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…

Question

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

Aicharya · Accepted Answer

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 $=f_o{ }^{\prime}+f_e{ }^{\prime}$ $\begin{aligned} & =2 f_o+2 f_e=2\left(f_o+f_eight) \ & =2(L)=2(16)=32 \mathrm{~cm}\end{aligned}$

The objective and eyepiece of an astronomical telescope are double convex lenses with refractive index 1.5.…

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