A ray of light refracts from medium 1 into a | vedclass

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

Question

(C) Given: Refractive indices $\mu_1 = 1.6$, $\mu_2 = 1.8$, $\mu_3 = 1.3$.At the interface between medium $2$ and $3$, the ray is incident at the crit

Vedclass · Accepted Answer

$\sin ^{-1}\left(\frac{13}{16}\right)$

Vedclass · Answer

(C) Given: Refractive indices $\mu_1 = 1.6$, $\mu_2 = 1.8$, $\mu_3 = 1.3$.At the interface between medium $2$ and $3$, the ray is incident at the critical angle $C$. Therefore, $\sin C = \frac{\mu_3}{\mu_2} = \frac{1.3}{1.8}$.Let $r$ be the angle of refraction in medium $2$. Since the ray is incident at the critical angle at the interface of $2$ and $3$, the angle of refraction $r$ at the first interface is equal to the critical angle $C$ (i.e.,$r = C$).Applying Snell's Law at the interface between medium $1$ and $2$:$\mu_1 \sin 	heta = \mu_2 \sin r$Since $r = C$, we have $\sin r = \sin C = \frac{1.3}{1.8}$.Substituting the values:$1.6 	imes \sin 	heta = 1.8 	imes \left(\frac{1.3}{1.8}ight)$$1.6 	imes \sin 	heta = 1.3$$\sin 	heta = \frac{1.3}{1.6} = \frac{13}{16}$$	heta = \sin ^{-1}\left(\frac{13}{16}ight)$

$A$ ray of light refracts from medium $1$ into a thin layer of medium $2$, crosses the layer, and is incident at the critical angle on the interface between medium $2$ and $3$ as shown in the figure. If the angle of incidence of the ray is $\theta$, the value of $\theta$ is

Similar Questions

$A$ light ray falls on a glass surface of refractive index $\sqrt{3}$,at an angle of $60^{\circ}$. The angle between the refracted and reflected rays would be ....... $^o$.

$A$ fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is $\frac{4}{3}$ and the fish is $12 \ cm$ below the surface,the radius of this circle in $cm$ is

In an optical fibre, is the refractive index of the core greater than that of the cladding, or smaller?

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

The figure shows a ray incident at an angle $i = \pi /3.$ If the plot drawn shows the variation of $|r - i|$ versus $\frac{\mu_1}{\mu_2} = k,$ ($r =$ angle of refraction) choose the wrong alternative.

Vedclass Test Series

Exam Paper Generator

Online Exam Module