Single Choice

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

A$${ 45 }^{ o };\dfrac { 1 }{ \sqrt { 2 } } $$
B$${ 45 }^{ o };\dfrac { 1 }{ \sqrt { 2 } } $$
Correct Answer
C$${ 45 }^{ o };\sqrt { 2 } $$
D$${ 30 }^{ o };\dfrac { 1 }{ \sqrt { 2 } } $$

Solution

Angle of incidence, $$i=45^0$$

Angle of prism, $$A=60^0$$

Angle of minimum deviation, $$\delta_m =2i-A=30^0$$

$$\Rightarrow \mu=\dfrac{sin \left ( \dfrac{A+\delta_m}{2}\right )}{sin \dfrac{A}{2}}=\dfrac{sin\ 45^0}{sin\ 30^0}=\dfrac{2}{\sqrt{2}}=\sqrt{2}$$


SIMILAR QUESTIONS

Optics

In an experiment for determination of refractive index of glass of a prism by $$i-\delta\ plot$$ it was found that a ray incident at angle $${ 35 }^{ \circ }$$ suffers a deviation of $${ 40 }^{ \circ }$$ and that it emerges at angle $${ 79 }^{ \circ }.$$ In that case which of the following is closest to the maximum possible value of the refractive index?

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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:

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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 :

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The maximum refractive index of a prism which permits the passage of light through it, when the refractive angle of the prism is $${90^0}$$, is

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Calculate the refractive index of glass with respect to water. It is given that refractive indices of glass and water with respect to air are $$\dfrac {3}{2}$$ and $$\dfrac {4}{3}$$ respectively.

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The refractive index of the material of prism, if a thin prism of angle $$A=6^o$$, produces a deviation $$\delta =3^o$$, is :

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