A rectangular block is composed of three diff | vedclass

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

Question

(C) Let the ray be incident normally on the left face. It enters the first prism (refractive index $\mu_1$) without deviation.At the interface between

Vedclass · Accepted Answer

$\mu_1^2+\mu_3^2=2 \mu_2^2$

Vedclass · Answer

(C) Let the ray be incident normally on the left face. It enters the first prism (refractive index $\mu_1$) without deviation.At the interface between prism $\mu_1$ and $\mu_2$,the angle of incidence is $i = 45^{\circ}$. By Snell's law: $\mu_1 \sin 45^{\circ} = \mu_2 \sin 	heta$,where $	heta$ is the angle of refraction.Thus,$\mu_1 \frac{1}{\sqrt{2}} = \mu_2 \sin 	heta \implies \sin 	heta = \frac{\mu_1}{\sqrt{2} \mu_2} \quad \dots(1)$At the interface between prism $\mu_2$ and $\mu_3$,the ray strikes at an angle of incidence $r_2 = \alpha - 	heta$. From the geometry,$\alpha = 90^{\circ}$. So,$r_2 = 90^{\circ} - 	heta$.The ray emerges normal to the right face,meaning the angle of refraction at the second interface is $0^{\circ}$. By Snell's law: $\mu_2 \sin(90^{\circ} - 	heta) = \mu_3 \sin 0^{\circ}$ is incorrect; rather,the ray emerges normal to the right face of the block,so the angle of incidence at the $\mu_2-\mu_3$ interface is $r_2$ and the angle of refraction is $i_2 = 45^{\circ}$.Applying Snell's law at the second interface: $\mu_2 \sin(90^{\circ} - 	heta) = \mu_3 \sin 45^{\circ} \implies \mu_2 \cos 	heta = \mu_3 \frac{1}{\sqrt{2}} \implies \cos 	heta = \frac{\mu_3}{\sqrt{2} \mu_2} \quad \dots(2)$Squaring and adding equations $(1)$ and $(2)$:$\sin^2 	heta + \cos^2 	heta = \left(\frac{\mu_1}{\sqrt{2} \mu_2}ight)^2 + \left(\frac{\mu_3}{\sqrt{2} \mu_2}ight)^2$$1 = \frac{\mu_1^2}{2 \mu_2^2} + \frac{\mu_3^2}{2 \mu_2^2} \implies 2 \mu_2^2 = \mu_1^2 + \mu_3^2$.

$A$ rectangular block is composed of three different glass prisms (with refractive indices $\mu_1, \mu_2$ and $\mu_3$) as shown in the figure below. $A$ ray of light incident normal to the left face emerges normal to the right face. Then,the refractive indices are related by

Similar Questions

The refractive index of water and glass with respect to air is $1.3$ and $1.5$ respectively. Then the refractive index of glass with respect to water is

$A$ light wave has a frequency of $4 \times 10^{14} \,Hz$ and a wavelength of $5 \times 10^{-7} \,m$ in a medium. The refractive index of the medium is

From the figure shown,establish a relation between $\mu_1, \mu_2, \mu_3$.

If the wavelength of light in vacuum is $\lambda$,the wavelength in a medium of refractive index $n$ will be

Vedclass Test Series

Exam Paper Generator

Online Exam Module