Question

In YDSE, having slits of equal width, let $$\beta$$ be the fringe width and $I_0$ be the maximum intensity. At a distance $x$ from the central bright fringe, the intensity will be
A. $I_0\cos \left({\displaystyle \frac{x}{\beta }}\right)$
B. $I_0{\cos }^2{\displaystyle \frac{2\pi x}{\beta }}$
C. $I_0{\cos }^2{\displaystyle \frac{\pi x}{\beta }}$
D. ${\displaystyle \frac{I_0}{4}}{\cos }^2{\displaystyle \frac{\pi x}{\beta }}$

Answer 1

Hint: We can solve this problem with the Young’s double slit experiments. This experiment is the proof of the dual nature of light. In this experiment wave theory of light is explained. In this experiment, we use a screen with two slits and an optical screen at which we get interference patterns.

Complete answer:
In YDSE, we break a single monochromatic light source into two coherent sources by placing the screen having two slits in front of a single light source and the optical screen is “D” distance away from the silts. If we take a point $S$ on the optical screen $x$ distance from the centre then waves from the both silts have to travel different paths to reach the point due to which path difference $\mathrm{\Delta }x$ is created.
Diagram for YDSE is given below,

Constructive and destructive interference is formed on the screen so alternate dark and bright fringe appears on the screen. Fringe width is the distance between the two adjacent fringes.
Its formula is : $\beta ={\displaystyle \frac{\lambda D}{d}}$ where $\lambda$is the wavelength of light and $d$ is the distance between the slits.
Intensity of light at the any point on the screen can be calculate by this formula:
$I_S=I_1+I_2+2\sqrt{I_1{I}_2}\cos \mathrm{\Delta }\varphi$ where $I_1,I_2$ is the intensity from the slits source and $\mathrm{\Delta }\varphi$ is the phase difference.
For the given question, we know that $\mathrm{\Delta }\varphi ={\displaystyle \frac{2\pi }{\lambda }}\mathrm{\Delta }x$ and $\mathrm{\Delta }x={\displaystyle \frac{xd}{D}}$ then,
$\mathrm{\Delta }\varphi ={\displaystyle \frac{2\pi }{\lambda }}\times {\displaystyle \frac{xd}{D}}={\displaystyle \frac{2\pi x}{\beta }}$
From the formula of the intensity of interference, we get intensity at any point that can also be shown as: $I=I_0{\cos }^2\mathrm{\Delta }\varphi /2$ where $I_0$ is the maximum intensity.
Hence, $I_S=I_0{\cos }^2{\displaystyle \frac{\pi x}{\lambda }}.{\displaystyle \frac{d}{D}}=I_0{\cos }^2{\displaystyle \frac{\pi x}{\beta }}$
At a distance $x$ from the central bright fringe, the intensity will be $I_0{\cos }^2{\displaystyle \frac{\pi x}{\beta }}$.

So, the correct answer is “Option C”.

Note:
In YDSE, brightness and darkness of fringes depends upon the many things such as size of both slits should be same and not be larger, the light sources should be monochromatic. From these experiments, it is proved that the light has wave and particle nature.