Question

In YDSE setup maximum and minimum intensity obtained on screen are $I_0$ and zero respectively. If a thin film is kept in front of one the slit then intensity at center of screen becomes $\frac{I_0}{2}$. Find the minimum path difference caused by the film.

• A. $\frac{\lambda}{2}$
• B. $\frac{\lambda}{4}$
• C. $\frac{\lambda}{3}$
• D. $\frac{\lambda}{6}$

Answer 1

Answer: (B)

(B)

Answer 2

The intensity at any point in a Young's Double Slit Experiment (YDSE) is given by: $I = I_{max} \cos^2\left(\frac{\phi}{2}\right)$ where $I_{max}$ is the maximum intensity and $\phi$ is the phase difference between the waves from the two slits at that point.

Given that the maximum intensity is $I_0$, the formula becomes: $I = I_0 \cos^2\left(\frac{\phi}{2}\right)$

Initially, at the center of the screen (without the film), the path difference is zero, so the phase difference $\phi = 0$. This results in maximum intensity: $I_{center, initial} = I_0 \cos^2\left(\frac{0}{2}\right) = I_0 \cos^2(0) = I_0 \times 1 = I_0$.

When a thin film is kept in front of one of the slits, it introduces an additional optical path difference. Let this path difference caused by the film be $\Delta x_{film}$. At the center of the screen, the geometric path difference is zero, so the total path difference at the center of the screen becomes $\Delta x_{film}$. The corresponding phase difference introduced by the film at the center of the screen is: $\phi = \frac{2\pi}{\lambda} \Delta x_{film}$

The problem states that the intensity at the center of the screen becomes $\frac{I_0}{2}$ after placing the film. Using the intensity formula: $\frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right)$

Dividing both sides by $I_0$: $\frac{1}{2} = \cos^2\left(\frac{\phi}{2}\right)$

Taking the square root of both sides: $\cos\left(\frac{\phi}{2}\right) = \pm \frac{1}{\sqrt{2}}$

For the minimum path difference, we consider the smallest positive phase difference. This occurs when: $\frac{\phi}{2} = \frac{\pi}{4}$

Therefore, the phase difference $\phi$ is: $\phi = 2 \times \frac{\pi}{4} = \frac{\pi}{2}$

Now, we relate this phase difference to the path difference caused by the film: $\phi = \frac{2\pi}{\lambda} \Delta x_{film}$ $\frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x_{film}$

To find $\Delta x_{film}$, we rearrange the equation: $\Delta x_{film} = \frac{\lambda}{2\pi} \times \frac{\pi}{2}$ $\Delta x_{film} = \frac{\lambda}{4}$

Thus, the minimum path difference caused by the film is $\frac{\lambda}{4}$.