Browse by Stream
QnA
Home
QnA
Go To Your Study Dashboard

Get Answers to all your Questions

header-bg qa
  • Home
  • NCERT
  • A jar of height h is filled with a transparent liquid of refractive index mu (Fig. 9.6). At the center of the jar on the bottom surface is a dot.

A jar of height h is filled with a transparent liquid of refractive index \mu (Fig. 9.6). At the center of the jar on the bottom surface is a dot. Find the minimum diameter of a disc, such that when placed on the top surface symmetrically about the center, the dot is invisible.

 

Answers (1)

Key concept :

In the above figure, ray 1 strikes the surface at an angle less than critical angle c and gets refracted in the rarer medium. Ray 2 strikes the surface, making an angle greater than the critical angle and gets internally reflected. All the rays strike at the circle of illuminance gets refracted in a rarer medium. The observer in this field will only see the light coming out of from the C.O.I.

In the above figure, O is the small dot at the bottom of the jar. The ray from the dot emerges out of the circular patch of water surface of diameter AB till the angle of incidence for the rays OA and AON exceeds the critical angle.

Rays of light incident at an angle greater than i_{C} are totally reflected with in water and consequently cannot emerge out of the water surface.

As \ sin \; i_{C}=\frac{1}{\mu }\Rightarrow \tan\; i_{C}=\frac{1}{\sqrt{\mu ^{2}-1}}

Now, \frac{d}{2h}=\tan\; i_{e}

\Rightarrow d=\frac{2h}{\sqrt{\mu ^{2}-1}}

This is the required expression of d.

 

Posted by

infoexpert23

View full answer

Similar Questions

Latest Question