Question

In a Young's double slit experiment, the intensity at a point where the path difference is \( \frac{\lambda}{6} \) (\( \lambda \) being the wavelength of light used) is \( I \). If \( I_{0} \) denotes the maximum intensity, \( I/I_{0} \) is equal to:

• A. 1/2
• B. \(\sqrt{3}/2\)
• C. 3/4
• D. 1/4
• E.

Hint:

Always remember: Path difference of \(\lambda\) corresponds to a phase difference of \(2\pi\). This makes it easy to visualize: \(\lambda/2 \to \pi\) (minima), \(\lambda/4 \to \pi/2\), etc.

Answer 1

The Correct Option is C

To solve this problem, we first need to understand the concept of intensity in Young's double slit experiment. The intensity at a point on the interference pattern depends on the path difference between the two waves arriving at that point. The general formula for the intensity in an interference pattern is given by:

\(I = I_0 \cos^2\left(\frac{\phi}{2}\right)\)

where \(\phi\) is the phase difference, and \(I_0\) is the maximum intensity. The phase difference \(\phi\) is related to the path difference \(\Delta x\) by the formula:

\(\phi = \frac{2\pi}{\lambda} \Delta x\)

Given, \(\Delta x = \frac{\lambda}{6}\). Substituting this in, we get:

\(\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3}\)

Now, substituting \(\phi = \frac{\pi}{3}\) into the intensity formula:

\(I = I_0 \cos^2\left(\frac{\pi}{6}\right)\)

We know that \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). Therefore:

\(I = I_0 \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \times \frac{3}{4}\)

Thus, the ratio of the intensity at the point to the maximum intensity is:

\(\frac{I}{I_0} = \frac{3}{4}\)

Hence, the correct answer is option 3/4.