In Young's double slit experiment using monochromatic light of wavelength ' $\lambda$ ', the maximum intensity of light at a point on the screen is ' $K$ ' units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$, is $\left(\cos 60^{\circ}=\sin 30^{\circ}=0.5, \sin…

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 ' $K$ ' units. The intensity of light at a point where the path difference is $\frac{\lambda}{6}$, is $\left(\cos 60^{\circ}=\sin 30^{\circ}=0.5, \sin 60^{\circ}=\cos 30^{\circ}=\sqrt{3} / 2\right)$.

Aicharya · Accepted Answer

The intensity is given by $\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}...(i)$
Maximum intensity $\mathrm{K}=4 \mathrm{I}_0$ when $\phi=0$ when path difference is $\frac{\lambda}{6}$, $\begin{aligned} & \phi=\frac{2 \pi}{\lambda} \times \text { path difference }=\frac{\pi}{3} ...(ii)\\ \therefore \quad & I=K \cos ^2\left(\frac{\pi}{6}\right)=K\left(\frac{3}{4}\right)=\frac{3 K}{4} \end{aligned}$ ...[From(i) and (ii)]

In Young's double slit experiment using monochromatic light of wavelength ' $\lambda$ ', the maximum…

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