Question

A horizontal, parallel-sided plate of glass having a refractive index of 1.56 is in contact with the surface of water in a tank. A ray coming from above in air makes an angle of incidence of 33.0 ∘ with the normal to the top surface of the glass. (A) What angle does the ray refracted into the water make with the normal to the surface?Use 1.33 for the index of refraction of water.(B) What is the dependence of this angle on the refractive index of the glass?

Answer 1

Answer:

(a) 36 degree

(b) It is directly proportional to the refractive index of glass

Explanation:

refractive index, ng = 1.56

refractive index of water, nw = 1.33

Angle of incidence in glass, i = 30 degree

Let the angle of refraction in glass is r.

(a) Use Snell's law on the glass water boundary

refractive index of water with respect to glass

$n_{g}^{w}=\frac{n_{w}}{n_{g}}= \frac{Sini}{Sinr}$

$\frac{1.33}{1.56} = \frac{Sin30}{Sinr}$

Sin r = 0.5865

r = 36 degree

(b)

According to Snell's law

Refractive index of water / refractive index of glass = Sin i / Sin r

Sin r  is directly proportional to the refractive index of glass.