Question

If the intensities of the two interfering beams in Young's double-slit experiment are ${I_1}\& {I_2}$ then the contrast between the maximum and minimum intensities is good when

(a) $\left| {{I_1} - {I_2}} \right|$ is large
(b) $\left| {{I_1} - {I_2}} \right|$ is small
(c) Either ${I_1}$ OR ${I_2}$ ${I_2}$
(d) ${I_1} = {I_2}$

Answer 1

As we know due to maxima we get bright spots and due to minima we get dark spots. Increasing the light intensity increases the kinetic energy.

Complete Step by Step Solution:
In Young's double slit experiment the minima and maxima are caused by the interference of the two coherent waves at a particular phase.
There are some conditions for interference and that's for good contrast
(A) The amplitude of two sources equal or nearly equal
(B) Slit must be narrow as much as possible
By interference of light we understand that it’s the phenomenon of redistribution of light energy in a medium on account of superposition of light waves from two coherent sources.
Constructive interference creates bright fringe ${I_{\max }} = K{(\sqrt {{I_1}} + \sqrt {{I_2}} )^2}$
Destructive interference creates dark fringe ${I_{\min }} = K{(\sqrt {{I_1}} - \sqrt {{I_2}} )^2}$
Due to maxima Bright spot are detected that is $0,\dfrac{{\lambda D}}{d},\dfrac{{2\lambda D}}{d}$
Due to minima Dark spot are detected that is $\dfrac{{\lambda D}}{{2d}},\dfrac{{3\lambda D}}{{2d}},\dfrac{{5\lambda D}}{{2d}}$
Contrast will be perfect when both beams intensities will be equal.
So can be concluded that intensity of bright fringes is maximum, while those of dark fringes is minimum which result is zero

Hence, option (d) is correct

Note
Fringe width which is the distance between two consecutive dark or bright fringes that is $\beta = {\beta _1} = {\beta _2}$
Which is $\dfrac{{\lambda D}}{d}$
And intensity of a wave is proportional to the square of its amplitude
These expressions require that θ be very small .Hence, yD needs to be very small.