The maximum intensity in Young's double slit experiment is `I_(0)` . What will be the intensity of light in front of one the slits on a screen where path difference is `(lambda)/(4)` ?

Options

A. `(I_(0))/(2)`
B. `(3)/(4)I_(0)`
C. `I_(0)`
D. `(I_(0))/(4)`

Answer

To find the intensity of light in front of one of the slits on a screen where the path difference is \( \frac{\lambda}{4} \), we can follow these steps:

### Step 1: Understand the maximum intensity
In Young's double slit experiment, the maximum intensity \( I_0 \) occurs when the two waves from the slits are in phase. The maximum intensity is given as \( I_0 = 4I \), where \( I \) is the intensity from each slit.

### Step 2: Calculate the phase difference
The path difference \( \Delta x \) given is \( \frac{\lambda}{4} \). The phase difference \( \Delta \phi \) can be calculated using the formula:
\[ \Delta \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \( \Delta x = \frac{\lambda}{4} \):
\[ \Delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} \]

### Step 3: Use the intensity formula
The resultant intensity \( I \) at a point where there is a phase difference can be calculated using the formula:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos(\Delta \phi) \]
Here, \( I_1 \) and \( I_2 \) are the intensities from each slit. Since both slits have the same intensity \( I \), we can write:
\[ I_1 = I_2 = I \]
Thus, the formula becomes:
\[ I = I + I + 2\sqrt{I \cdot I} \cos\left(\frac{\pi}{2}\right) \]
Since \( \cos\left(\frac{\pi}{2}\right) = 0 \), the equation simplifies to:
\[ I = 2I + 0 = 2I \]

### Step 4: Substitute the value of \( I \)
From the maximum intensity relation \( I_0 = 4I \), we can express \( I \) as:
\[ I = \frac{I_0}{4} \]
Substituting this into the intensity equation:
\[ I = 2 \left(\frac{I_0}{4}\right) = \frac{I_0}{2} \]

### Conclusion
The intensity of light in front of one of the slits on the screen where the path difference is \( \frac{\lambda}{4} \) is:
\[ \boxed{\frac{I_0}{2}} \]