A beam of light with intensity 10 - 3   N   m - 2 and cross sectional area 20   cm 2 is…

A beam of light with intensity 10-3 N m-2 and cross sectional area 20 cm2 is incident on a fully reflective surface at angle 45°. Then the force exerted by the beam on the surface is
  1. 2.3×10-15 N
  2. 1.33×10-14 N
  3. 6.67×10-15 N
  4. 9.4×10-15 N

Solution

Force exerted by the beam on the surface is given by F=∆p∆t=2IAcosθc

Given here: I=10-3 N m-2, A=20 cm2 and θ=45°.

Putting the values, we get

F=2×10-3×20×10-4×cos45°3×108=9.4×10-15 N

Asked in: AP EAMCET 2022 (04 Jul Shift 1)

Practice more Dual Nature of Matter and Radiation questions on Aicharya