Question

In a young’ double-slit experiment, the intensity at the central maximum is ${I_0}$. The intensity at a distance $\dfrac{\beta }{4}$ from the central maximum is ($\beta $ is fringe width).
A) ${I_0}$
B) $\dfrac{{{I_0}}}{2}$
C) $\dfrac{{{I_0}}}{{\sqrt 2 }}$
D) $\dfrac{{{I_0}}}{4}$

Answer 1

Hint: We know that intensity is directly proportional to the square of the amplitude. Amplitude is the max displacement of the particle from its mean position. fringe width depends on the wavelength of light used, the distance between the slits, and the distance between the screen and slit.

Complete step by step solution:
Given, the intensity at a distance $\dfrac{\beta }{4}$ from the central maximum then, we can write
$x = \dfrac{\beta }{4}$ …………………..(1)
Where $\beta $ is fringe width and x is the distance from the central maximum.
We know that path difference between two waves is given by,
$\Delta = \dfrac{{xd}}{D}$ ……………………….. (2)
Where $d$ is the distance between the slits, $D$ is the distance between the screen and slits.
Fringe width can be calculated using a formula,
$\beta = \dfrac{{\lambda D}}{d}$ ………………………. (3)
Now substitute equation (1) and (3) in (2) we get,
$\Delta = \left( {\dfrac{{\lambda D}}{{4d}}} \right)\dfrac{d}{D}$
After simplification,
$\Delta = \dfrac{\lambda }{4}$ ……………….. (4)
Now let us calculate phase difference between the waves,
$\phi = \dfrac{{2\pi }}{\lambda }\Delta $
Now substitute equation (4) in above equation we get,
$\phi = \dfrac{{2\pi }}{\lambda } \times \dfrac{\lambda }{4} = \dfrac{\pi }{2}$
From this value we can calculate intensity.
Intensity at any point is proportional to the square of the amplitude. That is, $I\propto {a^2}$
We have the amplitude of the resultant wave, $R = 2a\cos \dfrac{\phi }{2}$
Therefore, the resultant intensity of light at the given point due to both slits is given by,
$I\propto {R^2}$
Then, $I\propto 4{a^2}{\cos ^2}\dfrac{\phi }{2}$ ……………….. (5)
Now we will consider this above equation to solve this problem.
Let us consider, the intensity at the central maximum is ${I_0}$ and The intensity at a distance $\dfrac{\beta }{4}$ from the central maximum be $I$ then,
${I_0}\propto {R^2}_{\max }$
${I_0}\propto 4{a^2}$ ………………………. (6)
Now, Substitute $\phi = \dfrac{\pi }{2}$ in equation (5),
$I\propto 4{a^2}{\cos ^2}(45)$
$I\propto 4{a^2} \times \dfrac{1}{2}$
$I\propto 2{a^2}$ ………………………………… (7)
Now divide equation (7) by (6) we get,
$\dfrac{I}{{{I_0}}} = \dfrac{{2{a^2}}}{{4{a^2}}} = \dfrac{1}{2}$

Hence, the intensity $I = \dfrac{{{I_0}}}{2}$ is at a distance $\dfrac{\beta }{4}$ from the central maximum. Thus, correct option is (B).

Additional information:
Condition for constructive interference (bright fringe): when the crest of one superimposes with the crest of another or the trough of one wave superimposes with the trough of another then maximum amplitude is formed and there is a formation of bright fringe is called constructive interference.
Phase difference $\phi = 2n\pi $ where n=0,1,2,3…..
Path difference, $\Delta = n\lambda $
Destructive interference (dark fringe): When the crest of one wave superimposed with the trough of another wave then the resultant amplitude is minimum, and there is dark fringe is formed. This is called Destructive interference.
Phase difference $\phi = (2n - 1)\pi $ where $n=0,1,2,3…..$
Path difference, $\Delta = (2n - 1)\dfrac{\lambda }{2}$

Note:
Two sources must be coherent sources means two sources must emit waves of the same frequency or wavelength having zero phase difference.
From the young’ double-slit experiment, we can conclude that all bright and dark fringes are equally spaced.
Fringe width is defined as the distance between any two consecutive dark and bright fringe.