The diameter of the objective of a telescope is $1 \ m$. Its resolving limit for light of wavelength $4538 \ \text{Å}$ will be:

  • A
    $5.54 \times 10^{-7} \ \text{rad}$
  • B
    $2.54 \times 10^{-4} \ \text{rad}$
  • C
    $6.54 \times 10^{-7} \ \text{rad}$
  • D
    None of these

Explore More

Similar Questions

Resolving power of a telescope increases when

The correct relation between limit of resolution and resolving power is

The diameter of the objective lens of a telescope is $250\, cm$. For light of wavelength $600\, nm$ coming from a distant object, the limit of resolution of the telescope is close to:

The average distance between the earth and moon is $38.6 \times 10^4 \ km$. The minimum separation between the two points on the surface of the moon that can be resolved by a telescope whose objective lens has a diameter of $5 \ m$ with $\lambda = 6000 \ \mathring{A}$ is ...... $m$.

Difficult
View Solution

Wavelengths of light used in an optical instrument are $\lambda_1 = 4000 \; \mathring{A}$ and $\lambda_2 = 5000 \; \mathring{A}$. The ratio of their respective resolving powers (corresponding to $\lambda_1$ and $\lambda_2$) is:

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo