A rectangular glass block is placed on a prin | vedclass

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

Question

(A) The situation is shown in the figure. Light will not emerge from the vertical side $BC$ if the angle of incidence $i$ at the side $BC$ is greater

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(A) The situation is shown in the figure. Light will not emerge from the vertical side $BC$ if the angle of incidence $i$ at the side $BC$ is greater than the critical angle $	heta_c$.Condition for total internal reflection: $i > 	heta_c$ or $\sin i > \sin 	heta_c$.Since $\sin 	heta_c = \frac{1}{\mu}$,we have $\sin i > \frac{1}{\mu} \quad ......(i)$Applying Snell's law at point $O$ on the bottom surface: $1 	imes \sin 	heta = \mu \sin r$.In $	riangle OPR$,$r + 90^\circ + i = 180^\circ$,so $r + i = 90^\circ$,which means $r = 90^\circ - i$.Therefore,$\sin 	heta = \mu \sin(90^\circ - i) = \mu \cos i$,which gives $\cos i = \frac{\sin 	heta}{\mu}$.Using $\sin i = \sqrt{1 - \cos^2 i}$,we get $\sin i = \sqrt{1 - \left(\frac{\sin 	heta}{\mu}ight)^2} \quad ......(ii)$.Substituting $(ii)$ into $(i)$:$\sqrt{1 - \frac{\sin^2 	heta}{\mu^2}} > \frac{1}{\mu} \implies 1 - \frac{\sin^2 	heta}{\mu^2} > \frac{1}{\mu^2} \implies 1 > \frac{1 + \sin^2 	heta}{\mu^2}$.$\mu^2 > 1 + \sin^2 	heta$.For the letters to be invisible for all angles $	heta$,we consider the maximum value of $\sin^2 	heta$,which is $1$.$\mu^2 > 1 + 1 = 2 \implies \mu > \sqrt{2}$.Thus,the minimum refractive index is $\mu_{\min} = \sqrt{2}$.

$A$ rectangular glass block is placed on a printed paper lying on a horizontal surface. Find the minimum refractive index of the glass for which the letters on the paper are not visible from any of the vertical sides of the block.

Similar Questions

Calculate the critical angle for a glass-air surface,if a ray of light incident in air on the glass surface is deviated by $15^{\circ}$ when the angle of incidence is $45^{\circ}$. (in $^{\circ}$)

The phenomenon by which light travels in an optical fibre is

The critical angle of light passing from glass to air is minimum for which wavelength?

$A$ small bulb is placed at the bottom of a tank containing water to a depth of $80 \; cm$. What is the area (in $m^2$) of the surface of water through which light from the bulb can emerge out? Refractive index of water is $1.33$. (Consider the bulb to be a point source.)

Vedclass Test Series

Exam Paper Generator

Online Exam Module