Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

The angle of incidence for a ray of light at a refracting surface of a prism is $45^\circ$ . The angle of prism is $60^\circ$ . If the ray suffers minimum deviation through the prism, the angle of deviation and refractive index of the material of the prism respectively are:

$30^\circ, \sqrt{2}$

$45^\circ, \sqrt{2}$

$30^\circ, \frac{1}{\sqrt{2}}$

$45^\circ, \frac{1}{\sqrt{2}}$

Step-by-Step Solution

Condition for Minimum Deviation: When a prism is in the position of minimum deviation, the angle of incidence ( $i$ ) is equal to the angle of emergence ( $e$ ), i.e., $i = e$ . Also, the angle of refraction inside the prism at both surfaces is the same ( $r_1 = r_2 = r = A/2$ ).
Calculate Angle of Deviation ( $\delta_m$ ): The relationship between the prism angle ( $A$ ), deviation ( $\delta$ ), incidence ( $i$ ), and emergence ( $e$ ) is given by $A + \delta = i + e$ . At minimum deviation: $A + \delta_m = 2i$ . Given: $i = 45^\circ$ and $A = 60^\circ$ . $60^\circ + \delta_m = 2(45^\circ) = 90^\circ$ .

$\delta_m = 90^\circ - 60^\circ = 30^\circ$ .

Calculate Refractive Index ( $\mu$ ): Using the prism formula: $\mu = \frac{\sin(\frac{A + \delta_m}{2})}{\sin(A/2)}$ . Substituting values: $\mu = \frac{\sin(\frac{60^\circ + 30^\circ}{2})}{\sin(60^\circ/2)} = \frac{\sin(45^\circ)}{\sin(30^\circ)}$ .

$\mu = \frac{1/\sqrt{2}}{1/2} = \frac{2}{\sqrt{2}} = \sqrt{2}$ .

Conclusion: The angle of deviation is $30^\circ$ and the refractive index is $\sqrt{2}$ .

Practice Mode Available

Master this Topic on Sushrut

Join thousands of students and practice with AI-generated mock tests.

Get Started