Question

Chapter 26, Problem 027 A glass is half-full of water, with a layer of oil (n = 1.46) floating on top. A ray of light traveling downward through the oil is incident on the water at an angle of 73.7°. Take the refractive index of water to be 1.333. (a) Determine the critical angle for the oil-water interface. (b) Will the ray penetrate into the water

Answer 1

Answer:

a) angle is  65.63 degree

b) Ray cannot be penetrate. it will reflect form oi and water surface

Explanation:

Refractive index for oil , $n_1 = 1.46$

and Refractive index for water, $n_2 = 1.333$

when light passes from oil to water then we have following relation

        $n_1\times  sin i = n_2 \times sin r$

For i = critical angle, r is  90 degree

therefore

         $n_1. sin c = n_2$

$sin c = \frac{n_2}{n_1}$

$c =sin^{-1} \frac{1.33}{1.46} = 65.63 degree$

the angle of incidence is i 73.7 which is greater than 65.63 therefore , ray cannot be penetrate. it will reflect form oi and water surface