JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Geometrical Optics

A planar structure of length L and width W is made of two different optical media of refractive indices

n_{1} = 1.5

and

n_{2} = 1.44

as shown in figure. If L >> W, a ray entering from end AB will emerge from end CD only if the total internal reflection condition is met inside the structure. For

L = 9.6 m

, if the incident angle

θ

is varied, the maximum time taken by a ray to exit the plane CD is

t \times 1 0^{- 9} s

, where t is ____.
[Speed of light

c = 3 \times 1 0^{8} m / s

]

Answer: 50

Solution

For maximum time, the ray must undergo TIR at all surfaces at the critical angle.

For TIR at the interface:

n_{1} sin θ_{C} = n_{2}

sin θ_{C} = \frac{1.44}{1.5} = 0.96

The path length in the medium:

D = \frac{L}{sin θ _{C}} = \frac{9.6}{0.96} = 10 m

Speed of light in medium with refractive index

n_{1}

v = \frac{c}{n _{1}} = \frac{3 \times 1 0 ^{8}}{1.5} = 2 \times 1 0^{8} m/s

Time taken:

t = \frac{D}{v} = \frac{10}{2 \times 1 0 ^{8}} = 50 \times 1 0^{- 9} s

Therefore,

t = 50

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Numerical answer

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JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Geometrical Optics

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