A monochromatic beam of light is incident at | vedclass

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

Question

(B) Applying Snell's Law on the $1^{	ext{st}}$ surface: $\sin(60^{\circ}) = n \sin(r_1) \Rightarrow \sin(r_1) = \frac{\sqrt{3}}{2n}$.For an equilater

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Applying Snell's Law on the $1^{	ext{st}}$ surface: $\sin(60^{\circ}) = n \sin(r_1) \Rightarrow \sin(r_1) = \frac{\sqrt{3}}{2n}$.For an equilateral prism,the prism angle $A = 60^{\circ}$,so $r_1 + r_2 = 60^{\circ}$,which implies $r_2 = 60^{\circ} - r_1$.Applying Snell's Law on the $2^{	ext{nd}}$ surface: $n \sin(r_2) = \sin(	heta)$.Substituting $r_2$: $\sin(	heta) = n \sin(60^{\circ} - r_1) = n [\sin(60^{\circ}) \cos(r_1) - \cos(60^{\circ}) \sin(r_1)]$.Since $\sin(r_1) = \frac{\sqrt{3}}{2n}$,then $\cos(r_1) = \sqrt{1 - \frac{3}{4n^2}} = \frac{\sqrt{4n^2 - 3}}{2n}$.Substituting these into the equation for $\sin(	heta)$:$\sin(	heta) = n [\frac{\sqrt{3}}{2} \cdot \frac{\sqrt{4n^2 - 3}}{2n} - \frac{1}{2} \cdot \frac{\sqrt{3}}{2n}] = \frac{\sqrt{3}}{4} (\sqrt{4n^2 - 3} - 1)$.Differentiating with respect to $n$: $\cos(	heta) \frac{d	heta}{dn} = \frac{\sqrt{3}}{4} \cdot \frac{1}{2\sqrt{4n^2 - 3}} \cdot 8n = \frac{\sqrt{3} n}{\sqrt{4n^2 - 3}}$.At $n = \sqrt{3}$,$	heta = 60^{\circ}$,so $\cos(60^{\circ}) \frac{d	heta}{dn} = \frac{\sqrt{3} \cdot \sqrt{3}}{\sqrt{4(3) - 3}} = \frac{3}{\sqrt{9}} = 1$.$\frac{1}{2} \cdot \frac{d	heta}{dn} = 1 \Rightarrow \frac{d	heta}{dn} = 2$. Thus,$m = 2$.

Similar Questions

$A$ glass prism $A$ deviates the red and blue rays through $10^{\circ}$ and $12^{\circ}$ respectively. $A$ second prism $B$ deviates them through $8^{\circ}$ and $10^{\circ}$ respectively. The ratio of their dispersive powers is ($A$ to $B$):

$A$ ray of light suffers minimum deviation when incident on a prism having an angle of the prism equal to $60^{\circ}$. The refractive index of the prism material is $\sqrt{2}$. The angle of incidence (in degrees) is . . . . . . .

$A$ prism is made up of a material with a refractive index of $\sqrt{3}$. The angle of the prism is $A$. If the angle of minimum deviation is equal to the angle of the prism, then the value of $A$ is: (in $^{\circ}$)

For a glass prism $(\mu = \sqrt{3})$,the angle of minimum deviation is equal to the angle of the prism. Find the angle of the prism. (in $^{\circ}$)

The angle of prism is $5^o$ and its refractive indices for red and violet colours are $1.5$ and $1.6$ respectively. The angular dispersion produced by the prism is.....$^o$

Vedclass Test Series

Exam Paper Generator

Online Exam Module