A beam of monochromatic light is incident at | vedclass

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

Question

(B) The deviation produced by a prism is given by the formula: $\delta = i + e - A$.For an equilateral prism,the angle of the prism is $A = 60^{\circ}

Vedclass · Accepted Answer

$< 30^{\circ}$

Vedclass · Answer

(B) The deviation produced by a prism is given by the formula: $\delta = i + e - A$.For an equilateral prism,the angle of the prism is $A = 60^{\circ}$.Given $i = 50^{\circ}$ and $e = 40^{\circ}$,the deviation $\delta$ is:$\delta = 50^{\circ} + 40^{\circ} - 60^{\circ} = 30^{\circ}$.We know that the angle of minimum deviation $\delta_m$ occurs when $i = e$. For any other values of $i$ and $e$ (where $i 
eq e$),the deviation $\delta$ is always greater than the minimum deviation $\delta_m$.Since $i = 50^{\circ}$ and $e = 40^{\circ}$ $(i 
eq e)$,the calculated deviation $\delta = 30^{\circ}$ must be greater than the minimum deviation $\delta_m$.Therefore,$\delta_m Thus,the correct option is $B$.

$A$ beam of monochromatic light is incident at $i = 50^{\circ}$ on one face of an equilateral prism. If the angle of emergence is $e = 40^{\circ}$,then the angle of minimum deviation $\delta_m$ is:

Similar Questions

If the angle of minimum deviation produced by an equilateral prism is equal to the angle of the prism, then the refractive index of the material of the prism is nearly

$A$ prism of refractive index $1.53$ is placed in water of refractive index $1.33$. If the angle of prism is $60^{\circ}$,the angle of minimum deviation in water will be ........$^{\circ}$ $(\sin 35.1^{\circ} = 0.575)$

$A$ ray of light is incident normally on one face of a $30^{\circ}-60^{\circ}-90^{\circ}$ prism of refractive index $5/3$ immersed in water of refractive index $4/3$ as shown in the figure.

For refraction through a small angled prism,the angle of deviation:

Vedclass Test Series

Exam Paper Generator

Online Exam Module