Name: Angular Magnification and Magnifying Power
Uploaded: 2019-07-19
Duration: 7 min 43 s
Description: Angular magnification can be used to assess the size of the object as it appears to the eye. To learn more about angular magnification produced by an optical...

Question

An astronomical telescope has an objective of focal length $200 cm$ and an eye-piece of focal length $10 cm$ . An observer adjusts the distance of the eyepiece from the objective to obtain the image of the sun on a screen $40 cm$ behind the eye-piece. The diameter of the sun’s image is $6.0 cm$ . Find the diameter of the sun. The average earth-sun distance is $1 .5 \times 10^{11} m$ .

Accepted Answer

Given, focal length of eye- piece, $f_{e} = 10 cm .$

Diameter of the sun's image = $6.0 cm$

Focal length of objective, $1 .5 \times 10^{11} m$

Image of sun on the screen, $v_{e} = 40 cm$

Applying lens formula foe the eyepiece , $\frac{1}{v_{e}} - \frac{1}{u_{e}} = \frac{1}{f_{e}} or \frac{1}{u_{e}} = \frac{1}{40} - \frac{1}{10} or u_{e} = \frac{40}{3} cm$

So, magnification produced by eye piece, $m_{e} = \frac{v_{e}}{u_{e}} = \frac{40}{\frac{40}{3}} = 3$

Therefore, the size of image formed by objective = $\frac{object size}{m_{e}} = 6 / 3 = 2$

The angle subtended by the sun on its objective, $α = \frac{diameter (D) of the sun}{distance of the sun - earth} = \frac{D}{(1.5 \times 10^{11})}$

The angle subtended by the image on the eye piece $α = \frac{I m a g e s i z e}{f_{o}} = \frac{2}{200} o r \frac{1}{100}$

Therefore, we have $α = \frac{1}{100} = \frac{D}{1.5 \times 10^{11}} or D = 1.5 \times 10^{9} m$

Hence the diameter of the sun is $1.5 \times 10^{9} m$ .

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