Question 360: Waves Optics

Question

In Young's double slit experiment, the intensity at a point where path difference is $$\frac{\lambda}{6}$$ ($$\lambda$$ being the wavelength of light used) is $$I^{\prime}$$. If '$$I_0$$' denotes the maximum intensity, then $$\frac{I}{I_0}$$ is equal to $$\left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{\lambda}\right)$$

Accepted Answer

If $$\mathrm{I}_0$$ is the intensity of each wave, then the resultant intensity at a point is given by
$$\mathrm{I}=4 \mathrm{I}_0 \cos ^2 \frac{\phi}{2}$$
At point P, path difference is zero, hence phase difference $$\phi=0$$
$$	herefore \mathrm{I}_{\mathrm{p}}=4 \mathrm{I}_0 \cos ^2 0=4 \mathrm{I}_0$$
At point Q, path difference is $$\frac{\lambda}{6}$$, hence phase differences is $$\frac{2 \pi}{6}$$ or $$\frac{\pi}{3}$$.
$$\begin{aligned}
& 	herefore \mathrm{I}_Q=4 \mathrm{I}_0 \cos ^2 \frac{\pi}{6} \quad=4 \mathrm{I}_0 \frac{3}{4} \
& 	herefore \frac{\mathrm{I}_{\mathrm{p}}}{\mathrm{I}_Q}=\frac{4}{3}
\end{aligned}$$

Answer

$$\frac{\sqrt{3}}{2}$$

Answer

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

Answer

$$\frac{1}{\sqrt{2}}$$

Answer

$$\frac{1}{2}$$

Waves Optics

In Young's double slit experiment, the intensity at a point where path difference is $\frac{\lambda}{6} (\lambda being the wavelength of light used) is I^{\prime}. If 'I_0' denotes the maximum intensity, then \frac{I}{I_0} is equal to \left(\cos 0^{\circ}=1, \cos 60^{\circ}=\frac{1}{\lambda}\right)

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