The maximum intensity in young's double-slit experiment is `I_(0)`. Distance between the slit is `d = 5 lambda`, where `lambda` is the wavelength of monochromatic light used in the experiment. What will be the intensity of light in front of one of the slits on a screen at a distance `D = 10 d`?

Options

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

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the setup of Young's double-slit experiment
In Young's double-slit experiment, two slits (S1 and S2) are separated by a distance \(d\), and light is projected onto a screen at a distance \(D\). The maximum intensity on the screen is given as \(I_0\).

### Step 2: Identify the given values
- Maximum intensity, \(I_{\text{max}} = I_0\)
- Distance between the slits, \(d = 5\lambda\)
- Distance from the slits to the screen, \(D = 10d\)

### Step 3: Determine the position of interest
We want to find the intensity at a point directly in front of one of the slits (let's say S1). The distance from S1 to the screen is \(D\), and the distance from S1 to S2 is \(d\).

### Step 4: Calculate the path difference at point P
The path difference (\(\Delta x\)) at point P (in front of S1) can be calculated using the formula:
\[ \Delta x = \frac{d \cdot y}{D} \]
Where \(y\) is the distance from the central maximum to point P. Since point P is directly in front of S1, the distance \(y\) is half of the distance between the slits:
\[ y = \frac{d}{2} = \frac{5\lambda}{2} \]
Substituting the values:
\[ \Delta x = \frac{5\lambda \cdot \frac{5\lambda}{2}}{10 \cdot 5\lambda} = \frac{5\lambda \cdot \frac{5}{2}}{50\lambda} = \frac{25}{100} \lambda = \frac{\lambda}{4} \]

### Step 5: Calculate the phase difference
The phase difference (\(\phi\)) corresponding to the path difference can be calculated using:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Substituting \(\Delta x = \frac{\lambda}{4}\):
\[ \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{2\pi}{4} = \frac{\pi}{2} \]

### Step 6: Calculate the intensity at point P
The intensity at any point in the interference pattern is given by:
\[ I = I_{\text{max}} \cos^2\left(\frac{\phi}{2}\right) \]
Substituting \(\phi = \frac{\pi}{2}\):
\[ I = I_0 \cos^2\left(\frac{\pi}{4}\right) \]
Since \(\cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}}\):
\[ I = I_0 \left(\frac{1}{\sqrt{2}}\right)^2 = I_0 \cdot \frac{1}{2} = \frac{I_0}{2} \]

### Final Answer
The intensity of light in front of one of the slits on the screen is:
\[ \boxed{\frac{I_0}{2}} \]