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Question: A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel t...

A ray of light is incident at the glass–water interface at an angle i, it emerges finally parallel to the surface of water, then the value of μg\mu _ { g } would be

A

(4/3) sin I

B

1/ sin I

C

4/ 3

D

1

Answer

1/ sin I

Explanation

Solution

For glass water interface gμω=sin⁡isin⁡r{ } _ { g } \mu _ { \omega } = \frac { \sin i } { \sin r } ......(i)

and For water-air interface ωμa=sin⁡rsin⁡90{ } _ { \omega } \mu _ { a } = \frac { \sin r } { \sin 90 } .....(ii)

∴\therefore gμω×ωμa=sin⁡i{ } _ { g } \mu _ { \omega } \times _ { \omega } \mu _ { a } = \sin i ⇒\Rightarrow μg=1sin⁡i\mu _ { g } = \frac { 1 } { \sin i }