Question

TWO COHERENT MONOCHROMATIC LIGHT BEAMS OF INTENSOTES I & 4I ARE SUPERIMPOSED .THE MAXIMUM & MINIMUM POSSIBLE INTENSITIES IN THE RESULTING BEAM ARE

Answer 1

The maximum intensity is given as

= ( √Intensity1 + √Intensity2)^2
= ( √I + √4I)^2 =( √I + 2√I)^2 = { 3√I }^2 = 9I

Minimum Intensity
= ( √Intensity1 - √Intensity2)^2
= ( √4I - √I)^2 = ( √I)^2 = I

Hope this helps you (

Answer 2

The maximum and minimum possible intensities in the resulting beam are 9I and I respectively.

• If $I_{1}$ and $I_{2}$ are the intensities of two coherent monochromatic light beams, then

                     $I_{max}=(\sqrt{I_{1}}+\sqrt{I_{2}})^{2}$

                    $I_{min}=(\sqrt{I_{1}}-\sqrt{I_{2}})^{2}$

• Here,

                     $I_{1}=I,I_{2}=4I$

          Maximum intensity:

                     $I_{max}=(\sqrt{I}+\sqrt{4I})^{2}$

                     $I_{max}=\sqrt{I}^{2} +\sqrt{4I}^{2}+2\sqrt{I}\sqrt{4I}$

                     $I_{max}=I+4I+4I$

                      $I_{max}=9I$

          Minimum intensity:

                      $I_{min}=(\sqrt{I}-\sqrt{4I})^{2}$

                      $I_{min}=\sqrt{I}^{2} +\sqrt{4I}^{2}-2\sqrt{I}\sqrt{4I}$

                      $I_{min}=I+4I-4I$

                      $I_{min}=I$