Question

The objective lens of a telescope is a symmetric glass lens with $100cm$radii of curvature. The eyepiece lens is also a symmetric glass lens. What are the radii of curvature of the eyepiece lens if the telescope’s magnification is $20\times$?

Short answer

The radii of curvature of the eyepiece lens is $R_{eye}=5cm$.

Step-by-step solution

Given Information.

We have given a symmetric glass lens with$100cm$radii of curvature.

We need to find the radii of curvature.

Calculation.

As we know that:

$$\begin{gathered} M=20 \\ {R}_{eye}=2f_{eye} \\ {f}_{obj}=\frac{R}{2}=\frac{100}{2}=50 \end{gathered}$$

Magnification is related to:

$$\begin{gathered} M=\frac{f_{obj}}{f_{eye}} \\ {f}_{eye}=\frac{f_{obj}}{M} \\ {f}_{eye}=\frac{50cm}{20cm} \\ {f}_{eye}=\frac{5}{2}cm \end{gathered}$$

Here $M$is magnification, $f_{obj}$is the focal length of the object and$f_{eye}$is the focal length of the eye.

The radius of the eyepiece $R_{eye}$is :

$$\begin{gathered} R_{eye}=2f_{eye} \\ {R}_{eye}=2\times \frac{5}{2} \\ {R}_{eye}=5cm \end{gathered}$$