A ray of light is incident on the first face | 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 that the emergent ray grazes the adjacent face,the angle of emergence $e =

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 that the emergent ray grazes the adjacent face,the angle of emergence $e = 90^{\circ}$.From the condition of refraction at the second surface,$r_2 = C$,where $C$ is the critical angle.Using Snell's law at the second surface: $\mu \sin r_2 = 1 \sin e \implies \sqrt{2} \sin r_2 = \sin 90^{\circ} = 1$.Thus,$\sin r_2 = 1/\sqrt{2}$,which means $r_2 = 45^{\circ}$.Since $A = r_1 + r_2$,we have $60^{\circ} = r_1 + 45^{\circ}$,so $r_1 = 15^{\circ}$.Applying Snell's law at the first surface: $1 \sin i = \mu \sin r_1$.$\sin i = \sqrt{2} \sin 15^{\circ}$.Therefore,$i = \sin^{-1}(\sqrt{2} \sin 15^{\circ})$.

Similar Questions

The refractive indices for the light of violet and red colours of any material are $1.66$ and $1.64$ respectively. If the angle of the prism made of this material is $10^o$,then the angular dispersion will be......$^o$.

$A$ triangular prism of glass is shown in the figure. $A$ ray incident normally to one face is totally reflected,if $\theta = 45^o$. The index of refraction of glass is

$A$ ray of light passes through an equilateral prism such that the angle of incidence is equal to the angle of emergence. If the angle of incidence is $45^o$,the angle of deviation will be.......$^o$.

Angle of deviation $(\delta)$ by a prism (refractive index = $\mu$ and supposing the angle of prism $A$ to be small) can be given by

If a prism having refractive index $\sqrt{3}$ has an angle of minimum deviation equal to the refracting angle of the prism,then the refracting angle of the prism is......$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module