Question

In young's double slit experiment $\beta$ is the fringe width and I${}_o$ is the intensity of the central bright fringe. What will be the intensity at a distance x from the central bright fringe?

• A. I${}_o$ cos $\left(\frac{x}{\beta }\right)$
• B. I${}_o$ cos${}^2\left(\frac{x}{\beta }\right)$
• C. I${}_o$ cos${}^2\left(\frac{\pi x}{\beta }\right)$
• D. $\left(\frac{I_o}{4}\right)$ cos${}^2\frac{\pi x}{\beta }$

Answer 1

Answer: (C)

I${}_o$ cos${}^2\left(\frac{\pi x}{\beta }\right)$

If $\delta$ is the phase difference between two waves with intensities $I_1$ and $I_2$, then the total intensity is given by

$I=I_1+I_2+2\sqrt{I_1{I}_2}cos\delta$

At the central fringe, the phase difference, $\delta =0$. Also given $I_o$ as the intensity of central bright fringe.

Hence we have,

$I=I_1+I_2+2\sqrt{I_1{I}_2}$

Also for the points between maxima and minima, $I_1=I_2=I_1$

Therefore $I_o=I_1+I_1+2I_1=4I_1$

$⟹I_1=\frac{I_o}{4}$

Now at distance $x$ from the center, the path difference = $\frac{\pi x}{D}$

Therefore phase difference, $\delta =\frac{2\pi }{\lambda }\frac{\pi x}{D}$

Hence intensity at that point is given by

$I^{\prime }=I_1+I_1+2I_{1}cos\frac{2\pi xd}{\lambda D}$

i.e, $I^{\prime }=\frac{I_o}{4}+\frac{I_o}{4}+\frac{2I_o}{4}cos\frac{2\pi xd}{\lambda D}$

$⟹I^{\prime }=\frac{I_o}{2}[1+cos\frac{2\pi xd}{\lambda D}]$

$⟹I^{\prime }=I_{o}co{s}^2\frac{\pi xd}{\lambda D}$ $[2co{s}^{2}x=1+cos2x]$

Also we have, fringe width $\beta =\frac{\lambda D}{d}$

Hence we have, $I^{\prime }=I_{o}co{s}^2\frac{\pi x}{\beta }$