Question

A flat piece of glass covers the top of a vertical cylinder that is completely filled with water. If a ray of light traveling in the glass is incident on the interface with the water at an angle of ?a = 36.0 ? , the ray refracted into the water makes an angle of 49.8 ? with the normal to the interface.

a)What is the smallest value of the incident angle ?a for which none of the ray refracts into the water? Express your answer with the appropriate units.

Answer 1

All you have to find is the critical angle for the glass-water pair. At that angle of incidence, the angle of refraction is 90º. Means, no light will enter the water. It will travel along the edge of the Glass plate just above the water.

Find refractive index of water with respect to glass.

N = sin i / sin r = [sin 36 / sin 49.8] (Snell's Law)

n = 0.769558

now, since for 90º refraction, the ray should pass from denser medium to rarer medium,

i.e. you need Refractive index of glass with respect to water = n = 1/N = 1.299

Now, n=1/sin c, where 'c' is the critical angle.

So, 1.299 = 1/sin c;

c = arcsin (1/1.299) = 50.31º (ans)

So, if the angle of incidence is 50.31º, no ray will refract into the water.

Hope this helps :)