NEET Class neet Answered

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Asked by jhajuhi19 | 19 Jun, 2019, 02:34: AM

Expert Answer

when light ray enters in the structure so that angle of incidence at the interface of two optical media is critical angle as shown in figure,

the ray gets total internal reflection throughout the structure and exit at otherside. This particular path for light ray takes maximum time

to travel the length L of structure.

when the angle of incidence at interface is critical angle c , if d is the width of medium that has refractive index n₁ = 1.5 ,

total optical path length l travelled by light ray is given by,

begin mathsize 14px style l space equals space fraction numerator L over denominator d space tan left parenthesis c right parenthesis end fraction cross times space d space s e c left parenthesis c right parenthesis space equals space fraction numerator L over denominator sin left parenthesis c right parenthesis end fraction end style

..........................(1)

speed of light in medium that has refractive index n₁ = 1.5 is, ( 3×10⁸)/1.5 = 2×10⁸ m/s

Critical angle c is defined by, sin(c) = n₂/n₁ = 1.44/1.5 = 0.96 ..............................(2)

Maximum time taken for the light ray to travel from entrance to exit of the given structure is obtained

by dividing the optical length given in eqn.(1) by speed of light.

By substituting L = 0.96 m and sin(c) =0.96 as in eqn.(2),

we get maximum time as ,

begin mathsize 14px style fraction numerator L over denominator sin left parenthesis c right parenthesis cross times 2 cross times 10 to the power of 8 end fraction space equals space fraction numerator 9.6 over denominator 0.96 cross times 2 cross times 10 to the power of 8 end fraction space equals space 50 cross times 10 to the power of negative 9 end exponent space s end style

Hence as per the question, t = 50

Answered by Thiyagarajan K | 19 Jun, 2019, 06:49: PM

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