In Young's double slit experiment, the intensity on the screen at a point where path difference is `lambda` is K. What will be the intensity at the point where path difference is `lambda//4`?

Options

A. `(K)/(4)`
B. `(K)/(2)`
C. `K`
D. `zero`

Answer

To solve the problem, we will follow these steps:

### Step 1: Understand the relationship between path difference and phase difference.
The phase difference (Φ) is related to the path difference (Δx) by the formula:
\[ \Phi = \frac{2\pi}{\lambda} \Delta x \]

### Step 2: Calculate the phase difference for a path difference of λ.
Given that the path difference is λ, we can substitute this into the formula:
\[ \Phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]

### Step 3: Find the intensity at this phase difference.
The intensity (I) at a point in Young's double slit experiment is given by:
\[ I = 4I_0 \cos^2\left(\frac{\Phi}{2}\right) \]
Substituting Φ = 2π:
\[ I = 4I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0 \cdot (-1)^2 = 4I_0 \]
Since we are told that this intensity is K, we have:
\[ K = 4I_0 \]

### Step 4: Calculate the phase difference for a path difference of λ/4.
Now, we calculate the phase difference for a path difference of λ/4:
\[ \Phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

### Step 5: Find the intensity at this new phase difference.
Substituting Φ = π/2 into the intensity formula:
\[ I = 4I_0 \cos^2\left(\frac{\pi/2}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{4}\right) \]
We know that:
\[ \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \]
Thus:
\[ I = 4I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = 4I_0 \cdot \frac{1}{2} = 2I_0 \]

### Step 6: Relate this intensity back to K.
Since we established that \( K = 4I_0 \), we can express \( I \) in terms of \( K \):
\[ I = 2I_0 = \frac{2}{4}K = \frac{K}{2} \]

### Final Answer:
The intensity at the point where the path difference is λ/4 is:
\[ \boxed{\frac{K}{2}} \]