Question

In the phenomenon of interference, energy is:
A. destroyed at bright fringes
B. created at dark fringes
C. conserved, but it is redistributed
D. same at all points

Answer 1

Hint: We know that for light waves, energy is considered in terms of light intensity. For solving this question, we will consider the total light intensities of two light sources before and after the interference. Here, we will also use the relation between light intensity and its amplitude.

Formulas used:
\[I = {A^2}\], where, \[I\]is the intensity of light and \[A\] is the amplitude of the light wave
\[{I_{\max }} = {\left( {{A_1} + {A_2}} \right)^2}\], where, \[{I_{\max }}\]is maximum intensity of the light due to interference, \[{A_1}\] is the amplitude of the first light source and \[{A_2}\] is the amplitude of the second light source
\[{I_{\min }} = {\left( {{A_1} - {A_2}} \right)^2}\], where, \[{I_{\min }}\]is minimum intensity of the light due to interference, \[{A_1}\] is the amplitude of the first light source and \[{A_2}\] is the amplitude of the second light source
\[{I_{avg}} = \dfrac{{{I_{\max }} + {I_{\min }}}}{2}\], where, \[{I_{avg}}\]is the average intensity of the light due to interference, \[{I_{\max }}\]is maximum intensity of the light due to interference and \[{I_{\min }}\]is minimum intensity of the light due to interference

Complete step by step answer:
Let us consider two sources of light having intensities of ${I_1}$and ${I_2}$, and they emit the waves of amplitude \[{A_1}\] and \[{A_2}\]respectively.Now, initially there is no interference between these two waves. At this time, the intensity of light at any point will be
$I = {I_1} + {I_2}$
But, we know that \[I = {A^2}\]
$I = {A_1}^2 + {A_2}^2$
Now, we will consider the time when these two light sources interfere. Here, there will be maximum intensity and minimum intensity of light.
The maximum intensity is given by \[{I_{\max }} = {\left( {{A_1} + {A_2}} \right)^2}\]and minimum intensity is given by \[{I_{\min }} = {\left( {{A_1} - {A_2}} \right)^2}\]
Therefore, average intensity of light because of interference is given by
\[{I_{avg}} = \dfrac{{{I_{\max }} + {I_{\min }}}}{2}\]
\[ \Rightarrow {I_{avg}} = \dfrac{{{{\left( {{A_1} + {A_2}} \right)}^2} + {{\left( {{A_1} - {A_2}} \right)}^2}}}{2} = {A_1}^2 + {A_2}^2\]
Thus, it can be clearly observed that total intensity before and after interference remains the same and is simply being redistributed. Thus, here no energy is created or destroyed which obeys the law of conservation of energy.

Hence, option C is the right answer.

Note:Here, we have concluded that in the phenomenon of interference energy is conserved but redistributed. In this redistribution at the time of interference, the light energy which disappears at the regions of destructive interference appears at the regions of constructive interference. Thus, the average intensity of light remains constant.