A ray of light incident parallel to the base | vedclass

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

Question

(A) For total internal reflection $(TIR)$ to occur at the faces,the angle of incidence at each face must be greater than the critical angle $	heta_c$

Vedclass · Accepted Answer

$\mu > \sqrt{5}$

Vedclass · Answer

(A) For total internal reflection $(TIR)$ to occur at the faces,the angle of incidence at each face must be greater than the critical angle $	heta_c$.From the geometry of the prism,the angles of incidence at the two faces are $(45^{\circ} + r)$ and $(45^{\circ} - r)$,where $r$ is the angle of refraction at the first surface.For $TIR$ to occur at both faces:$45^{\circ} + r > 	heta_c$ and $45^{\circ} - r > 	heta_c$.The more restrictive condition is $45^{\circ} - r > 	heta_c$.Since $\sin r = \frac{\sin 45^{\circ}}{\mu} = \frac{1}{\mu\sqrt{2}}$,we have $r = \arcsin\left(\frac{1}{\mu\sqrt{2}}ight)$.Substituting this into $45^{\circ} - r > 	heta_c$,we get $45^{\circ} - 	heta_c > r$.Taking the sine of both sides: $\sin(45^{\circ} - 	heta_c) > \sin r$.Using the identity $\sin(A-B) = \sin A \cos B - \cos A \sin B$ and knowing $\sin 	heta_c = \frac{1}{\mu}$ and $\cos 	heta_c = \frac{\sqrt{\mu^2-1}}{\mu}$:$\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{\mu^2-1}}{\mu} - \frac{1}{\sqrt{2}} \cdot \frac{1}{\mu} > \frac{1}{\mu\sqrt{2}}$.Multiplying by $\mu\sqrt{2}$,we get $\sqrt{\mu^2-1} - 1 > 1$,which simplifies to $\sqrt{\mu^2-1} > 2$.Squaring both sides: $\mu^2 - 1 > 4$,so $\mu^2 > 5$,or $\mu > \sqrt{5}$.

Similar Questions

$A$ well-cut diamond appears bright because

Light travels from an optically denser medium $A$ into the optically rarer medium $B$ with speeds $1.8 \times 10^{8} \ m/s$ and $2.7 \times 10^{8} \ m/s$ respectively. The critical angle between them is ($\mu_{A}$ and $\mu_{B}$ are the refractive indices of media $A$ and $B$ respectively).

The velocity of light in a medium is half its velocity in air. If a ray of light emerges from such a medium into air,the angle of incidence,at which it will be totally internally reflected,is ......... $^o$

The minimum area of the disc needed to cut off all the light coming out of water $\left(\mu_{w} = \frac{4}{3}\right)$ from a point source of light placed $7 \ m$ below the water surface is: (in $\pi \ m^2$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module