Two monochromatic beams A and B of equal inte | vedclass

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(B) Intensity of light $I$ is given by the formula $I = \frac{n h f}{A t}$,where $n$ is the number of photons,$h$ is Planck's constant,$f$ is the freq

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$f_A = f_B / 2$

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(B) Intensity of light $I$ is given by the formula $I = \frac{n h f}{A t}$,where $n$ is the number of photons,$h$ is Planck's constant,$f$ is the frequency,$A$ is the area,and $t$ is the time.Let $n' = \frac{n}{A t}$ be the number of photons incident on unit area per unit time.Thus,$I = n' h f$.Since the intensity $I$ is the same for both beams,we have $n'_A f_A = n'_B f_B$.Given that the number of photons from beam $A$ is twice that of beam $B$,we have $n'_A = 2 n'_B$.Substituting this into the equation: $(2 n'_B) f_A = n'_B f_B$.Simplifying,we get $2 f_A = f_B$,or $f_A = \frac{f_B}{2}$.Therefore,the frequency of beam $A$ is half the frequency of beam $B$.

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 $B$. What inference can you make about their frequencies?

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