Monochromatic light of wavelength 6000 Å is i | vedclass

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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

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 $0.24^{\circ}$.

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