Question

In a double-slit experiment, at a certain point on the screen, the path difference between the two interfering waves is $\frac{1}{8}$ of a wavelength. The ratio of the intensity of light at that point to that at the center of a bright fringe is?

• A. $0.568$
• B. $0.853$
• C. $0.672$
• D. $0.760$

Answer 1

The correct option is B

$0.853$

Step 1:Given data:

In a double-slit experiment, the path difference between the two interfering waves is given as $\frac{1}{8}$ of a wavelength.

Hence, $\mathrm{\Delta x}=\frac{\mathrm{\lambda }}{8}$

Where $\mathrm{\Delta x}=$Path difference

$\mathrm{\lambda }=$wavelength of the two interfering waves.

Step 2: Calculating Phase difference :

The phase difference between two waves can be given as:
Phase difference$\left(\mathrm{\Delta \varphi }\right)=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}.\mathrm{\Delta x}$

$$\begin{gathered} \Rightarrow \mathrm{\Delta \varphi }=\frac{2\mathrm{\pi }}{\mathrm{\lambda }}.(\frac{\mathrm{\lambda }}{8}) \\ \Rightarrow \mathrm{\Delta \varphi }=\frac{\mathrm{\pi }}{4} \end{gathered}$$​

Step 3: Calculating the intensity ratio:

According to the Young's double-slit experiment, we know that at any point, intensity can be given as:

$\mathrm{I}={\mathrm{I}}_{\max }{\cos }^2(\frac{\mathrm{\Delta \varphi }}{2})$

Where, I = Intensity of light at the given point

${\mathrm{I}}_{\max }=$The maximum intensity of light at the center of a bright fringe.

Hence, putting all the given values we can calculate the ratio of the intensities of light as:

$$\begin{gathered} \mathrm{I}={\mathrm{I}}_{\max }{\cos }^2(\frac{\mathrm{\pi }/4}{2}) \\ \frac{\mathrm{I}}{{\mathrm{I}}_{\max }}={\left[{\cos }^2\right(\frac{\mathrm{\pi }}{8}\left)\right]}^{} \\ \frac{\mathrm{I}}{{\mathrm{I}}_{\max }}={(0.9238)}^2 \\ \frac{\mathrm{I}}{{\mathrm{I}}_{\max }}=0.853 \end{gathered}$$

Thus, the ratio of the intensity of light at the given point to that at the center of a bright fringe is $0.853$. Hence, option B is correct.