JEE Advanced · Physics · 24. Ray Optics

Paragraph:

Light guidance in an optical fiber can be understood by considering a structure comprising of thin solid glass cylinder of refractive index $n_{1}$ surrounded by a medium of lower refractive index $n_{2}$. The light guidance in the structure takes place due to successive total internal reflections at the interface of the media $n_{1}$ and $n_{2}$ as shown in the figure. All rays with the angle of incidence $i$ less than a particular value $i_{m}$ are confined in the medium of refractive index $n_{1}$. The numerical aperture (NA) of the structure is defined as $\sin i_{m}$.

Question:

For two structures namely $S_{1}$ with $n_{1}=\sqrt{45} / 4$ and $n_{2}=3 / 2$, and $S_{2}$ with $n_{1}=8 / 5$ and $n_{2}=7 / 5$ and taking the refractive index of water to be $4 / 3$ and that of air to be $1$ , the correct option(s) is(are)

A NA of $S_{1}$ immersed in water is the same as that of $S_{2}$ immersed in a liquid of refractive index $\frac{16}{3 \sqrt{15}}$
B NA of $S_{1}$ immersed in liquid of refractive index $\frac{6}{\sqrt{15}}$ is the same as that of $S_{2}$ immersed in water
C NA of $S_{1}$ placed in air is the same as that of $S_{2}$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$
D NA of $S_{1}$ placed in air is the same as that of $S_{2}$ placed in water

Verified Solution

Answer & Solution

Correct Answer

(C) NA of $S_{1}$ placed in air is the same as that of $S_{2}$ immersed in liquid of refractive index $\frac{4}{\sqrt{15}}$

Step-by-step Solution

Detailed explanation

Let the whole structure is placed in a medium of refractive index

n^{'}

, then

n^{'} \sin i = n_{1} \cos (90 - θ)

n^{'} \sin i = n_{1} c o s θ

.....(i)
Here for

i_{m}; θ = C

and

\sin C = \frac{n_{2}}{n_{1}}

From equation (i),

n^{'} \sin i_{m} = n_{1} \sqrt{\frac{{1 - n}_{2}^{2}}{n_{1}^{2}}} = \sqrt{n_{1}^{2} - n_{2}^{2}}

\Rightarrow \sin i_{m} = \frac{\sqrt{n_{1}^{2} - n_{2}^{2}}}{n^{'}}

Now, for (i)

{(N A)}_{s}_{1} = \frac{3}{4} \sqrt{\frac{45}{16} - \frac{9}{4}} = \frac{3}{4} \times \frac{3}{4} = \frac{9}{16}

{(N A)}_{s_{2}} = \frac{3 \sqrt{15}}{16} \sqrt{\frac{64}{25} - \frac{49}{25}} = \frac{3 \sqrt{15}}{16} \frac{1}{5} \sqrt{15} = \frac{9}{16}

For (ii)

{(N A)}_{s}_{1} = \frac{\sqrt{15}}{6} \times \frac{3}{4} = \frac{\sqrt{15}}{8}

{(N A)}_{s}_{2} = \frac{3}{4} = \frac{\sqrt{15}}{5}

Not equal
For (iii)

{(N A)}_{s}_{1} = 1 \times \frac{3}{4} = \frac{3}{4}

{(N A)}_{s}_{2} = \frac{\sqrt{15}}{4} \times \frac{\sqrt{15}}{5} = \frac{15}{4 \times 5} = \frac{3}{4}

For (iv)

{(N A)}_{s_{1}} = \frac{3}{4}

{(N A)}_{s}_{2} = \frac{3}{4} \frac{\sqrt{15}}{5}

Not equal

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

The image of an object, formed by a plano - convex lens at a distance of 8 m behind the lens, is real and is one-third the size of the object. The wavelength of light inside the lens is $\frac{2}{3}$ times the wavelength in free space. The radius of the curved surface of the lens is

JEE Advanced 2013 Medium

Answer the following by appropriately matching the lists based on the information given in the paragraph.
In a thermodynamics process on an ideal monatomic gas, the infinitesimal heat absorbed by the gas is given by

T Δ X,

where

T

is temperature of the system and

Δ X

is the infinitesimal change in a thermodynamic quantity

X

of the system. For a mole of monatomic ideal gas

X = \frac{3}{2} R l n (\frac{T}{T_{A}}) + R l n (\frac{V}{V_{A}}) .

Here,

R

is gas constant,

V

is volume of gas,

T_{A}

and

V_{A}

are constants.
The List-I below gives some quantities involved in a process and List-II gives some possible values of these quantities.

	Column - I		Column - II
(a)	Work done by the system in process $1 \to 2 \to 3$	(p)	$\frac{1}{3} R T_{0} l n 2$
(b)	Change in internal energy in process $1 \to 2 \to 3$	(q)	$\frac{1}{3} R T_{0}$
(c)	Heat absorbed by the system in process $1 \to 2 \to 3$	(r)	$R T_{0}$
(d)	Heat absorbed by the system in process $1 \to 2$	(s)	$\frac{4}{3} R T_{0}$
		(t)	$\frac{1}{3} R T_{0} (3 + l n 2)$
		(u)	$\frac{5}{6} R T_{0}$

If the process on one mole of monatomic ideal gas is as shown in the

P V

-diagram with

P_{0} V_{0} = \frac{1}{3} R T_{0},

the correct match is

JEE Advanced 2019 Medium

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