Question

Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.48 × 106 m, and the radius of lunar orbit is 3.8 × 108 m.

Answer 1

The diameter of the image of the Moon formed by the objective lens is 13.7 cm

Given:

Object distance = u = 3.8 × 10⁸ m

Focal length = f = 15 m

Diameter of the moon/Height of the object = h = 3.48 × 10⁶ m

To find:

Diameter of the image = h' = ?

Formula:

The lens formula is:

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Solution:

Since, the object is very larger than focal length, then the object distance is taken as infinity for easier calculation.

$\frac{1}{15}=\frac{1}{v}+\frac{1}{\infty}$

$\because \frac{1}{\infty}=0$

$\frac{1}{15}=\frac{1}{v}$

Therefore, v = 15 m

We know that,

$\frac{h'}{h}=-\frac{v}{u}$

$\frac{h'}{3.48 \times 10^{6} \ \mathrm{m}}=-\frac{15 \ \mathrm{m}}{3.8 \times 10^{8} \ \mathrm{m}}$

$h'=-0.137 \ \mathrm{m}$

$\therefore h' =-13.7 \ \mathrm{cm}$

The negative sign indicates the image is inverted.