Ray Optics and Optical Instruments - Result Question 59

62. For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive index :

(a) lies between $\sqrt{2}$ and 1

[2012M]

(b) lies between 2 and $\sqrt{2}$

(c) is less than 1

(d) is greater than 2

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Answer:

Correct Answer: 62. (b)

Solution:

  1. (b)

(b)

The angle of minimum deviation is given as $\delta _{\text{min }}=i+e-A$

for minimum deviation

$\delta _{\text{min }}=A$ then

$2 A=i+e$

in case of $\delta _{\min }, i=e$

$ \begin{aligned} 2 A & =2 i \quad r_1=r_2=\frac{A}{2} \\ i & =A=90^{\circ} \end{aligned} $

from Snell’s law

$1 \sin i=n \sin r_1$

$\sin A=n \sin \frac{A}{2}$

$2 \sin \frac{A}{2} \cos \frac{A}{2}=n \sin \frac{A}{2}$

$2 \cos \frac{A}{2}=n$

when $A=90^{\circ}=i _{\text{min }}$

then $\quad n _{\text{min }}=\sqrt{2}$

$i=A=0 \quad n _{\text{max }}=2$