The aperture diameter of a telescope is 5m . The separation between the moon and the earth is 4 × 10 ^ 5 km . With light of wavelength of 5500Å , the minimum separation between objects on the surface of moon, so that

Correct answer: A. 60m θ = 1.22(λ)/(a) Distance = O _ 1 O _ 2 = dθ = 1.22(λ)/(a)d Distance = O _ 1 O _ 2 = 1.22 × 5893 × 10 ^ - 10 × 4 × 10 ^ 8 5 57.5m ∴ answer from options = 60m (minimum distance)

JEE Main 20209 JanuaryMorning Shift

Exam: JEE Main
Year: 2020
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Wave Optics

JEE Main 2020 · 9 January · Morning Shift · Physics · Wave Optics

The aperture diameter of a telescope is 5m . The separation between...

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 of

5500Å

, the minimum separation between objects on the surface of moon, so that they are just resolved, is close to: The aperture diameter of a telescope is 5m . The separation between the moon and the earth is 4 × 10 ^ 5 km . With

The aperture diameter of a telescope is 5m . The separation between the moon and the earth is 4 × 10 ^ 5 km . With

Options

A $60m$
B $20m$
C $200m$
D $600m$

Correct answer

A. $60m$

Step-by-step solution

\theta = 1.22\dfrac{\lambda}{a}

Distance

= {O}_{1}{O}_{2} = d\theta

= 1.22\dfrac{\lambda}{a}d

Distance

= {O}_{1}{O}_{2} = \dfrac{1.22 \times 5893 \times {10}^{- 10} \times 4 \times {10}^{8}}{5} \approx 57.5m

∴

answer from options

= 60m

(minimum distance)

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