JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Geometrical Optics

A point source

S

is placed at the bottom of a transparent block of height

10 mm

and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54

mm

on the top of the block. The refractive index of the liquid is

(A)1.21
(B)1.30
(C)1.36✓ Correct
(D)1.42

Answer: (C)

Solution

The circular bright spot indicates total internal reflection at the critical angle. The radius of the spot is

r = 11.54/2 = 5.77

mm, and the height is

h = 10

mm.

From geometry:

tan θ = \frac{5.77}{10}

, so

sin θ = \frac{5.77}{5.7 7 ^{2} + 100} = \frac{11.54}{( 11.54 ) ^{2} + 400}

.

At the critical angle, using Snell's law:

n_{block} sin θ_{c} = n_{liquid} sin 9 0^{\circ}

.

Therefore:

2.72 \times \frac{11.54}{( 11.54 ) ^{2} + 400} = μ

.

Calculating:

μ = 2.72 \times \frac{11.54}{133.17 + 400} = 2.72 \times \frac{11.54}{23.1} \approx 1.36

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option

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JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Geometrical Optics

Solution

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