In young's double slit experiment using monochromatic light of wavelengths `lambda`, the intensity of light at a point on the screen with path difference `lambda` is M units. The intensity of light at a point where path difference is `lambda//3` is

Options

A. `(M)/(2)`
B. `(M)/(4)`
C. `(M)/(8)`
D. `(M)/(16)`

Explanation

To solve the problem, we will follow these steps:

### 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:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]

### Step 2: Calculate the phase difference for a path difference of \( \lambda \)
Given that the path difference is \( \lambda \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \lambda = 2\pi \text{ radians} \]

### Step 3: Use the intensity formula for the case of path difference \( \lambda \)
The intensity \( I \) at a point on the screen can be expressed as:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\phi) \]
For \( \phi = 2\pi \), \( \cos(2\pi) = 1 \):
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \]
This simplifies to:
\[ I = (I_1 + I_2 + 2\sqrt{I_1 I_2}) = 4I_1 \quad \text{(assuming \( I_1 = I_2 \))} \]

### Step 4: Relate the intensity at path difference \( \lambda \) to \( m \)
We know from the problem that the intensity at this point is \( m \):
\[ 4I_1 = m \implies I_1 = \frac{m}{4} \]

### Step 5: Calculate the phase difference for a path difference of \( \frac{\lambda}{3} \)
Now, we need to find the intensity at a point where the path difference is \( \frac{\lambda}{3} \):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} \text{ radians} \]

### Step 6: Substitute the phase difference into the intensity formula
Using the intensity formula again:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos\left(\frac{2\pi}{3}\right) \]
Since \( \cos\left(\frac{2\pi}{3}\right) = -\frac{1}{2} \):
\[ I = I_1 + I_2 - \sqrt{I_1 I_2} \]
Assuming \( I_1 = I_2 \):
\[ I = 2I_1 - \sqrt{I_1^2} = 2I_1 - I_1 = I_1 \]

### Step 7: Substitute \( I_1 \) into the equation
Since \( I_1 = \frac{m}{4} \):
\[ I = \frac{m}{4} \]

### Step 8: Final answer
Thus, the intensity at the point where the path difference is \( \frac{\lambda}{3} \) is:
\[ \boxed{\frac{m}{3}} \]