In Young's double-slit experiment, the intensity at a point P on the screen is half the maximum intensity in the interference pattern. If the wavelength of light used is `lambda` and d is the distance between the slits, the angular separation between point P and the center of the screen is

Options

A. `sin^(-1)((lambda)/(d))`
B. `sin^(-1)((lambda)/(2d))`
C. `sin^(-1)((lambda)/(3d))`
D. `sin^(-1)((lambda)/(4d))`

Answer

To solve the problem step by step, we will analyze the situation in Young's double-slit experiment where the intensity at point P on the screen is half the maximum intensity.

### Step 1: Understand the relationship between intensity and phase difference
In Young's double-slit experiment, the intensity \( I \) at a point on the screen can be expressed in terms of the maximum intensity \( I_0 \) and the phase difference \( \phi \) as follows:
\[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Given that the intensity at point P is half the maximum intensity:
\[ I = \frac{I_0}{2} \]

### Step 2: Set up the equation
Substituting the expression for intensity into the equation gives:
\[ \frac{I_0}{2} = I_0 \cos^2\left(\frac{\phi}{2}\right) \]
Dividing both sides by \( I_0 \) (assuming \( I_0 \neq 0 \)):
\[ \frac{1}{2} = \cos^2\left(\frac{\phi}{2}\right) \]

### Step 3: Solve for the phase difference \( \phi \)
Taking the square root of both sides:
\[ \cos\left(\frac{\phi}{2}\right) = \frac{1}{\sqrt{2}} \]
This implies:
\[ \frac{\phi}{2} = \frac{\pi}{4} \quad \text{or} \quad \frac{\phi}{2} = \frac{3\pi}{4} \]
Thus, we can find \( \phi \):
\[ \phi = \frac{\pi}{2} \quad \text{or} \quad \phi = \frac{3\pi}{2} \]
Since we are interested in the first order maximum, we take:
\[ \phi = \frac{\pi}{2} \]

### Step 4: Relate phase difference to path difference
The phase difference \( \phi \) can also be expressed in terms of the path difference \( \Delta x \) as:
\[ \phi = \frac{2\pi}{\lambda} \Delta x \]
Setting this equal to our earlier result:
\[ \frac{\pi}{2} = \frac{2\pi}{\lambda} \Delta x \]

### Step 5: Solve for the path difference \( \Delta x \)
Rearranging gives:
\[ \Delta x = \frac{\lambda}{4} \]

### Step 6: Relate path difference to angular separation
In the double-slit experiment, the path difference \( \Delta x \) can also be expressed in terms of the angle \( \theta \) and the slit separation \( d \):
\[ \Delta x = d \sin \theta \]
Setting this equal to our expression for \( \Delta x \):
\[ d \sin \theta = \frac{\lambda}{4} \]

### Step 7: Solve for \( \sin \theta \)
Rearranging gives:
\[ \sin \theta = \frac{\lambda}{4d} \]

### Step 8: Find the angle \( \theta \)
Finally, we can find the angle \( \theta \):
\[ \theta = \sin^{-1}\left(\frac{\lambda}{4d}\right) \]

### Final Answer
The angular separation between point P and the center of the screen is:
\[ \theta = \sin^{-1}\left(\frac{\lambda}{4d}\right) \]