Question

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

Answer 1

relation between refractive index of a substance and it's critical angle is ,,,, refractive index of water = 1÷ sin C (here C is the critical angle ) hence ,,,refractive index of water= 1÷sin(48.5)=1÷0.981= 1.01=1 (approx) therefore the absolute refractive index of water is 1

Answer 2

The absolute refractive index of water is 1.33.

Explanation:

The critical angle is 48.5 degrees. The relation between the refractive index and the critical angle is given by :

$sin\theta_c=\dfrac{1}{n}$

$n=\dfrac{1}{sin\theta_c}$

$n=\dfrac{1}{sin(48.5)}$

n = 1.33

So, the absolute refractive index of water is 1.33. Hence, this is the required solution.

Learn more,

Critical angle

https://brainly.in/question/8835212