SATHEE: Ray Optics and Optical Instruments

Ray Optics and Optical Instruments - Result Question 62

65. The refractive index of the material of the prism is $\sqrt{3}$ ; then the angle of minimum deviation of the prism is

(a) $30^{\circ}$

(b) $45^{\circ}$

(d) $75^{\circ}$

[1999]

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Answer:

Correct Answer: 65. (c)

Solution:

$\begin{aligned} μ = \frac{\sin (\frac{A + δ_{m}}{2})}{\sin (\frac{A}{2})} \Rightarrow \sqrt{3} = \frac{\sin (\frac{60^{\circ} + δ_{m}}{2})}{\sin (\frac{60^{\circ}}{2})} \\ \Rightarrow \sin (30^{\circ} + \frac{δ_{m}}{2}) = \frac{\sqrt{3}}{2} \Rightarrow 30^{\circ} + \frac{δ_{m}}{2} = 60^{\circ} \\ \Rightarrow δ_{m} = 60^{\circ} . \end{aligned}$

Ray Optics and Optical Instruments - Result Question 62

65. The refractive index of the material of the prism is $\sqrt{3}$ ; then the angle of minimum deviation of the prism is

Answer:

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Ray Optics and Optical Instruments - Result Question 62

65. The refractive index of the material of the prism is 3; then the angle of minimum deviation of the prism is

Answer:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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65. The refractive index of the material of the prism is $\sqrt{3}$ ; then the angle of minimum deviation of the prism is