JEE Main 2022 · 28 Jun · Shift 1 · Physics Q43

PhysicsSingle correct+4 / -1Question 43 of 90

The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 Å is used to see the object will be :

8.1\times10^6

10.0\times10^7

8.2\times10^5

1.0\times10^{-8}

Answer and worked solution

Official answer: C

Resolving power. $\dfrac D{1.22\lambda}=\dfrac{0.244}{1.22\times2.44\times10^{-7}}\approx8.2\times10^5$ .

(C)

Topics:Resolving power

JEE Main 2022 · 28 Jun · Shift 1, Physics question 43