Question

Two monochromatic beams $A$ and $B$ of equal intensity $I$, hit a screen. The number of photons hitting the screen by beam $A$ is twice that by beam $B$. Then what inference can you make about their frequencies?

Answer 1

$\text{Formula used: Intensity}=\frac{\text{Power}}{\text{Area}}$

Given,

The number of photons hitting the screen by beam $A$ is twice that by beam $B$,

$n_A=2n_B$

Intensity of $A$ is equal to Intensity of $B$.

Suppose that $n_A$ is the number of photons falling per second for beam $A$ and $n_B$ is the number of photons falling per second for beam $B$.

Energy of falling photon of beam

$A=h{v}_A$

Energy of falling photon of beam

$B=h{v}_B$

Now,

$I=n_A{v}_A=n_B{v}_B$

$\Rightarrow \frac{v_A}{v_B}=\frac{n_B}{n_A}=\frac{n_B}{2n_B}=\frac{1}{2}$

$\Rightarrow v_B=2v_A$

$\text{Final Answer}:v_B=2v_A$