Question 366: Waves Optics

Question

In Young's double slit experiment using monochromatic light of wavelength '$$\lambda$$', the maximum intensity of light at a point on the screen is $$\mathrm{K}$$ units. The intensity of light at point where the path difference is $$\frac{\lambda}{3}$$ is
$$\left[\cos 60^{\circ}=\sin 30^{\circ}=\frac{1}{2}\right]$$

Accepted Answer

The intensity is given by
$$\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}$$
Maximum intensity $$\mathrm{K}=4 \mathrm{I}_0$$ when $$\phi=0$$ when path difference is $$\frac{\lambda}{3}, \phi=\frac{2 \pi}{3}$$
$$	herefore \quad \mathrm{I}=\mathrm{k} \cos ^2 \frac{2 \pi}{3}=\mathrm{k}\left(-\frac{1}{2}ight)^2=\frac{\mathrm{k}}{4}$$

Answer

$$\frac{K}{4}$$

Answer

$$\frac{3 K}{4}$$

Answer

K

Answer

$$\frac{\mathrm{K}}{2}$$

Waves Optics

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