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NEETPhysicsDual Nature of Matter

An isotropic point source emits light of wavelength at a constant power P . A small detector of area A is placed at a distance r from the source, oriented such that the light falls normally on it. Which of the following expressions gives the number of photons striking the detector per second? (Assume h is Planck's constant and c is the speed of light in vacuum)

Options

  1. AP A 4 r^2 h c
  2. BP A r^2 h c
  3. CP A 4 r^2 h c
  4. DP 4 r^2 h c A

Correct answer

A. P A 4 r^2 h c

Step-by-step solution

The energy of a single photon is given by E = hc . The total number of photons emitted by the source per second is N = P E = P hc . Since the source is isotropic, the emitted photons spread uniformly over a spherical surface area of 4 r^2 at a distance r . The fraction of these photons intercepted by the detector of area A is A 4 r^2 . Thus, the number of photons striking the detector per second is N' = N ( A 4 r^2 ) = P A 4 r^2 h c . Answer: P A 4 r^2 h c

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