First slide

The optical instruments

Question

In an astronomical telescope in normal adjustment, a straight black line of length L is drawn on the objective lens. The eyepiece forms a real image of this line. The length of this image is l. The magnification of the telescope is

Easy

A

$\frac{L}{l}$
B

$\frac{L}{l} + 1$
C

$\frac{L}{l} - 1$
D

$\frac{L + l}{L - l}$

Solution

Here we treat the line on the objective as the object and the eyepiece as the lens.
Hence $u = - (f_{o} + f_{e})$ and $f = f_{e}$
Now $\frac{1}{v} - \frac{1}{- (f_{o} + f_{e})} = \frac{1}{f_{e}}$
Solving we get $v = \frac{(f_{o} + f_{e}) f_{e}}{f_{o}}$
Magnification $= |\frac{v}{u}| = \frac{f_{e}}{f_{o}} = \frac{Image size}{Object size} = \frac{l}{L}$
$∴$ Magnification of telescope in normal adjustment
$= \frac{f_{o}}{f_{e}} = \frac{L}{l}$

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