Q47P - University Physics with Modern Physics

Question

A thin layer of ice   floats on the surface of water  in a bucket. A ray of light from the bottom of the bucket travels upward through the water. (a) What is the largest angle with respect to the normal that the ray can make at the ice–water interface and still pass out into the air above the ice? (b) What is this angle after the ice melts?

Step-by-Step Solution

Answer

 the largest incident angle is 48.60 deg.

(b) The incident angle will be the same 48.6 Degree  

Step 1: About refractive index

It is one of the principle defining characteristics of an optical material. In simplified terms, the refractive index defines to what degree light is bent or “refracted” when crossing the boundary between two mediums. 

Step 2 : Determine the largest angle with respect to the normal that the ray can make at the ice–water interface

a) The refractive index of an optical material is expressed by n and represents the speed of light in the vacuum divided by

the speed of light in the materiaL Snell's law is given by equation (33-4) in the form

 

We Want to calculate 0,, but we missed 01,. For the refraction betWeen the ice and the air We can get the critical angle 0cm}

by

”

 

In the ice, the refracted beam gives us 01, = 06”}. So, we can use this value into equation (1) to get an as next

 

Therefore  the largest incident angle is 48.60 deg.   

This is the largest incident angle.

Step 3 : Determine the angle after the ice melts

(b) After the ice is melted, the refractive index of the air and the water will give us a critical angle equals the critical angle in

part (a). Hence, the incident angle will be the same

 

Therefore 

(b) The incident angle will be the same 48.6 Degree