Question 331: Waves Optics

Question

In Young's double slit experiment the intensities at two points, for the path difference $$\frac{\lambda}{4}$$ and $$\frac{\lambda}{3}$$ ($$\lambda=$$ wavelength of light used) are $$I_1$$ and $$I_2$$ respectively. If $$\mathrm{I}_0$$ denotes the intensity produced by each one of the individual slits then $$\frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}$$ is equal to $$\left(\cos 60^{\circ}=0.5, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$$

Accepted Answer

Phase difference, $$\phi=\frac{2 \pi}{\lambda} \Delta l$$
For first point, $$\phi_1=\frac{2 \pi}{\lambda}\left(\frac{\lambda}{4}ight)$$
$$\begin{array}{l}
	herefore \quad \phi_1 =\frac{\pi^{\mathrm{c}}}{2} \
\mathrm{I}_1 =2 \mathrm{I}_0\left(1+\cos \phi_1ight) \
	herefore \quad \mathrm{I}_1 =2 \mathrm{I}_0 \quad \ldots\left[\because \cos \phi_1=\cos (\pi / 2)=0ight]
\end{array}$$
Similarly, for second point, $$\phi_2=\frac{2 \pi^{\mathrm{c}}}{3}$$
$$\begin{aligned}
& 	herefore \quad \mathrm{I}_2=2 \mathrm{I}_0\left(1+\cos \phi_2ight) \
& 	herefore \quad \mathrm{I}_2=2 \mathrm{I}_0\left(1-\frac{1}{2}ight) \quad \ldots\left[\because \cos \left(\frac{2 \pi}{3}ight)=\frac{-1}{2}ight] \
& 	herefore \quad \mathrm{I}_2=\mathrm{I}_0 \
\end{aligned}$$
Hence, $$\frac{\mathrm{I}_1+\mathrm{I}_2}{\mathrm{I}_0}=\frac{2 \mathrm{I}_0+\mathrm{I}_0}{\mathrm{I}_0}=3$$

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