Question 248: Waves Optics

Question

In Young's double slit experiment, the light of wavelength ' $\lambda$ ' is used. The intensity at a point on the screen is ' T ' where the path difference is $\lambda \frac{-}{4}$. If ' $\mathrm{I}_0$ ' denotes the maximum intensity then the ratio of ' $\mathrm{I}_0$ ' to ' I ' is $\left(\cos 45^{\circ}=1 / \sqrt{2}\right)$

Accepted Answer

$$ \begin{aligned} &\frac{I}{I_0}=I_0 \cos ^2\left(\frac{\phi}{2}ight)\ &	ext { Phase difference }=\frac{2 \pi}{\lambda} 	imes 	ext { path difference }\ &\begin{aligned} & Q=\frac{2 \pi}{\lambda} \Delta x=\frac{2 \pi}{\lambda} 	imes \frac{\lambda}{4}=\frac{\pi}{2} \ & I=I_0 \cos ^2\left(\frac{\pi}{4}ight)=\frac{I_0}{2} \end{aligned} \end{aligned} $$
$$ 	herefore  \frac{I_0}{I}=\frac{I_0}{I_0 / 2}=2 $$

Answer

$2: 1$

Answer

$4: 1$

Answer

$8: 1$

Answer

$12: 1$

Waves Optics

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