Question

Find the absolute refractive index of the water if its critical angle is 48.5 degrees?

Answer 1

$\huge{Answer}$

Given,

Critical angle C = 48.5°

------>Sin C = 0.75 = 3/4

Absolute refractive index of water = n=?

Formula

$n = \frac{1}{sin \: c}$

$n = \frac{ \frac{1}{4} }{3}$

$= n = \frac{4}{3} = 1.33$

Answer 2

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plzzz refer to the attached file..

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