Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
0 votes

A point source of light is placed at the bottom of a tank containing a liquid $(R.I= \mu)$ upto a depth $h$. A bright circular spot is seen on the surface of the liquid. Then the radius of this bright spot.:

$(a)\;-h \\ (b)\;\frac{h}{\mu} \\ (c)\;\frac{h}{\sqrt {\mu-1}} \\ (d)\;\frac{h}{\sqrt{\mu^2-1}} $

Can you answer this question?

1 Answer

0 votes
Rays coming out of source and incident at an angle greater than $\theta_c$ will be reflected back into liquid therefore, the corresponding region on surface will appear dark.
As is obvious from the figure the radius of bright spot is
$R= h \tan \theta_c=\large\frac{h \sin \theta_c}{\cos \theta_c}$
or $R= \large\frac{h \sin \theta}{\sqrt {1- \sin ^2 \theta _c}}$
Since $\sin \theta _c =\large\frac{1}{\mu}$
$\therefore R= \large=\frac{h}{\sqrt {\mu^2-1}}$
Hence d is the correct answer.


answered Feb 14, 2014 by meena.p
edited Jul 22, 2014 by thagee.vedartham

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App