Calculate the dispersive power for crown glass from the given data array rll & μ_ V =1.523 and & μ_ R =1.5145 array

Correct answer: A. 0.01639 Here, μ_V=1.523 and μ_R=1.5145 Mean refractive index, μ=(1.523+1.5145)/(2)=1.51875 Dispersive power is given by, aligned ω & =((μ_V-μ_R))/((μ-1)) & =(1.523-1.5145)/((1.51875-1))=0.01639 aligned

JIPMER 2018

Exam: JIPMER
Year: 2018
Subject: Physics
Chapter: Ray Optics

JIPMER 2018 · Physics · Ray Optics

Calculate the dispersive power for crown glass from the given data...

Calculate the dispersive power for crown glass from the given data

\begin{array}{rll} & \mu_{\mathrm{V}}=1.523 \\ \text{ and } & \mu_{\mathrm{R}}=1.5145\end{array}

Options

A $0.01639$
B $1.05639$
C $0.05639$
D $2.05639$

Correct answer

A. $0.01639$

Step-by-step solution

Here,

\mu_V=1.523

and

\mu_R=1.5145

Mean refractive index,

\mu=\frac{1.523+1.5145}{2}=1.51875

Dispersive power is given by,

\begin{aligned} \omega & =\frac{\left(\mu_V-\mu_R\right)}{(\mu-1)} \\ & =\frac{1.523-1.5145}{(1.51875-1)}=0.01639\end{aligned}

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