Question

The focal length 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.45cm \\\ B.\,55cm \\\ C.\dfrac{{275}}{6}cm \\\ D.\dfrac{{325}}{6}cm \\\$

Answer 1

Length of telescope when image is formed at least distance of distinct vision (D) $= L = {f_0} + D{f_e}/D + {f_e}$where $D = 25cm$

Complete step by step answer:
The given values are
Focal length of the objective ${f_0} = 50\,cm$
Focal length of the eye-piece ${f_e} = 5\,cm$
We have to find
Length of telescope, L$= ?$
Now, Astronomical telescopes are refracting type telescopes. It has two lenses; objective and eye-piece.
The position of the eye-piece can be adjusted in two ways.
In the first way, the final image is formed at infinity. In the second way, the image is formed at least at a distance of distinct vision.
In our question, it is given that an image is formed at the least distance of distinct vision.
The formula for length of telescope in this case is
$L = {f_0} + \dfrac{{D{f_e}}}{{D + {f_e}}} ……….....(i)$
$D =$Least distance of distinct vision $= 25\,cm$
So putting D$= 25\,cm,\,{f_0} = 50\,cm,\,{f_e} = 5\,cm\,$in (i)
$L = 50 + \dfrac{{25 \times 5}}{{30}} \\\ = \dfrac{{325}}{6}\,cm \\\$

So, the correct answer is “Option D”.

Note:
The most critical thing to be noted in these questions is that where the image is formed. If final image is formed at infinity then formulae for length of telescope will change to $L = {f_0} + {f_e}$ otherwise it remains as $L = {f_0} + \dfrac{{D{f_e}}}{{D + {f_e}}}$.