JEE MAIN - Physics (2022 - 28th June Morning Shift - No. 13)
The aperture of the objective is 24.4 cm. The resolving power of this telescope, if a light of wavelength 2440 $$\mathop A\limits^o $$ is used to see th object will be :
8.1 $$\times$$ 106
10.0 $$\times$$ 107
8.2 $$\times$$ 105
1.0 $$\times$$ 10$$-$$8
Explicació
$$R.P. = {1 \over {1.22\,\lambda /a}}$$
$$ = {{24.4 \times {{10}^{ - 2}}} \over {1.22 \times 2440 \times {{10}^{ - 10}}}}$$
$$ = 8.2 \times {10^5}$$