For an isosceles prism of angle A and refract | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) At minimum deviation,$\delta_m = 2i - A = A$,which implies $i = A$. Since $i_1 = i_2 = i$ and $r_1 = r_2 = r$,we have $2r = A$,so $r = A/2$.Thus,$

Vedclass · Accepted Answer

$A, C, D$

Vedclass · Answer

(C) At minimum deviation,$\delta_m = 2i - A = A$,which implies $i = A$. Since $i_1 = i_2 = i$ and $r_1 = r_2 = r$,we have $2r = A$,so $r = A/2$.Thus,$r_1 = i_1/2$,making option [$A$] correct.Using Snell's law,$\sin i = \mu \sin r \implies \sin A = \mu \sin(A/2) \implies 2 \sin(A/2) \cos(A/2) = \mu \sin(A/2) \implies \mu = 2 \cos(A/2)$.Rearranging gives $\cos(A/2) = \mu/2$,so $A = 2 \cos^{-1}(\mu/2)$. Option [$B$] is incorrect.For the emergent ray to be tangential,$i_2 = 90^\circ$,so $r_2 = 	heta_c = \sin^{-1}(1/\mu)$. Then $r_1 = A - r_2 = A - \sin^{-1}(1/\mu)$.Using $\sin i_1 = \mu \sin r_1 = \mu \sin(A - \sin^{-1}(1/\mu)) = \mu [\sin A \cos(\sin^{-1}(1/\mu)) - \cos A \sin(\sin^{-1}(1/\mu))] = \mu [\sin A \sqrt{1 - 1/\mu^2} - \cos A (1/\mu)] = \sin A \sqrt{\mu^2 - 1} - \cos A$.Substituting $\mu = 2 \cos(A/2)$,$\mu^2 - 1 = 4 \cos^2(A/2) - 1$. Thus,$i_1 = \sin^{-1}[\sin A \sqrt{4 \cos^2(A/2) - 1} - \cos A]$. Option [$C$] is correct.At minimum deviation,the ray inside the prism is parallel to the base,which occurs when $i_1 = i_2 = A$. Option [$D$] is correct.

Similar Questions

$A$ ray of light suffers minimum deviation when incident on a prism having an angle of the prism equal to $60^{\circ}$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is . . . . . . .

$A$ ray is incident at an angle of $53^{\circ}$ on a prism and emerges at an angle of $37^{\circ}$ as shown. If the angle of incidence is changed to $50^{\circ}$,which of the following is a possible value of the angle of emergence (in $^{\circ}$)?

$A$ ray of light is incident at $60^{\circ}$ on one face of a prism of angle $30^{\circ}$ and the emergent ray makes $30^{\circ}$ with the incident ray. The refractive index of the prism is $(\sin 30^{\circ}=0.5, \sin 60^{\circ}=\sqrt{3}/2)$.

The variation of the refractive index of a crown glass thin prism with the wavelength of the incident light is shown. Which of the following graphs is the correct one,if $D_m$ is the angle of minimum deviation?

Vedclass Test Series

Exam Paper Generator

Online Exam Module