Question 241: Waves Optics

Question

In Young's double slit experiment, at two points $P$ and $Q$ on screen, waves from slits $S_1$ and $S_2$ have a path difference of 0 and $\frac{\lambda}{4}$ respectively. The ratio of intensities at point $P$ to that at $Q$ will be $\left(\cos 0^{\circ}=1, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)$

Accepted Answer

If $\mathrm{I}_0$ is the intensity of each wave and $\phi$ is the phase difference, then the resultant intensity is given by
$$ I=4 I_0 \cos ^2 \frac{\phi}{2} $$
When path difference $=0$, phase difference $\phi=0$, $\mathrm{I}_1=4 \mathrm{I}_0 \cos ^2 0=4 \mathrm{I}_0$
When path difference $=\frac{\lambda}{4}$, phase difference
$$ \begin{array}{ll}  & \phi=\frac{\pi}{2} \ 	herefore & I_2=4 I_0 \cos ^2 \frac{\pi}{4}=2 I_0 \ 	herefore & \frac{I_1}{I_2}=\frac{4 I_0}{2 I_0}=\frac{2}{1} \end{array} $$

Answer

$3: 2$

Answer

$2: 1$

Answer

$\sqrt{2}: 1$

Answer

$4: 1$

Waves Optics

In Young's double slit experiment, at two points P and Q on screen, waves from slits S_1 and S_2 have a path difference of 0 and \frac{\lambda}{4} respectively. The ratio of intensities at point P to that at Q will be \left(\cos 0^{\circ}=1, \cos 45^{\circ}=\frac{1}{\sqrt{2}}\right)

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