Optical Instruments: Practice Problem & Solution
In an astronomical telescope in normal adjustment a straight black line of length $L$ is drawn on inside part of objective lens. The eye-piece forms a real image of this line. The length of this image is $I$. The magnification of the telescope is: (2015 Pre)
Solution Explained:
To solve this problem, we apply the core principles of Optical Instruments. Understanding the underlying formula is key to arriving at the correct answer below:
Magnification $m = \frac{f_o}{f_e}$. The objective acts as an object for the eyepiece at distance $u = -(f_o + f_e)$. Using lens formula $\frac{1}{v} - \frac{1}{u} = \frac{1}{f_e}$, we get $v = \frac{f_e(f_o+f_e)}{f_o}$. Magnification of this image is $\frac{I}{L} = \left|\frac{v}{u}\right| = \frac{f_e}{f_o}$, so the telescope magnification is $\frac{L}{I}$.
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