JEE Main 2014 · 9 Apr · Shift 1 · Physics Q25

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The focal lengths of objective lens and eye lens of a Gallelian Telescope are respectively 30 cm and 3.0 cm. Telescope produces virtual, erect image of an object situated far away from it at least distance of distinct vision from the eye lens. In this condition, the Magnifying Power of the Gallelian Telescope should be :

+11.2

-11.2

-8.8

+8.8

Answer and worked solution

Official answer: D

For the final image at the least distance of distinct vision ( $D=25$ cm), with a diverging eye lens ( $f_e=3$ cm):
$M=\frac{f_o}{f_e}\left(1-\frac{f_e}{D}\right)=10\left(1-\frac3{25}\right)=+8.8.$
The image is erect, so $M$ is positive.

Topics:Optical instruments

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JEE Main 2014 · 9 Apr · Shift 1, Physics question 25