A rectangular glass slab ABCD,of refractive i | vedclass

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

Question

(A) For the ray to emerge only from the surface $CD$,it must undergo total internal reflection at the surface $AD$.Applying Snell's law at the surface

Vedclass · Accepted Answer

$\sin^{-1} \left[ \frac{n_1}{n_2} \cos \left( \sin^{-1} \frac{n_2}{n_1} \right) \right]$

Vedclass · Answer

(A) For the ray to emerge only from the surface $CD$,it must undergo total internal reflection at the surface $AD$.Applying Snell's law at the surface $AB$:$n_2 \sin \alpha_{max} = n_1 \sin r_1 \implies \alpha_{max} = \sin^{-1} \left( \frac{n_1}{n_2} \sin r_1 ight) \dots (i)$At the surface $AD$,the angle of incidence is $r_2$. For total internal reflection to occur,$r_2$ must be at least the critical angle $C$,where $\sin C = \frac{n_2}{n_1}$.Since the surface $AD$ is perpendicular to $AB$,we have $r_1 + r_2 = 90^\circ$,so $r_1 = 90^\circ - r_2$.To ensure the ray reaches $CD$,we set $r_2 = C = \sin^{-1} \left( \frac{n_2}{n_1} ight)$.Thus,$r_1 = 90^\circ - \sin^{-1} \left( \frac{n_2}{n_1} ight)$.Substituting this into equation $(i)$:$\alpha_{max} = \sin^{-1} \left[ \frac{n_1}{n_2} \sin \left( 90^\circ - \sin^{-1} \frac{n_2}{n_1} ight) ight]$Using the identity $\sin(90^\circ - 	heta) = \cos 	heta$:$\alpha_{max} = \sin^{-1} \left[ \frac{n_1}{n_2} \cos \left( \sin^{-1} \frac{n_2}{n_1} ight) ight]$.

Similar Questions

Light is incident normally on face $AB$ of a prism as shown in the figure. $A$ liquid of refractive index $\mu$ is placed on face $AC$ of the prism. The prism is made of glass of refractive index $\frac{3}{2}$. The limit of $\mu$ for which total internal reflection takes place on face $AC$ is

$A$ large glass slab $(\mu = 5/3)$ of thickness $8 \text{ cm}$ is placed over a point source of light on a plane surface. It is observed that light emerges from the top surface of the slab through a circular area of radius $R \text{ cm}$. What is the value of $R$?

Light is incident from a medium into air at two possible angles of incidence $(A) \ 20^o$ and $(B) \ 40^o$. In the medium,light travels $3.0 \ cm$ in $0.2 \ ns$. The ray will

$A$ ray of light travels in the fashion as shown in the figure. After passing through water,the ray grazes along the water-air interface. The value of $\mu_g$ in terms of $i$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module