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A telescope has focal length of objective and eye-piece as 200 cm and 5 cm respectively. What is magnification of telescope?
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Verified Answer
The correct answer is:
$40$
In relaxed eye state of telescope or when final image is formed at infinity, the magnification of telescope is given by
$|M|=\frac{f_o}{f_e}$
where $\begin{aligned} f_o & =\text { focal length of objective }=200 \mathrm{~cm} \\ f_e & =\text { focal length of eye-piece }=5 \mathrm{~cm}\end{aligned}$
Hence, $|M|=\frac{200}{5}=40$
$|M|=\frac{200}{5}=40$
Note : The aperture of the objective lens is kept large so that more and more rays coming from the heavenly body may enter the telescope and a bright image is formed by the objective lens. The aperture of the eye lens is kept comparatively small so that all the rays may enter the eye.
$|M|=\frac{f_o}{f_e}$
where $\begin{aligned} f_o & =\text { focal length of objective }=200 \mathrm{~cm} \\ f_e & =\text { focal length of eye-piece }=5 \mathrm{~cm}\end{aligned}$
Hence, $|M|=\frac{200}{5}=40$
$|M|=\frac{200}{5}=40$
Note : The aperture of the objective lens is kept large so that more and more rays coming from the heavenly body may enter the telescope and a bright image is formed by the objective lens. The aperture of the eye lens is kept comparatively small so that all the rays may enter the eye.
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