Question 297: Waves Optics

Question

In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is $\lambda(\lambda=$ wavelength $)$ is $\beta$. The intensity at a point where the path difference is $\lambda / 3$, will be $\left.\cos \frac{\pi}{3}=1 / 2\right]$

Accepted Answer

In Young's double slit experiment, the intensity at a point on the screen is determined by the path difference between the light waves. When the path difference is equal to the wavelength $\lambda$, the intensity is given as $\beta$. We want to find the intensity at a point where the path difference is $\lambda/3$.
The intensity $I$ can be calculated using the formula:
$ I = 4 I_0 \cos^2 \frac{\phi}{2} $
Here, $I_0$ is the intensity of each slit and $\phi$ is the phase difference between the two waves. Maximum intensity $\beta = 4I_0$ occurs when the phase difference $\phi = 0$.
For a path difference of $\lambda/3$, the phase difference $\phi$ is:
$ \phi = \frac{2\pi}{3} $
Now, calculate the intensity using:
$ I = \beta \cos^2 \frac{2\pi}{3} $
Since
$ \cos \frac{2\pi}{3} = -\frac{1}{2} $
we have:
$ I = \beta \left(-\frac{1}{2}\right)^2 = \frac{\beta}{4} $
Therefore, the intensity at the point where the path difference is $\lambda/3$ is $\frac{\beta}{4}$.

Answer

$\beta$

Answer

$\beta / 2$

Answer

$\frac{\beta}{4}$

Answer

$\beta / 8$

Waves Optics

In the Young's double slit experiment, the intensity at a point on the screen, where the path difference is \lambda(\lambda= wavelength ) is \beta. The intensity at a point where the path difference is \lambda / 3, will be \left.\cos \frac{\pi}{3}=1 / 2\right]

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