The 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:

  • A
    $16 : 25$
  • B
    $9 : 1$
  • C
    $4 : 5$
  • D
    $5 : 4$

Explore More

Similar Questions

$A$ plane passing through the points $(0, -1, 0)$ and $(0, 0, 1)$ and making an angle $\frac{\pi}{4}$ with the plane $y - z + 5 = 0$,also passes through the point:

The cause of the potential barrier in a $P-N$ junction diode is

An $\alpha$-particle of energy $E$ is liberated during the decay of a nucleus of mass number $236$. The total energy released in this process is

If $g$ is the acceleration due to gravity on the earth's surface,the gain in the potential energy of an object of mass $m$ raised from the surface of earth to a height equal to the radius of the earth $R$,is

Assertion : Both $Be$ and $Al$ can form complexes such as $BeF_4^{2-}$ and $AlF_6^{3-}$ respectively,$BeF_6^{3-}$ is not formed.
Reason : In case of $Be$,no vacant $d-$orbitals are present in its outermost shell.

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