JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Wave Optics

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength

2440 \overset{˚}{A}

is used to see the object will be

(A) $8.1 \times 1 0^{6}$
(B) $10.0 \times 1 0^{7}$
(C) $8.2 \times 1 0^{5}$ ✓ Correct
(D) $1.0 \times 1 0^{- 8}$

Answer: (C)

Solution

The resolving power of a telescope is given by

R . P = \frac{d}{1.22 λ}

, where

d

is the aperture diameter and

λ

is the wavelength. Substituting:

R . P = \frac{24.4 \times 1 0 ^{- 2}}{1.22 \times 2440 \times 1 0 ^{- 10}} = \frac{24.4 \times 1 0 ^{- 2}}{2.977 \times 1 0 ^{- 7}} \approx 8.2 \times 1 0^{5}

. The answer is (C).

JEE Main · 2022 · 28 Jun 2022 · Shift 1 · Single correct option

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