The cross-section of a prism is an equilatera | vedclass

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

Question

(D) For an equilateral prism,the prism angle $A = 60^{\circ}$.Given that minimum deviation occurs when the angle of incidence $i = A = 60^{\circ}$.The

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(D) For an equilateral prism,the prism angle $A = 60^{\circ}$.Given that minimum deviation occurs when the angle of incidence $i = A = 60^{\circ}$.The refractive index $\mu$ of the prism is given by $\mu = \frac{\sin(\frac{\delta_{min} + A}{2})}{\sin(\frac{A}{2})}$.At minimum deviation,$i = e = 60^{\circ}$,so $\delta_{min} = 2i - A = 2(60^{\circ}) - 60^{\circ} = 60^{\circ}$.Thus,$\mu = \frac{\sin(60^{\circ})}{\sin(30^{\circ})} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$.The speed of light in the prism is $v = \frac{c}{\mu} = \frac{3 	imes 10^8}{\sqrt{3}} \, m/s$.The distance $AP$ is the height of the equilateral triangle with side $a = 10 \, cm = 0.1 \, m$.$AP = a \sin 60^{\circ} = 0.1 	imes \frac{\sqrt{3}}{2} = 0.05\sqrt{3} \, m$.The time taken $t = \frac{AP}{v} = \frac{0.05\sqrt{3}}{3 	imes 10^8 / \sqrt{3}} = \frac{0.05 	imes 3}{3 	imes 10^8} = 0.05 	imes 10^{-8} = 5 	imes 10^{-10} \, s$.Therefore,the value is $5$.

Similar Questions

Monochromatic light is incident on a glass prism of angle $A$. If the refractive index of the material of the prism is $\mu$,a ray,incident at an angle $\theta$ on the face $AB$,would get transmitted through the face $AC$ of the prism provided:

The refractive index of a glass is $1.520$ for red light and $1.525$ for blue light. Let $D_1$ and $D_2$ be the angles of minimum deviation for red and blue light,respectively,in a prism of this glass. Then,

The figures represent three cases of a ray passing through a prism of angle $A$. The case corresponding to minimum deviation is

$A$ thin prism $P_1$ with angle $4^o$ and made from glass of refractive index $1.54$ is combined with another thin prism $P_2$ made from glass of refractive index $1.72$ to produce dispersion without deviation. The angle of prism $P_2$ is......$^o$

$A$ light ray is incident on a prism of angle $A = 30^{\circ}$ at an angle of incidence of $60^{\circ}$. If the emergent ray makes an angle of $30^{\circ}$ with the incident ray,then find the refractive index of the prism $:-$

Vedclass Test Series

Exam Paper Generator

Online Exam Module