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A parallel beam of monochromatic light of wavelength 500 nm is incident normally on a perfectly absorbing surface. The power through any cross–section of the beam is 10 W. Find (a) the number of photons absorbed per second by the surface and (b) the force exerted by the light beam on the surface.


3 years ago

Arun
24737 Points
							(a) Energy of each photon isE = hc / $\lambda$  = 1242 eV – nm/ 500 nm = 2.48 eVIn one second 10 J energy passes through any cross section of the beam,Thus the no. of photons crossing a cross section is =  n = 10 J / 2.48 eVn = 2.52 * 1019 (b)total momentum of all the photons =  p = nh/ $\lambda$ = nh $\nu$/ c = 10 J / 3*108 = 3.33 * 10 –8 N-sec hence total force F = dP/dt = 3.33 *10 –8 Newton

3 years ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions