Question 282: Waves Optics

Question

In Young's double slit experiment, the intensity of light at a point on the screen where the path difference is $\lambda$ is x units, $\lambda$ being the wavelength of light used. The intensity at a point where the path difference is $\frac{\lambda}{4}$ will be $\left(\cos 2 \pi=1, \cos \frac{\pi}{2}=0\right)$

Accepted Answer

When path difference $=\lambda$
Phase difference
$$\begin{aligned}
& \Delta \phi=\frac{2 \pi}{\lambda} 	imes \lambda=2 \pi \
& I=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}ight) \
& x=4 \mathrm{I}_0 \cos ^2\left(\frac{2 \pi}{2}ight)=4 \mathrm{I}_0 \
	herefore \quad & x=4 \mathrm{I}_0
\end{aligned}$$
When path difference $=\frac{\lambda}{4}$
Phase difference
$$\begin{aligned}
& \Delta \delta=\frac{2 \pi}{\lambda} 	imes \frac{\lambda}{4}=\frac{\pi}{2} \
& I=4 I_0 \cos ^2\left(\frac{\frac{\pi}{2}}{2}ight)=2 I_0
\end{aligned}$$
Intensity, $\mathrm{x}^{\prime}=\frac{\mathrm{x}}{2}$

Answer

$\frac{x}{4}$

Answer

$\frac{x}{2}$

Answer

$x$

Answer

zero

Waves Optics

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