JEE MAIN - Physics (2020 - 9th January Morning Slot - No. 23)
The aperture diameter of a telescope is 5m. The
separation between the moon and the earth is
4 × 105 km. With light of wavelength of
5500 $$\mathop A\limits^o $$, the minimum separation between
objects on the surface of moon, so that they are
just resolved, is close to :
20 m
200 m
600 m
60 m
Explanation
_9th_January_Morning_Slot_en_23_1.png)
$$\theta $$ = 1.22$${\lambda \over a}$$
Distance O1O2 = ($$\theta $$)d
= (1.22$${\lambda \over a}$$)d
= $${{\left( {1.22} \right)\left( {5500 \times {{10}^{ - 10}}} \right)\left( {4 \times {{10}^5}} \right) \times {{10}^3}} \over 5}$$
= 5368 × 10–2 m
= 53.68 m
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