Two monochomatic beams of A& 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 about their frequencies?

Two monochomatic beams of A& B of equal intensity I, hit a screen. The

1.	Two monochomatic beams of A& 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 about their frequencies?
Answer» Solution :GIVEN `I_1=I_2=I` and `n1=2n_2 ` Where `n_1` & `n_2` are the numbers of photons falling PER second of beam A & B respectively. Energy of incident photon of beam A is hV. Energy of incident photon of beam is `hV_2` As `I=n_1 hV1=n_2hV_2`or `V_1/V_2=n_2/n_1=n_2/(2n_2)=1/2 `i.e, `V_2=2V_1`

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Two monochomatic beams of A& 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 about their frequencies?

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