Question

Two coherent monochromatic light beams of intensities I and 9I are superimposed. The maximum and the minimum intensities of the resultant beam are:
(A) $10I$ \[and\] zero
(B) \[10I{\text{ }}and{\text{ }}8I\]
(C) \[10I{\text{ }}and{\text{ }}4I\]
(D) \[16I{\text{ }}and{\text{ }}41\]

Answer 1

HINT Intensity of the light beam depends on the amplitude of light. When two coherent light beams merge it can have different types of superpositions. The maximum intensity occurs when they are in phase and minimum superposition occurs when they are completely out of phase.

Complete step by step solution
Superposition of two light beams occurs when two light beams converge at a position. The resultant light beam has an amplitude which depends on the amplitude of the constituent waves and the phase of these waves. The intensity of light waves depends on the amplitude of the wave in this manner.
\[\mathop {I \propto A}\nolimits^2 \]
When two light waves converge, we cannot add the intensity of the light waves. we need to find the amplitude of the waves and then calculate the resultant amplitude of the new superimposed wave.
Let’s assume the amplitude of the wave that has intensity I, is A.
As intensity depends on the square of the amplitude, we can write
$\mathop {I = KA}\nolimits^2 $
Hence, for the other light wave amplitude can be given by,
$\mathop {9I = K\left( {3A} \right)}\nolimits^2 $
If the two light waves converge in phase, the amplitude can be added.
Hence, the resultant amplitude will be $ = \left( {A + 3A} \right) = 4A$
When the two light waves converge out of phase, the amplitude will be$\left( {3A - A} \right) = 2A$
Hence, the intensity of the resultant light will be,
\[\mathop I\nolimits^{''} = \mathop {K(2A)}\nolimits^2 = \mathop {4KA}\nolimits^2 = \mathop {4I}\nolimits^{} \]
So, the maximum intensity will be $16I$and the minimum value will be $4I$

The correct answer is option D.

NOTE: there is another formula to calculate the maximum and minimum intensity of the light wave.
The maximum intensity can be given by,
$I = \mathop {\left( {\sqrt {\mathop I\nolimits_1 } + \sqrt {\mathop I\nolimits_2 } } \right)}\nolimits^2 $
The minimum intensity can be given by,
$I = \mathop {\left( {\sqrt {\mathop I\nolimits_1 } - \sqrt {\mathop I\nolimits_2 } } \right)}\nolimits^2 $
The intensity of the superposition can be given by$I = \mathop {\left( {\mathop I\nolimits_1 + \mathop I\nolimits_2 + 2\sqrt {\mathop I\nolimits_1 \mathop I\nolimits_2 } \cos \theta } \right)}\nolimits^2 $
$\mathop I\nolimits_1 $= intensity of one light beam
$\mathop I\nolimits_2 $= intensity of second light beam
$\theta $= phase difference between two light beams.