10}} = \left( {{{{f_2} + {f_1} - 2} \over {{f_1}{f_2}}}} \right)$

$ \Rightarrow {f_1}{f_2} = 10{f_2} + 10{f_1} - 20$

$ \Rightarrow 10{f_1} + 10{f_2} - {f_1}{f_2} = + 20$

For f₁ = 18 cm and f₂ = 20 cm, the above equation satisfies.

The given optical situation is depicted in the following ray diagram:

In this case, the image forms 55 cm behind the convex mirror and then the reflection takes place due to the mirror image that forms at a distance (v) behind the mirror. Now, this image acts as an object for lens and the final image forms at a distance of 20 m from the lens if the focal length of convex mirror is as follows:

${{55} \over 2} = 27.5$ cm

The refractive index is

$\mu = {{{\mathop{\rm Real}\nolimits} \,depth} \over {Apparent\,depth}} = {{{\mathop{\rm Reading}\nolimits} \,3 - Reading\,1} \over {{\mathop{\rm Reading}\nolimits} \,3 - Reading\,2}}$

Therefore, the minimum of three readings are required.

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