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

The phenomenon utilised in an optical fibre is

$A$ ray of light enters from a rarer to a denser medium. The angle of incidence is $i$. Then the reflected and refracted rays are mutually perpendicular to each other. The critical angle for the pair of media is

$A$ parallel beam of light is incident on a glass prism in the shape of a quarter cylinder of radius $R=0.05 \ m$ and refractive index $n=1.5$,placed on a horizontal table as shown in the figure. Beyond the cylinder,a patch of light is found whose nearest distance $x$ from the cylinder is

The reason for the shining of an air bubble in water is

$A$ beam of light consisting of red,green and blue colours is incident on a right-angled prism on face $AB$. The refractive indices of the material for the above red,green and blue colours are $1.39, 1.44$ and $1.47$ respectively. $A$ person looking on surface $AC$ of the prism will see

Vedclass Test Series

Exam Paper Generator

Online Exam Module