An astronomical telescope uses two lenses of power 10 D and 2 D. The distance between the objective and the eye lens in normal adjustment would be
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( 60\,\mathrm{cm} \)
The focal length of a lens is the reciprocal of its power, \( f = 1/P \), taken in metres.
\( \displaystyle f_1 = \frac{1}{10\,\mathrm{D}} = 0.10\,\mathrm{m} = 10\,\mathrm{cm} \)
\( \displaystyle f_2 = \frac{1}{2\,\mathrm{D}} = 0.50\,\mathrm{m} = 50\,\mathrm{cm} \)
In normal adjustment the final image is at infinity, so the focal point of the objective coincides with that of the eye lens. The distance between the lenses is the sum of the two focal lengths.
\( \displaystyle L = f_o + f_e = 50\,\mathrm{cm} + 10\,\mathrm{cm} = 60\,\mathrm{cm} \)