Question 52: Wave Optics

Question

In interference experiment the path difference between two interfering waves at a point $A$ on the screen is $\lambda / 3$, where $\lambda$ is the wavelength of these waves, and at another point $B$ the path difference is $\lambda / 6$. The ratio of intensities at points $A$ and $B$ is $\_\_\_\_$ .

Accepted Answer

The relation between phase difference $(\phi)$ and path difference $(\Delta x)$ is:
$$ \phi=\frac{2 \pi}{\lambda} \Delta x $$
For two waves of equal intensity $\mathrm{I}_0$, the resultant intensity I at a point is:
$$ \mathrm{I}=4 \mathrm{I}_0 \cos ^2\left(\frac{\phi}{2}ight) $$
At point A the path difference is $\Delta \mathrm{x}_{\mathrm{A}}=\frac{\lambda}{3}$.
$$ \phi_{\mathrm{A}}=\frac{2 \pi}{\lambda} 	imes \frac{\lambda}{3}=\frac{2 \pi}{3} $$
So, the resultant intensity at A is,
$$ I_A=4 I_0 \cos ^2\left(\frac{2 \pi / 3}{2}ight)=4 I_0 \cos ^2\left(\frac{\pi}{3}ight)=4 I_0\left(\frac{1}{2}ight)^2=I_0 $$
At point B the path difference is $\Delta \mathrm{x}_{\mathrm{B}}=\frac{\lambda}{6}$.
$$ \phi_{\mathrm{B}}=\frac{2 \pi}{\lambda} 	imes \frac{\lambda}{6}=\frac{\pi}{3} $$
So, the resultant intensity at B is,
$$ I_B=4 I_0 \cos ^2\left(\frac{\pi / 3}{2}ight)=4 I_0 \cos ^2\left(\frac{\pi}{6}ight)=4 I_0\left(\frac{\sqrt{3}}{2}ight)^2=3 I_0 $$
So, the ratio of intensities at A and B is;
$$ \frac{\mathrm{I}_{\mathrm{A}}}{\mathrm{I}_{\mathrm{B}}}=\frac{\mathrm{I}_0}{3 \mathrm{I}_0}=\frac{1}{3} $$
Thus, the correct option is (C).

Answer

3

Answer

4

Answer

1/3

Answer

1/4

Wave Optics

In interference experiment the path difference between two interfering waves at a point A on the screen is \lambda / 3, where \lambda is the wavelength of these waves, and at another point B the path difference is \lambda / 6. The ratio of intensities at points A and B is \_\_\_\_ .

Choose the correct answer: