The diameter of the objective lens of a microscope makes an angle $\beta$ at the focus of the microscope. Further,the medium between the object and the lens is an oil of refractive index $n$. Then the resolving power of the microscope

  • A
    increases with decreasing value of $n$
  • B
    increases with decreasing value of $\beta$
  • C
    increases with increasing value of $n \sin \beta$
  • D
    increases with increasing value of $\frac{1}{n \sin \beta}$

Explore More

Similar Questions

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:

In telescopes, for a given wavelength, the objectives with large aperture are used for

In an optical instrument,the wavelengths of light used are $\lambda_1 = 4000 \ \mathring{A}$ and $\lambda_2 = 5000 \ \mathring{A}$. What is the ratio of their resolving powers?

The diameter of the objective of a telescope is $200 \text{ cm}$. What is the resolving power of the telescope? Take the wavelength of light $\lambda = 5000 \text{ \AA}$.

Explain the resolving power of a microscope.

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