In Young's double-slit experimetn, the intensity of light at a point on the screen, where the path difference is `lambda`,is `I` . The intensity of light at a point where the path difference becomes `lambda // 3` is

Options

A. `(I)/(4)`
B. `(I)/(3)`
C. `(I)/(2)`
D. `I`

Explanation

To solve the problem, we need to determine the intensity of light at a point on the screen where the path difference is \( \frac{\lambda}{3} \), given that the intensity at a point where the path difference is \( \lambda \) is \( I \).

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

### Step 2: Calculate the phase difference for path difference \( \lambda \)
For the point where the path difference is \( \lambda \):
\[ \Delta x = \lambda \]
Substituting this into the equation for phase difference:
\[ \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \]

### Step 3: Calculate the intensity at this point
The intensity \( I \) at a point in Young's double-slit experiment can be expressed as:
\[ I = 4I_0 \cos^2\left(\frac{\phi}{2}\right) \]
At \( \phi = 2\pi \):
\[ I = 4I_0 \cos^2\left(\frac{2\pi}{2}\right) = 4I_0 \cos^2(\pi) = 4I_0 \cdot 1 = 4I_0 \]
Given that this intensity is \( I \), we can equate:
\[ 4I_0 = I \quad \Rightarrow \quad I_0 = \frac{I}{4} \]

### Step 4: Calculate the phase difference for path difference \( \frac{\lambda}{3} \)
Now, we need to find the intensity at the point where the path difference is \( \frac{\lambda}{3} \):
\[ \Delta x = \frac{\lambda}{3} \]
Substituting this into the phase difference formula:
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \]

### Step 5: Calculate the intensity at this new point
Now we can find the intensity at this point:
\[ I_Q = 4I_0 \cos^2\left(\frac{\phi}{2}\right) = 4I_0 \cos^2\left(\frac{2\pi/3}{2}\right) = 4I_0 \cos^2\left(\frac{\pi}{3}\right) \]
Since \( \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \):
\[ I_Q = 4I_0 \left(\frac{1}{2}\right)^2 = 4I_0 \cdot \frac{1}{4} = I_0 \]

### Step 6: Substitute \( I_0 \) back to find \( I_Q \)
Now substituting \( I_0 = \frac{I}{4} \):
\[ I_Q = \frac{I}{4} \]

### Final Result
The intensity of light at the point where the path difference is \( \frac{\lambda}{3} \) is:
\[ \boxed{\frac{I}{4}} \]