Question

In a Young’s double-slit experiment, the path difference, at a certain point on the screen between two interfering waves is ${(1/8)}^{th}$ of wavelength. The ratio of the intensity at this point to that at the centre of a bright fringe is close to

• A. $0.74$
• B. $0.80$
• C. $0.85$
• D. $0.94$

Answer 1

The correct option is C

$0.85$

Step 1: Given

Path difference, $∆x=\frac{\lambda }{8}$

Where $\lambda$ is the wavelength.

Step 2: Determine the phase difference

The phase difference is given as:

$$\begin{gathered} \Delta \varphi =\frac{2\mathrm{\pi }}{\lambda }\Delta x \\ =\frac{2\mathrm{\pi }}{\lambda }\times \frac{\lambda }{8} \\ =\frac{\mathrm{\pi }}{4} \end{gathered}$$

Step 3: Determine the intensity ratio

Let $I_{max}$ be the intensity in the central fringe. In Young’s double-slit experiment, the intensity is given by

$$\begin{gathered} I=I_{max}{\cos }^2\left(\frac{\Delta \varphi }{2}\right) \\ =I_{max}{\cos }^2\left(\frac{\mathrm{\pi }/4}{2}\right) \\ \frac{I}{I_{max}}={\cos }^2\left(\frac{\mathrm{\pi }}{8}\right) \\ \frac{I}{I_{max}}=0.{9238}^2 \\ \frac{I}{I_{max}}=0.853 \end{gathered}$$

Therefore, option C is the correct option.