If one of the slits of a standard Young's double-slit exper- iment is covered by a thin parallel sided glass slab so that it transmits only one-half the light intensity of the other, then

Options

A. The fringe paliern will get shifted toward the covered slit
B. The fringe pattern will get shifted away from the covered slit
C. The bright fringes will become less bright and the dark ones will become more bright
D. The fringe width will remain unchanged

Answer

To solve the problem of what happens when one of the slits in Young's double-slit experiment is covered by a thin parallel-sided glass slab that transmits only one-half the light intensity of the other slit, we can follow these steps:

### Step 1: Understand the Setup
In a standard Young's double-slit experiment, we have two slits (S1 and S2) that produce an interference pattern on a screen. The intensity of light from both slits is initially equal, denoted as \( I_0 \).

**Hint:** Remember that the interference pattern is created due to the superposition of light waves from both slits.

### Step 2: Analyze the Effect of the Glass Slab
When a thin glass slab is placed in front of slit S1, the intensity of light from this slit becomes half, so we have:
- Intensity from S1, \( I_1 = \frac{I_0}{2} \)
- Intensity from S2, \( I_2 = I_0 \)

**Hint:** Consider how the change in intensity affects the overall interference pattern.

### Step 3: Calculate the New Maximum Intensity
The maximum intensity of the interference pattern can be calculated using the formula:
\[ I_{\text{max}} = \left( \sqrt{I_1} + \sqrt{I_2} \right)^2 \]
Substituting the values:
\[ I_{\text{max}} = \left( \sqrt{\frac{I_0}{2}} + \sqrt{I_0} \right)^2 = \left( \sqrt{I_0} \left( \frac{1}{\sqrt{2}} + 1 \right) \right)^2 \]

**Hint:** Simplify the expression to see how the maximum intensity changes.

### Step 4: Calculate the New Minimum Intensity
The minimum intensity can be calculated using:
\[ I_{\text{min}} = \left( \sqrt{I_1} - \sqrt{I_2} \right)^2 \]
Substituting the values:
\[ I_{\text{min}} = \left( \sqrt{\frac{I_0}{2}} - \sqrt{I_0} \right)^2 = \left( \sqrt{I_0} \left( \frac{1}{\sqrt{2}} - 1 \right) \right)^2 \]

**Hint:** This will show how the dark fringes are affected by the change in intensity.

### Step 5: Determine the Shift in the Fringe Pattern
Since the intensity from S1 is reduced, the path length difference will also affect the fringe pattern. The fringe pattern will shift towards the covered slit (S1) because the effective path length from S2 to the screen is greater than that from S1.

**Hint:** Think about how the path lengths affect the position of the fringes on the screen.

### Step 6: Conclusion
1. The fringe pattern will shift towards the covered slit (S1).
2. The bright fringes will become less bright due to the reduced intensity from S1.
3. The dark fringes will become less dark (or more bright) because the minimum intensity is no longer zero.
4. The fringe width remains unchanged as it depends on the wavelength and distance between the slits, which are not affected by the glass slab.

**Final Answer:**
- The fringe pattern shifts towards the covered slit.
- Bright fringes become less bright.
- Dark fringes become more bright.
- Fringe width remains unchanged.