Question

Define magnifying power of a telescope. Write its expression. A small telescope has an objective lens with a focal length of 150 cm and an eyepiece with a focal length of 5 cm. If this telescope is used to view a 100 m high tower 3 km away, find the height of the final image when it is formed 25 cm away from the eyepiece.

Answer 1

Step 1: Given data:

$f0=150cm,fe=5cm,D=25cm$

Step 2: Magnification:

The magnifying power of the telescope is the ratio of the angle subtended at the eye by the image to the angle subtended at the unaided eye by the object.

$Magnifyingpower,m=\frac{f_o}{fe}\frac{{}_{}{(1+fe)}_{}}{D}$

$$\begin{gathered} f_0=focallengthoftheobject \\ fe=focallengthoftheeyepiece \\ D=leastdis\tan ceofthedistinctvision \end{gathered}$$

Step 3: Calculation the height of the final image.

Using the lens equation for an object lens

$Magnifyingpower,m=\frac{f_o}{fe}\frac{{}_{}{(1+fe)}_{}}{D}$

$\frac{-150}{5}\left(1+\frac{5}{25}\right)=-36$

Magnification due to the object lens

$$\begin{gathered} M=\alpha /\beta \\ =\tan \alpha /\tan \beta (Asanglesaresmall) \\ \tan \alpha =H/uwhereH=heightoftheobjectandu=dis\tan ceoftheobjectfromtheobjective. \\ =100/3000 \\ =1/30 \\ M=\tan \beta /(1/30) \\ =-36/30 \\ =H^{,}/D;H\text{'}=heightoftheimageandD=dis\tan ceoftheimageformation \\ Thus,H\prime =-36\times 25/30=-30cm \end{gathered}$$

Negative sign indicate that we get an inverted image.