According to Abbe,in the formula for the resolving power of a microscope,the numerical aperture is represented by:
- A$\frac{2 \mu \sin \alpha}{\lambda}$
- B$\frac{2 \sin \alpha}{\mu \lambda}$
- C$\mu \sin \alpha$
- D$\frac{\lambda}{2 \mu \sin \alpha}$
Explore More
Similar Questions
Assuming the human pupil to have a radius of $0.25 \ cm$ and a comfortable viewing distance of $25 \ cm$,the minimum separation between two objects that the human eye can resolve at $500 \ nm$ wavelength is..... $\mu m$.
To increase both the resolving power and magnifying power of a telescope:
Easy
View SolutionIf $f_{0} = 5 \, cm$,$\lambda = 6000 \, \mathring{A}$,and $a = 1 \, cm$ for a microscope,what will be its resolving power?
MediumAIIMS 2019
View SolutionExplain the resolving power of optical instruments and explain the resolving power of telescopes.
Medium
View SolutionOur eyes are sensitive to the wavelength range of .......
Easy
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo