A ray of light incident on one face of an equ | vedclass

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

Question

(A) For an equilateral prism, the angle of the prism $A = 60^{\circ}$.Given refractive index $\mu = \sqrt{2}$.The emergent ray grazes the adjacent fac

Vedclass · Accepted Answer

$\sin^{-1} (\sqrt{2} \sin 15^{\circ})$

Vedclass · Answer

(A) For an equilateral prism, the angle of the prism $A = 60^{\circ}$.Given refractive index $\mu = \sqrt{2}$.The emergent ray grazes the adjacent face, which means the angle of emergence $e = 90^{\circ}$.At the second face, by Snell's law: $\mu \sin r_2 = 1 \cdot \sin e$.$\sqrt{2} \sin r_2 = \sin 90^{\circ} = 1 \implies \sin r_2 = 1/\sqrt{2} \implies r_2 = 45^{\circ}$.Since $r_1 + r_2 = A$, we have $r_1 + 45^{\circ} = 60^{\circ} \implies r_1 = 15^{\circ}$.Applying Snell's law at the first face: $1 \cdot \sin i = \mu \sin r_1$.$\sin i = \sqrt{2} \sin 15^{\circ} \implies i = \sin^{-1} (\sqrt{2} \sin 15^{\circ})$.

$A$ ray of light incident on one face of an equilateral glass prism having refractive index $\sqrt{2}$, produces the emergent ray which just grazes along the adjacent face. The value of angle of incidence is $(\sin 90^{\circ} = 1, \sin 30^{\circ} = 0.5, \sin 45^{\circ} = 1/\sqrt{2})$.

Similar Questions

The angle of a prism is $30^o$ and its refractive index is $\sqrt{2}$. One of its surfaces is silvered. At what angle of incidence should a ray be incident on the other surface so that after reflection from the silvered surface,it retraces its path?

Write the equation for the refractive index of a prism in terms of the angle of minimum deviation $(\delta_m)$ and the angle of the prism $(A)$.

When a glass prism of refracting angle $60^{\circ}$ is immersed in a liquid, its angle of minimum deviation is $30^{\circ}$. The critical angle of glass with respect to the liquid medium is (in $^{\circ}$)

Light rays from a source are incident on a glass prism of refractive index $\mu$ and prism angle $\alpha$. At near-normal incidence,the angle of deviation of the emerging rays is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module