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The objective and eyepiece of an astronomical telescope are double convex lenses with refractive index 1.5. When the telescope is adjusted to infinity, the separation between the two lenses is $16 \mathrm{~cm}$. If the space between the lenses is now filled with water and again telescope is adjusted for infinity, then the present separation between the lenses is
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The correct answer is:
$32 \mathrm{~cm}$
From the lens maker formula
$\begin{array}{cc} & \frac{\mu_3}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R_1}+\frac{\mu_3-\mu_2}{R_2} \\ \Rightarrow \quad & \frac{4}{3 f_o{ }^{\prime}}-\frac{1}{\infty}=\frac{1.5-1}{R}+\frac{\frac{4}{3}-1.5}{-R} \\ f_o{ }^{\prime}=R=2 f\end{array}$
Now, length $=f_o{ }^{\prime}+f_e{ }^{\prime}$
$\begin{aligned} & =2 f_o+2 f_e=2\left(f_o+f_e\right) \\ & =2(L)=2(16)=32 \mathrm{~cm}\end{aligned}$
$\begin{array}{cc} & \frac{\mu_3}{v}-\frac{\mu_1}{u}=\frac{\mu_2-\mu_1}{R_1}+\frac{\mu_3-\mu_2}{R_2} \\ \Rightarrow \quad & \frac{4}{3 f_o{ }^{\prime}}-\frac{1}{\infty}=\frac{1.5-1}{R}+\frac{\frac{4}{3}-1.5}{-R} \\ f_o{ }^{\prime}=R=2 f\end{array}$
Now, length $=f_o{ }^{\prime}+f_e{ }^{\prime}$
$\begin{aligned} & =2 f_o+2 f_e=2\left(f_o+f_e\right) \\ & =2(L)=2(16)=32 \mathrm{~cm}\end{aligned}$
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