Intensities of light due to the two slits of Young's double-slit experiment are I and 4I. How far from the centre maxima will the intensity be equal to the average intensity on the screen? (`beta` is the fringe width.)

Options

A. `(beta)/(2)`
B. `(beta)/(3)`
C. `(beta)/(6)`
D. `(beta)/(4)`

Answer

To solve the problem step by step, we need to find the distance from the central maximum where the intensity is equal to the average intensity on the screen in Young's double-slit experiment.

### Step 1: Calculate the Average Intensity
The average intensity \( I_{avg} \) can be calculated using the formula:
\[ I_{avg} = \frac{I_{max} + I_{min}}{2} \]
Given the intensities from the two slits are \( I \) and \( 4I \):
- The maximum intensity \( I_{max} = I + 4I + 2\sqrt{I \cdot 4I} = 5I + 2\sqrt{4I^2} = 5I + 4I = 9I \)
- The minimum intensity \( I_{min} = I - 4I = -3I \) (but since intensity cannot be negative, we consider the minimum intensity as \( 0 \) when they are completely out of phase)

Thus, the average intensity is:
\[ I_{avg} = \frac{9I + 0}{2} = \frac{9I}{2} \]

### Step 2: Relate Intensity to Phase Difference
The intensity at a point on the screen can be expressed as:
\[ I = I_{0} \cos^2 \phi \]
where \( I_{0} = 9I \) is the maximum intensity and \( \phi \) is the phase difference.

Setting \( I = I_{avg} \):
\[ \frac{9I}{2} = 9I \cos^2 \phi \]

### Step 3: Solve for Cosine of Phase Difference
Dividing both sides by \( 9I \):
\[ \frac{1}{2} = \cos^2 \phi \]
Taking the square root gives:
\[ \cos \phi = \frac{1}{\sqrt{2}} \quad \text{or} \quad \phi = 45^\circ \]

### Step 4: Relate Phase Difference to Path Difference
The phase difference \( \phi \) is related to the path difference \( \Delta x \) by:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \phi = 45^\circ = \frac{\pi}{4} \):
\[ \frac{\pi}{4} = \frac{2\pi}{\lambda} \Delta x \]
Solving for \( \Delta x \):
\[ \Delta x = \frac{\lambda}{8} \]

### Step 5: Calculate the Distance from the Central Maximum
The distance \( y \) from the central maximum can be calculated using the fringe width \( \beta \):
\[ \Delta x = y \cdot \frac{D}{d} \]
where \( D \) is the distance from the slits to the screen, and \( d \) is the distance between the slits. Thus:
\[ y = \Delta x \cdot \frac{d}{D} = \frac{\lambda}{8} \cdot \frac{d}{D} \]

### Final Answer
The distance from the central maximum where the intensity is equal to the average intensity on the screen is:
\[ y = \frac{\lambda}{8} \cdot \frac{d}{D} \]