Question

A liquid is placed in a hollow prism of angle $60^\circ $. If the angle of the minimum deviation is $30^\circ $, what is the refractive index of the liquid?
A) 1.41
B) 1.50
C) 1.65
D) 1.95

Answer 1

Hint: There two ways through which maximum deviation can be achieved one is when the angle is at 90 degrees. This is also known as the grazing incidence and another way is when the emergent light ray after leaving the prism grazes along the surface of the prism.

Complete step by step solution:
Step 1:
Find the refractive index of the liquid:
The condition of maximum deviation is given as:
$i = e;$ …(i = incident ray; e = emergent ray)
${\delta _{\min }} = 2i - A$;
Here:
${\delta _{\min }}$= Minimum Deviation;
i = Incident Angle;
A = Angle of hollow prism;
The formula for refractive index is given as:
$\mu = \dfrac{{\sin \left( {\dfrac{{{\delta _{\min }} + A}}{2}} \right)}}{{\sin \left( {\dfrac{A}{2}} \right)}}$;
Here:
$\mu $= Refractive Index.
${\delta _m}$= Minimum deviation.
$A$= Angle of hollow prism.
Here we have been given ${\delta _{\min }} = 30^\circ $; $A = 60^\circ $.

Step 2:
Put in the value of minimum deviation and angle of prism in the above equation and solve:
 $ \Rightarrow \mu = \dfrac{{\sin \left( {\dfrac{{30 + 60}}{2}} \right)}}{{\sin \left( {\dfrac{{60}}{2}} \right)}};$
$ \Rightarrow \mu = \dfrac{{\sin {{45}^\circ }}}{{\sin {{30}^\circ }}};$
Put in the value and solve:
$ \Rightarrow \mu = \dfrac{{\dfrac{1}{{\sqrt 2 }}}}{{\dfrac{1}{2}}};$
Do the needed calculation:
$ \Rightarrow \mu = \sqrt 2 ;$
The refractive index of the liquid is:
$ \Rightarrow \mu = 1.41;$

Final Answer: Option “A” is correct. The refractive index of the liquid is 1.41.

Note: We have been given the minimum deviation and the angle of the prism. Write the formula for the refractive index and put in the given value in the formula. The deviation reduces as the incidence angle increases until the deviation gets minimum. There is an increase in the angle of deviation as the angle of incidence increases.