Monochromatic light of wavelength 6000 Å is i | vedclass

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

Question

(A) For a small-angled prism,the angle of deviation $\delta$ is given by the formula: $\delta = (\mu - 1)A$,where $\mu$ is the refractive index and $A

Vedclass · Accepted Answer

$0.24$

Vedclass · Answer

(A) For a small-angled prism,the angle of deviation $\delta$ is given by the formula: $\delta = (\mu - 1)A$,where $\mu$ is the refractive index and $A$ is the angle of the prism.The angle of dispersion (angular dispersion) $	heta$ is the difference between the angles of deviation for violet and red light: $	heta = \delta_v - \delta_r$.Substituting the formula for deviation: $	heta = (\mu_v - 1)A - (\mu_r - 1)A = (\mu_v - \mu_r)A$.Given values: $\mu_v = 1.52$,$\mu_r = 1.48$,and $A = 6^{\circ}$.Calculation: $	heta = (1.52 - 1.48) 	imes 6^{\circ} = 0.04 	imes 6^{\circ} = 0.24^{\circ}$.Therefore,the angle of dispersion is $0.24^{\circ}$.

Similar Questions

$A$ graph is plotted between angle of deviation $(\delta)$ and angle of incidence $(i)$ for a prism. The nearly correct graph is

Obtain the relation between incidence angle,emergence angle,prism angle,and deviation angle for refraction through a prism.

$A$ 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 . . . . . . .

The angle of a prism is $6^{\circ}$ and its refractive index for green light is $1.5$. If a green ray passes through it,the angle of minimum deviation will be ........$^{\circ}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module