JEE PYQ: Ray Optics Question 30
Question 30 - 2020 (09 Jan Shift 1)
The aperture diameter of a telescope is 5m. The separation between the moon and the earth is $4 \times 10^5$ km. With light of wavelength 5500 $\text{\AA}$, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to:
(1) 60 m
(2) 20 m
(3) 200 m
(4) 600 m
Show Answer
Answer: (1)
Solution
Smallest angular separation between two distant objects here moon and earth, $\theta = 1.22\frac{\lambda}{a}$ $a$ = aperture diameter of telescope Distance $O_1 O_2 = (\theta)d$ Minimum separation between objects on the surface of moon, $= \left(1.22\frac{\lambda}{a}\right)d$ $= \frac{(1.22)(5500 \times 10^{-10}) \times 4 \times 10^5 \times 10^3}{5}$ $= 5368 \times 10^{-2}$ m $= 53.68$ m $\approx 60$ m
BETA
