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 cm
Given:
- Power of objective lens, $P_o = 2 \ \text{D}$
- Power of eye lens, $P_e = 10 \ \text{D}$
Formula: Focal length $f$ (in meters) is given by:
$f = \frac{100}{P}$
So,
$f_o = \frac{100}{2} = 50 \ \text{cm}$
$f_e = \frac{100}{10} = 10 \ \text{cm}$
In normal adjustment (final image at infinity), the distance between the objective and eye lens is:
$D = f_o + f_e = 50 + 10 = 60 \ \text{cm}$