Question

In young’s double slit experiment if the maximum intensity of light is $ {I_{\max }}, $ then the intensity at path difference $ \dfrac{\lambda }{2} $ will be
(A) $ {I_{\max }} $
(B) $ \dfrac{{{I_{\max }}}}{2} $
(C) $ \dfrac{{{I_{\max }}}}{4} $
(D) $ zero $

Answer 1

Hint:
In Young's modulus double slit experiment the maximum intensity of light is $ {I_{\max }} $ then the intensity of path difference will be varying according to the intensity. The intensity of light coming out of the slit is directly proportional to the width of the slit.

Complete step by step solution:
Intensity of the bright fringes decrease because the bright fringes will reduce in brightness due to the foremost important aspects of the sum of the amplitudes will not be the high as the number as before of the value of the intensity.
Example of destructive interference is the gravitational waves are the specimen of the destructive interference. Here, the light beams evaluated the destructive interference. The destructive interference of the moving electrons and the radio waves are also performing the task of destructive interference.
Destructive interference occurs only when the difference is an odd multiple of the value of $ \pi $ for the path difference of the value of the $ \dfrac{\lambda }{2} $ will be resulting as zero.
Therefore, the intensity of the path difference is zero.
Hence, from the above options option (D) is correct.

Note:
Destructive interference occurs because the maxima of two waves are 180 degrees out of phase. Here, a positive displacement of one wave is released definitely by a negative displacement of some of the waves. So, the amplitude of the resulting wave is zero.