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Question: The x-z plane separates two media A and B of refractive indices m<sub>1</sub> = 1.5 and m<sub>2</sub...

The x-z plane separates two media A and B of refractive indices m1 = 1.5 and m2 = 2. A ray of light travels from A to B. Its directions in the two media are given by unit vectors μ1\vec { \mu } _ { 1 } = ai^+bb^a \hat { i } + b \hat { b } , μ2\vec { \mu } _ { 2 }= ci^+dd^c \hat { i } + d \hat { d } , then –

A

ac=43\frac { \mathrm { a } } { \mathrm { c } } = \frac { 4 } { 3 }

B

C

D

Answer

ac=43\frac { \mathrm { a } } { \mathrm { c } } = \frac { 4 } { 3 }

Explanation

Solution

n1 sin i = n2 sin r

1.5 = 2 (μ2×j^)\left( \vec { \mu } _ { 2 } \times \hat { \mathrm { j } } \right)

1.5 [(×] = 2 [ () × j]

1.5 ak^\mathrm { a } \hat { \mathrm { k } } = 2c

= 21.5\frac { 2 } { 1.5 } = 43\frac { 4 } { 3 }