in youngs double - slit experiment the fringe pattern is observed on a screen placed at a distance d. the…

in youngs double - slit experiment the fringe pattern is observed on a screen placed at a distance d. the slits are illuminated by light of wavelength $lambda$. the distance from the central point where the intensity falls to half the maximum is 1) $\frac{lambda d}{3d}$ 2) $\frac{lambda d}{2d}$ 3) $\frac{lambda d}{d}$ 4) $\frac{lambda d}{4d}$

in youngs double - slit experiment the fringe pattern is observed on a screen placed at a distance d. the slits are illuminated by light of wavelength $lambda$. the distance from the central point where the intensity falls to half the maximum is 1) $\frac{lambda d}{3d}$ 2) $\frac{lambda d}{2d}$ 3) $\frac{lambda d}{d}$ 4) $\frac{lambda d}{4d}$

Answer

Explanation:

Step1: Recall intensity formula for double - slit

The intensity in Young's double - slit experiment is given by $I = I_0\cos^{2}\left(\frac{\pi d y}{\lambda D}\right)$, where $I_0$ is the maximum intensity, $d$ is the slit separation, $y$ is the distance from the central maximum, $\lambda$ is the wavelength of light and $D$ is the distance between the slits and the screen.

Step2: Set intensity condition

We want $I=\frac{I_0}{2}$. So, $\frac{I_0}{2}=I_0\cos^{2}\left(\frac{\pi d y}{\lambda D}\right)$. Then, $\cos^{2}\left(\frac{\pi d y}{\lambda D}\right)=\frac{1}{2}$, which implies $\cos\left(\frac{\pi d y}{\lambda D}\right)=\frac{1}{\sqrt{2}}$.

Step3: Solve for $y$

We know that when $\cos\theta=\frac{1}{\sqrt{2}}$, $\theta = \pm\frac{\pi}{4}$. So, $\frac{\pi d y}{\lambda D}=\frac{\pi}{4}$. Solving for $y$, we get $y = \frac{\lambda D}{4d}$.

Answer:

  1. $\frac{\lambda D}{4d}$