Question

The focal lengths of objective and eyepiece of a telescope are 200cm
and 10cm respectively. It is used to get an image of the Sun screen
placed 40 cm behind the eyepiece. The diameter of the image is 6cm.
What is the diameter of the Sun? Given, the distance from earth to the
Sun is 1.5x10^11m​

Answer 1

Solution_______________

$here \\ f_{0} = 200cm \\ f_{e} = 10cm \\ v_{e} = + 40cm \: (for \: real \: image)$

$as \\ \frac{1}{ v_{e} } - \frac{1}{ u_{e} } = \frac{1}{ f_{e} } \\ so. \\ \frac{1}{ u_{e} } = \frac{1}{ v_{e} } - \frac{1}{ f_{e} } \\ \\ = > \frac{1}{40} - \frac{1}{10} \\ \\ = > \frac{1 - 4}{40} = \frac{ - 3}{40} \\ \\ u_{e} = - \frac{40}{3}cm$

Magnification produced by eyepiece is

$m_{e} = \frac{ v_{e} }{ | u_{e} | } = \frac{40}{ \frac{40}{3} } = 3$

Diameter of the image formed by the objective is

$d = \frac{6}{3}cm = 2cm.$

If D is the diameter of the sun ( in m ), then the angle subtended by it on the objective will be

$\alpha = \frac{D}{1.5 \times {10}^{11} } rad$

Angle subtended by the image at the objective will be equal to this angle and is given by

$\alpha = \frac{size \: of \: image}{ f_{o} } = \frac{2}{200} = \frac{1}{100}rad \\ \\ = > \frac{D}{1.5 \times {10}^{11} } = \frac{1}{100} \\ \\ D = \frac{1.5 \times {10}^{11} }{100} = 1.5 \times {10}^{9} m.$

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