For the angle of minimum deviation of a prism | vedclass

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

Question

(B) The refractive index of a prism is given by $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_m = A

Vedclass · Accepted Answer

lies between $2$ and $\sqrt{2}$

Vedclass · Answer

(B) The refractive index of a prism is given by $\mu = \frac{\sin((A+\delta_m)/2)}{\sin(A/2)}$.Given that the angle of minimum deviation $\delta_m = A$,we substitute this into the formula:$\mu = \frac{\sin((A+A)/2)}{\sin(A/2)} = \frac{\sin A}{\sin(A/2)}$.Using the identity $\sin A = 2\sin(A/2)\cos(A/2)$,we get:$\mu = \frac{2\sin(A/2)\cos(A/2)}{\sin(A/2)} = 2\cos(A/2)$.For a physical prism,the angle of incidence $i$ must satisfy $0 If $A 	o 0^o$,then $\mu 	o 2\cos(0^o) = 2$.If $A 	o 90^o$,then $\mu 	o 2\cos(45^o) = 2 	imes (1/\sqrt{2}) = \sqrt{2}$.Therefore,the refractive index $\mu$ must lie between $\sqrt{2}$ and $2$.

For the angle of minimum deviation of a prism to be equal to its refracting angle,the prism must be made of a material whose refractive index

Similar Questions

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

For a small-angled prism,derive the expression for the angle of minimum deviation,${D_m} = A({n_{21}} - 1)$.

Yellow light is refracted through a prism producing minimum deviation. If $i_1$ and $i_2$ denote the angle of incidence and emergence for the prism,then ..........

Vedclass Test Series

Exam Paper Generator

Online Exam Module