The slits in a Young's double slit experiment have equal width and the source is placed symmetrically with respect to the slits. The Intensity at the central fringe is `I_0`. If one of the slits is closed, the intensity at this point will be

Options

A. `I_0`
B. `I_0//4`
C. `I_0//2`
D. `4I_0`

Answer

To solve the problem, we will analyze the situation step by step.

### Step 1: Understanding the Initial Conditions
In a Young's double slit experiment, we have two slits (let's call them S1 and S2) that are equally wide. The source of light is placed symmetrically with respect to these slits. The intensity at the central fringe (where the two waves from the slits constructively interfere) is given as \( I_0 \).

**Hint:** Remember that when both slits are open, the intensity at the central maximum is the result of constructive interference.

### Step 2: Expression for Resultant Intensity
The resultant intensity \( I \) at any point due to two coherent sources can be expressed as:
\[ I = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
where \( I_1 \) and \( I_2 \) are the intensities from each slit, and \( \phi \) is the phase difference between the two waves.

**Hint:** The phase difference \( \phi \) at the central maximum is zero because the path difference is zero.

### Step 3: Applying Symmetry
Since the source is symmetrically placed with respect to the slits, the intensities from both slits are equal, i.e., \( I_1 = I_2 = I_c \). Thus, the expression for intensity simplifies to:
\[ I = I_c + I_c + 2\sqrt{I_c I_c} \cos 0 = 2I_c + 2I_c = 4I_c \]
At the central maximum, this resultant intensity is given as \( I_0 \). Therefore, we have:
\[ 4I_c = I_0 \]
From this, we can solve for \( I_c \):
\[ I_c = \frac{I_0}{4} \]

**Hint:** This step shows how to relate the intensity at the central fringe to the individual slit intensities.

### Step 4: Closing One Slit
Now, if one of the slits (let's say S2) is closed, the intensity from that slit becomes zero: \( I_2 = 0 \). The resultant intensity when one slit is closed will be:
\[ I' = I_1 + I_2 + 2\sqrt{I_1 I_2} \cos \phi \]
Substituting \( I_2 = 0 \):
\[ I' = I_1 + 0 + 2\sqrt{I_1 \cdot 0} \cos \phi = I_1 \]
Since \( I_1 = I_c \) (because the source is still symmetrically placed), we have:
\[ I' = I_c \]

**Hint:** When one slit is closed, the intensity at the central maximum is simply the intensity from the open slit.

### Step 5: Final Calculation
Substituting the value of \( I_c \) that we found earlier:
\[ I' = \frac{I_0}{4} \]

### Conclusion
Thus, the intensity at the central point when one of the slits is closed is \( \frac{I_0}{4} \).

**Final Answer:** The intensity at this point will be \( \frac{I_0}{4} \).