Question

in the given figure ray incident on an interface would converge 2 cm below the interface if they continued to move in straight lines without bending. But due to refraction, the rays will bend and meet somewhere else. Find the distance of meeting point of refracted rays below the interface (in cm).(Assuming the rays to be making small angles with the normal to the interface)​

Answer 1

Answer:

The answer will be 5 cm.

Explanation:

sin i = x/ √2^2 + x^2

sin r = x/ √y^2 + x^2

sin i / sin r = 5/ 2 =  √y^2 + x^2/ √2^2 + x^2

In case of small i and r,

x <<< 2 and x <<< y

So,

5 / 2 = y / 2

10 = 2y

y = 5 cm