Ask Questions, Get Answers

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

A ray of light is incident on a prism of refractive index $\mu =\sqrt 2$ It is found that the deviation caused is $30^{\circ}$ The angle of prism is $60^{\circ}$ What will be the angle between the ray inside the prism and the base of the prism .

$(a)\;0^{\circ} \\ (b)\;45^{\circ} \\ (c)\;30^{\circ} \\ (d)\;60^{\circ} $

Can you answer this question?

1 Answer

0 votes
We know that $\mu =\large\frac{\sin \bigg( \Large\frac{A + \delta _m}{2}\bigg)}{\sin \large \frac{A}{2}}$
$\sqrt 2 =\large\frac{\sin \bigg( \Large\frac{60 + \delta _m}{2}\bigg)}{\sin 30}$
From this condition we observe that if $ \delta=30 ^{\circ} $ the expression hold c.
Also given that the deviation caused is $30^{\circ}$
$\therefore $ At minimum deviation condition the ray travels parallel to the base inside the prism.
The angle between the ray and base inside the prism is zero.
Hence a is the correct answer.


answered Jan 24, 2014 by meena.p
edited Jul 28, 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