In the given figure,the face AC of the equila | vedclass

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

Question

(A) According to Snell's law at the interface of the glass prism and the liquid at face $AC$:$\mu_{glass} \sin(i) = n \sin(r)$Given that the incident

Vedclass · Accepted Answer

$27$

Vedclass · Answer

(A) According to Snell's law at the interface of the glass prism and the liquid at face $AC$:$\mu_{glass} \sin(i) = n \sin(r)$Given that the incident angle $i = 60^{\circ}$ and the refracted light grazes along the face $AC$,the angle of refraction $r = 90^{\circ}$.Substituting the values:$1.5 	imes \sin(60^{\circ}) = n 	imes \sin(90^{\circ})$$1.5 	imes \frac{\sqrt{3}}{2} = n 	imes 1$$n = \frac{1.5 \sqrt{3}}{2} = \frac{3 \sqrt{3}}{4}$We are given $n = \frac{\sqrt{x}}{4}$.Equating the two expressions for $n$:$\frac{\sqrt{x}}{4} = \frac{3 \sqrt{3}}{4}$$\sqrt{x} = 3 \sqrt{3} = \sqrt{9 	imes 3} = \sqrt{27}$Therefore,$x = 27$.

Similar Questions

If the refractive angles of two prisms made of crown glass are $10^o$ and $20^o$ respectively,then the ratio of their angular dispersions will be

Three prisms $1, 2,$ and $3$ have a prism angle $A = 60^{\circ}$. Their refractive indices are $1.4, 1.5,$ and $1.6$ respectively. If $\delta_1, \delta_2,$ and $\delta_3$ are their angles of deviation,then:

$A$ light ray is incident on a prism at an angle of $60^{\circ}$ and the refracting angle of the prism is $30^{\circ}$. The emergent ray makes an angle of $30^{\circ}$ with the incident ray. The refractive index of the prism is ...

The refraction through the prisms is as shown. Pick out the wrong statement from the following. Path of the light ray in

$A$ prism of refractive index $\sqrt{2}$ has a refracting angle of $60^{\circ}$. In order for a ray to suffer minimum deviation,it should be incident at an angle of .......$^{\circ}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module