#### 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 xx 10^6`m, and the radius of the lunar orbit is `3.48 xx 10^8`m.

#### Solution

When the final image is formed at the least distance of distinct vision

Magnifying power,

`M = beta/alpha`

α and β are small

∴ `M = tan β/tan α` ...(i)

In ΔA'B'C_{2 ,}tan β = `("A"'"B"')/("C"_2"B"')`

In ΔA'B'C_{1, }tanα = `("A"'"B"')/("C"_1"B"')`

From equation (i), we have:

`M = ("A"'"B"')/("C"_2"B"') xx ("C"_1"B"')/("A"'"B"')`

`"M"= ("C"_1"B"')/("C"_2"B"')`

Here,

`C_1B' = +f_0`

`C_2B' = -u_e`

`M = f_0/-u_e` ...(ii)

Using the lens equation `(1/ν - 1/u = 1/f)` for the eyepieces, we get

`1/(-D) - 1/(-u_c) = 1/f_e`

`-1/D + 1/u_c = 1/f_e`

`1/u_e = 1/f_e + 1/D`

`f_0/u_e = f_0/f_e (1 + f_e/D)`

`(-f_0)/u_e = (-f_0)/f_e (1 + f_e/D)`

M = `-f_0/f_e (1 + f_e/D)`

Angular magnification is:

`m_0 = |(f_0)/(f_e)| = |1500/1| = 1500`

Where, `"f"_0` is the focal length of the objective lens and f_{e} is the focal length of the eye piece

Given, the diameter of the moon = `3.48 xx 10^6`m

The radius of the lunar orbit = `3.8 xx 10^8`m

The diameter of the image of the moon formed by the objective lens is given by d = af_{0}

d = `"Diameter of the moon"/"Radius of the lunar orbit" xx f_0`

d = `(3.48 xx 10^6)/(3.8 xx 10^8) xx 15 = 13.74` cm