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Get the answer to your homework problem. Try Numerade free for 7 days When light rays are incident on a prism at an angle of 45° , the minimum deviation is obtained. If refractive index of the material of the prism is 2, then the angle of prism will be
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The minimum deviation δm is related to refractive index μ and angle of prism A by the prism: \(Sin\frac{{\frac{{A + {\delta _m}}}{2}}}{{Sin\left( {\frac{A}{2}} \right)}} = \mu\) \(\frac{{A + {\delta _m}}}{2} = i = 45^\circ\) \(\frac{{\sin 45^\circ }}{{\sin \frac{A}{2}}} = \sqrt 2\) Sin A/2 = ½ India’s #1 Learning Platform Start Complete Exam Preparation
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When light rays are incident on a prism at an angle of 45 the minimum deviation is obtained?
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