Question

In Young’s double slit experiment, the ratio of intensities of bright and dark bands is $16$ which means
(A) The ratio of their amplitudes is $5$.
(B) Intensities of individual sources are $25$ and $9$ units respectively
(C) The ratio of their amplitudes is $4$
(D) Intensities of Individual sources are $4$ and $3$ units respectively

Answer 1

To solve this problem, Young’s double slit experiment process should be recalled and the mathematical calculations of ratio for bright and dark fringes are used to find the individual intensities.

Complete step by step answer:
According to the question let the intensities of individual sources be ${I_1}$ and ${I_2}$.
Let us assume the net intensity is ${I_{net}}$.
$\therefore {I_{net}} = {I_1} + {I_2} + 2\sqrt {{I_1}} \sqrt {{I_2}} \cos \phi$
The above equation for bright bands can be written as: ${I_{\max }} = {I_1} + {I_2} + 2\sqrt {{I_1}} \sqrt {{I_2}} {\text{ (because }}\cos \phi = 1)$
$\Rightarrow {I_{\max }} = {(\sqrt {{I_1}} + \sqrt {{I_2}} )^2}$
Similarly for dark bands it can be written as:
${I_{\min }} = {I_1} + {I_2} - 2\sqrt {{I_1}} \sqrt {{I_2}} {\text{ (because }}\cos \phi = - 1)$
$\Rightarrow {I_{\min }} = {(\sqrt {{I_1}} - \sqrt {{I_2}} )^2}$
Now, in the question it is given that the ratio of intensities of bright and dark bands is $16$ $\therefore \dfrac{{{I_{\max }}}}{{{I_{\min }}}} = \dfrac{{{{(\sqrt {{I_1}} + \sqrt {{I_2}} )}^2}}}{{{{(\sqrt {{I_1}} - \sqrt {{I_2}} )}^2}}} = 16$
Now, we will simplify the above equation
$\sqrt {{I_1}} + \sqrt {{I_2}} = 4\sqrt {{I_1}} - 4\sqrt {{I_2}}$
$\Rightarrow 5\sqrt {{I_2}} = 3\sqrt {{I_1}}$
After simplification we will get,
$\dfrac{{\sqrt {{I_1}} }}{{\sqrt {{I_2}} }} = \dfrac{5}{3}$
Now we know that Intensity of the E-M wave depends upon the Amplitude which in mathematical form is told as intensity is directly proportional to the square of the amplitude.
So we can write
$\dfrac{{{A_1}}}{{{A_2}}} = \dfrac{5}{3}$
Now as intensity is directly proportional to the square of the amplitude so,
$\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{25}}{9}$
Considering ${I_2} = 9units$ as units we get ${I_1} = 25units$ and ${I_2} = 9units$

So, option B is correct answer.

Note: The Young’s double slit experiment uses Monochromatic, Coherent sources. Where Monochromatic sources mean that they have frequencies, but those frequencies are constant with time and coherent sources means both the frequencies are same and also the phase difference is not varying with respect to time (provided if there is any phase difference). The net intensity calculation for maximum and minimum intensity should be done carefully. Also take the value of $\cos \phi$ wisely.