Question

The focal lengths of the lenses of an astronomical telescope are 50 cm and 5 cm. The length of the telescope when the image is formed at the least distance of distinct vision is

• A. 45 cm
• B. 55 cm
• C. $\frac{275}{6}cm$
• D. $\frac{325}{6}cm$

Answer 1

The correct option is D $\frac{325}{6}cm$

The length of the telescope, $L_D=f_0+u_e$, as the objective forms the image at its focus, as the rays are from infinity, and this image acts as the object for the eyepiece.
Now, using lens formula, and using the fact that the eyepiece forms image at near point, that is $v=D(=-25cm)$:
$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}.$
Applying sign convention, $\frac{1}{u_e}=-\frac{1}{D}-\frac{1}{f_e}.$
Or, considering only the modulus, $u_e=\frac{f_{e}D}{f_e+D}$.
Therefore, $L_D=f_0+u_e=f_0+\frac{f_{e}D}{f_e+D}=50+\frac{5\times 25}{(5+25)}=\frac{325}{6}cm$.