A ray of light travels from a denser medium t | vedclass

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

Question

(B) According to the law of reflection,the angle of incidence '$i$' is equal to the angle of reflection '$r$'. So,$i = r$.Given that the reflected and

Vedclass · Accepted Answer

$	an ^{-1}(\sin C)$

Vedclass · Answer

(B) According to the law of reflection,the angle of incidence '$i$' is equal to the angle of reflection '$r$'. So,$i = r$.Given that the reflected and refracted rays are perpendicular to each other,the sum of the angle of reflection and the angle of refraction is $90^{\circ}$.$i + r_1 = 90^{\circ} \implies r_1 = 90^{\circ} - i$.Using Snell's Law,the refractive index '$\mu$' of the denser medium with respect to the rarer medium is given by $\mu = \frac{\sin i}{\sin r_1}$.Substituting $r_1 = 90^{\circ} - i$,we get $\mu = \frac{\sin i}{\sin(90^{\circ} - i)} = \frac{\sin i}{\cos i} = 	an i$.We also know that the refractive index is related to the critical angle '$C$' by $\mu = \frac{1}{\sin C}$.Equating the two expressions for '$\mu$': $	an i = \frac{1}{\sin C}$.This implies $	an i = \csc C$,which is not among the options. Let us re-evaluate: $\mu = \frac{\sin r_1}{\sin i}$ is for light going from rarer to denser. Since light goes from denser to rarer,$\mu = \frac{\sin r_1}{\sin i} = \frac{\sin(90^{\circ}-i)}{\sin i} = \cot i$.Since $\mu = \frac{1}{\sin C}$,we have $\cot i = \frac{1}{\sin C}$,so $	an i = \sin C$.Therefore,$i = 	an^{-1}(\sin C)$.

$A$ ray of light travels from a denser medium to a rarer medium. The reflected and the refracted rays are perpendicular to each other. If '$i$' and '$r_1$' are the angle of incidence and angle of refraction respectively and '$C$' is the critical angle,then the angle of incidence is

Similar Questions

$A$ ray of light incident normally on an isosceles right-angled prism travels as shown in the figure. The least value of the refractive index of the prism must be

$PQR$ is a right-angled prism with other angles as $60^o$ and $30^o$. The refractive index of the prism is $1.5$. $PQ$ has a thin layer of liquid. Light falls normally on the face $PR$. For total internal reflection at the face $PQ$,the maximum refractive index of the liquid is

The critical angle for a medium is $60^o$. The refractive index of the medium is:

$A$ point source of light is placed $4 \; m$ below the surface of water of refractive index $5/3$. The minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is ... $m$.

$A$ ray of light travelling in a transparent medium of refractive index $n$ falls on a surface separating the medium from air at an angle of incidence of $45^{\circ}$. For which of the following values of $n$ can the ray undergo total internal reflection?

Vedclass Test Series

Exam Paper Generator

Online Exam Module