Question 328: Waves Optics

Question

The ratio of intensities of two points on a screen in Young's double slit experiment when waves from the two slits have a path difference of $$\frac{\lambda}{4}$$ and $$\frac{\lambda}{6}$$ is
$$\left(\cos 90^{\circ}=0, \cos 60^{\circ}=0.5\right)$$

Accepted Answer

The intensity at the point due to interference is given as $$I=I_1+I_2+2 \sqrt{I_1 I_2} \cos \phi$$ ..... (i)
For path difference $$\frac{\lambda}{4}$$, the phase difference is
$$\phi_1=\frac{2 \pi}{\lambda} 	imes \frac{\lambda}{4}=\frac{\pi}{2}$$
For path difference $$\frac{\lambda}{6}$$, the phase difference is
$$\phi_2=\frac{2 \pi}{\lambda} 	imes \frac{\lambda}{6}=\frac{\pi}{3}$$
Assuming equal intensity of the interfering waves i.e., $$I_1=I_2=I_0$$
Equation (i) becomes,
$$\begin{aligned}
& \mathrm{I}=\mathrm{I}_0+\mathrm{I}_0+2 \mathrm{I}_0 \cos \phi \
& \mathrm{I}=2 \mathrm{I}_0(1+\cos \phi)
\end{aligned}$$
For the given path difference, $$I_1=2 I_0\left(1+\cos \frac{\pi}{2}ight)$$, and $$I_2=2 I_0\left(1+\cos \frac{\pi}{3}ight)$$
$$	herefore \quad \frac{I_1}{I_2}=\frac{1+\cos \frac{\pi}{2}}{1+\cos \frac{\pi}{3}}$$
$$\begin{aligned}
& \frac{\mathrm{I}_1}{\mathrm{I}_2}=\frac{1+0}{1+0.5} \
	herefore \quad \frac{\mathrm{I}_1}{\mathrm{I}_2} & =\frac{1}{1.5}=\frac{2}{3}
\end{aligned}$$

Answer

$$2: 1$$

Answer

$$2: 3$$

Answer

$$3: 4$$

Answer

$$3: 5$$

Waves Optics

The ratio of intensities of two points on a screen in Young's double slit experiment when waves from the two slits have a path difference of $\frac{\lambda}{4} and \frac{\lambda}{6} is \left(\cos 90^{\circ}=0, \cos 60^{\circ}=0.5\right)

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