Three transparent media of refractive indices | vedclass

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(D) The correct option is $(D)$.$1$. At the first interface (between medium $1$ and medium $2$),the light ray bends towards the normal. According to S

Vedclass · Accepted Answer

$\mu_3 < \mu_1 < \mu_2$

Vedclass · Answer

(D) The correct option is $(D)$.$1$. At the first interface (between medium $1$ and medium $2$),the light ray bends towards the normal. According to Snell's law,when a ray bends towards the normal,it travels from a rarer medium to a denser medium. Therefore,$\mu_2 > \mu_1$.$2$. At the second interface (between medium $2$ and medium $3$),the light ray undergoes total internal reflection,as no light enters the third medium. Total internal reflection occurs only when light travels from a denser medium to a rarer medium. Therefore,$\mu_2 > \mu_3$.$3$. Since the light is totally internally reflected at the second interface,the angle of incidence must be greater than the critical angle for the pair of media $(2, 3)$. This implies that medium $2$ is optically denser than medium $3$. Combining these observations,we have $\mu_2 > \mu_1$ and $\mu_2 > \mu_3$. Furthermore,since the ray is bent towards the normal at the first interface and undergoes total internal reflection at the second,the refractive index $\mu_1$ must be greater than $\mu_3$ to satisfy the geometry of the path. Thus,the correct order is $\mu_3 < \mu_1 < \mu_2$.

Three transparent media of refractive indices $\mu_1, \mu_2, \mu_3$,respectively,are stacked as shown below. $A$ ray of light follows the path shown. No light enters the third medium. Then,

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