Question

In Young's double'sIit experiment, Let $\beta$ be the fringe width and $I_0$ be the intensity at the central bright fringe. At a distance 'x' from the central bright fringe, the intensity will be.

• A. $I_0\,\cos\left(\frac{x}{\beta}\right)$
• B. $I_0\,\cos^2\left(\frac{x}{\beta}\right)$
• C. $I_0\,\cos^2\left(\frac{\pi\,x}{\beta}\right)$
• D. $\frac{I_0}{4}\,\cos^2\left(\frac{\pi\,x}{\beta}\right)K-3$

Answer 1

Answer: (C)

$I_0\,\cos^2\left(\frac{\pi\,x}{\beta}\right)$

Answer 2

$\Delta=x \frac{d}{D}$ $\therefore$ phase difference $=\phi=\frac{2 \pi}{\lambda} \Delta$ Let $a =$ amplitude at the screen due to each slit. $\therefore I _{0}= k (2 a )^{2}=4 ka ^{2}$, where $k$ is a constant. For phase difference $\phi$, amplitude $= A =2 a \cos(\phi / 2)$. Intensity, I $= k A ^{2}= k \left(4 a ^{2}\right) \cos (\phi / 2)= I _{0} \cos ^{2}\left(\frac{\pi}{\lambda} \Delta\right)$ $I _{0} \cos ^{2}\left(\frac{\pi}{\lambda} \cdot \frac{ xd }{ D }\right)= I _{0} \cos ^{2}\left(\frac{\pi x }{\beta}\right)$