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A thin prism $P_1$ with an angle $6^{\circ}$ and made of glass of refractive index 1.54 is combined with another prism $\mathrm{P}_2$ made from glass of refractive index 1.72 to produce dispersion without average deviation. The angle of prism $\mathrm{P}_2$ is :
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Verified Answer
The correct answer is:
$4.5^{\circ}$
Angle of deviation for first prism
$\delta_1=A_1\left(\mu_1-1\right)=6(1.54-1)$
Angle of deviation for second prism
$\begin{aligned}
& \delta_2=A_2\left(\mu_2-1\right) \\
& =A_2(1.72-1)
\end{aligned}$
For dispersion without deviation
$\begin{aligned}
& \delta_1=\delta_2 \\
& \Rightarrow 6^{\circ}(1.54-1)=\mathrm{A}_2(1.72-1) \\
& \Rightarrow \mathrm{A}_2=\frac{6^{\circ} \times 0.54}{0.72}=\frac{18^{\circ}}{4}=4.5^{\circ}
\end{aligned}$
$\delta_1=A_1\left(\mu_1-1\right)=6(1.54-1)$
Angle of deviation for second prism
$\begin{aligned}
& \delta_2=A_2\left(\mu_2-1\right) \\
& =A_2(1.72-1)
\end{aligned}$
For dispersion without deviation
$\begin{aligned}
& \delta_1=\delta_2 \\
& \Rightarrow 6^{\circ}(1.54-1)=\mathrm{A}_2(1.72-1) \\
& \Rightarrow \mathrm{A}_2=\frac{6^{\circ} \times 0.54}{0.72}=\frac{18^{\circ}}{4}=4.5^{\circ}
\end{aligned}$
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