Two light beams of equal energies consist of $n_{1}$ and $n_{2}$ photons. If their frequencies are $v_{1}$ and $v_{2}$ respectively, then the correct relation is

\frac{n_{1}^{2}}{n_{2}^{2}} = \frac{v_{1}}{v_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n_{1}^{2}}{n_{2}^{2}} = \frac{v_{2}}{v_{1}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n_{1}}{n_{2}} = \frac{v_{1}}{v_{2}}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $\frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}}$
$n \times \frac{1}{v}$

$\frac{n_{1}}{n_{2}} = \frac{v_{2}}{v_{1}}$

Suggest Corrections

Similar questions

How is refractive index of a medium 2 given with respect to medium 1 related to speed of light?
where, $n_{1}$ and $n_{2}$ are the respective refractive indicies of medium 1 and medium 2. And, $c_{1}$ and $c_{2}$ are speed of light in medium 1 and medium 2 respectively.

What is the difference between n21 and n12 ? please give any difference other than v1/v2=n21 and v2/v1=n12

Photoelectric emission is observed from a metallic surface for frequencies $v_{1}$ and $v_{2}$ and of the incident light rays $(v_{1} > v_{2})$ . If the maximum values of kinetic energy of the photoelectrons emitted in the two cases are in the ratio of 1 : k, then the threshold frequency of the metallic surface is