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A thin symmetrical double convex lens of refractive index $\mu_2=1.5$ is placed between a medium of refractive index $\mu_1=1.4$ to the left and another medium of refractive index $\mu_3=1 \cdot 6$ to the right. Then, the system behaves as
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Verified Answer
The correct answer is:
a glass plate
For first refracting surface,
For parallel beam of light; $-\frac{\mu_1}{u}+\frac{\mu_2}{V}=\frac{\mu_2-1}{R}$
$\begin{aligned}
& \Rightarrow-\frac{1 \cdot 4}{-\infty}+\frac{1 \cdot 5}{V}=\frac{1 \cdot 5-1.4}{R} \\
& V=15 R\end{aligned}$
For second refracting surface,
$\begin{aligned}
& -\frac{\mu_2}{1.5 R}+\frac{\mu_3}{V^{\prime}}=\frac{\mu_3-\mu_2}{R} \\
& \Rightarrow-\frac{1.5}{15 R}+\frac{1.6}{V^{\prime}}=\frac{1.6-1.5}{R} \Rightarrow V^{\prime}=\infty\end{aligned}$
Hence, the combination behaves as glass plate.
For parallel beam of light; $-\frac{\mu_1}{u}+\frac{\mu_2}{V}=\frac{\mu_2-1}{R}$
$\begin{aligned}
& \Rightarrow-\frac{1 \cdot 4}{-\infty}+\frac{1 \cdot 5}{V}=\frac{1 \cdot 5-1.4}{R} \\
& V=15 R\end{aligned}$
For second refracting surface,
$\begin{aligned}
& -\frac{\mu_2}{1.5 R}+\frac{\mu_3}{V^{\prime}}=\frac{\mu_3-\mu_2}{R} \\
& \Rightarrow-\frac{1.5}{15 R}+\frac{1.6}{V^{\prime}}=\frac{1.6-1.5}{R} \Rightarrow V^{\prime}=\infty\end{aligned}$
Hence, the combination behaves as glass plate.
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