In a Young's double-slit expriment using identical slits, the intensity at a bright fringe is I_(0). If one of the slits is now covered, the intensity at any point on the screen will be

Options

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

Answer

To solve the problem, let's analyze the situation step by step.

### Step 1: Understanding the Initial Conditions
In a Young's double-slit experiment, when both slits are open, the intensity at a bright fringe is given as \( I_0 \). This intensity is the result of the interference of light waves coming from both slits.

**Hint:** Remember that the intensity at a bright fringe is the result of constructive interference from both slits.

### Step 2: Expressing the Intensity at Bright Fringe
The intensity at a bright fringe can be expressed in terms of the intensities from each slit. If we denote the intensity from each slit as \( I_1 \), then the total intensity at a bright fringe (where the phase difference \( \phi = 0 \)) can be written as:
\[ I_0 = I_1 + I_1 + 2\sqrt{I_1 I_1} \cos(0) \]
This simplifies to:
\[ I_0 = 2I_1 + 2I_1 = 4I_1 \]

**Hint:** Use the formula for intensity in terms of the contributions from both slits and remember that \( \cos(0) = 1 \).

### Step 3: Finding the Intensity from Each Slit
From the equation \( I_0 = 4I_1 \), we can solve for \( I_1 \):
\[ I_1 = \frac{I_0}{4} \]

**Hint:** Rearranging the equation can help you find the intensity from each slit.

### Step 4: Covering One Slit
Now, if one of the slits is covered, only one slit is contributing to the intensity at any point on the screen. Therefore, the intensity at any point on the screen will simply be equal to the intensity from the remaining open slit:
\[ I = I_1 = \frac{I_0}{4} \]

**Hint:** When one slit is covered, the intensity becomes equal to the intensity from the single open slit.

### Final Answer
Thus, the intensity at any point on the screen when one slit is covered is:
\[ \frac{I_0}{4} \]