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JEE · NEET Previous Year Question

$The aperture of the objective is 24.4 cm. The resolving power of this telescope. If a light of wavelength 2440 A is used to see the object will be$

Topic: JEE PYQ

Answer: C. Full step-by-step solution below.

A $8. 1 x 1 0^{6}$
B $10.0 x 1 0^{7}$
C $8. 2 x 1 0^{5}$
D $1. 0 x 1 0^{- 8}$

Correct answer: C

Solution

(c) R . P = \frac{d}{1.22 λ} = \frac{24.4 \times 1 0 ^{- 2}}{1.22 \times 2440 \times 1 0 ^{- 10}} = 8.2 \times 1 0^{5}

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