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

Options

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

Answer

To solve this problem, we need to analyze the effects of covering one of the slits in a Young's double-slit experiment with a thin glass slab that reduces the intensity of light passing through that slit.

### Step-by-Step Solution:

1. **Initial Setup**: In a standard Young's double-slit experiment, we have two slits (S1 and S2) that are illuminated by a coherent light source. Initially, both slits transmit equal intensity of light. Let the intensity from each slit be \( I_1 = I_2 = I \).

2. **Effect of the Glass Slab**: When one slit (let's say S2) is covered with a thin parallel-sided glass slab that transmits only half of the intensity, the new intensity from S2 becomes:
\[ I_2 = \frac{I}{2} \]
The intensity from S1 remains:
\[ I_1 = I \]

3. **Intensity at the Screen**: The intensity at any point on the screen due to interference from both slits can be calculated using the formula for the resultant intensity:
\[ I_{max} = (I_1 + I_2)^2 + 2 \sqrt{I_1 I_2} \cos(\phi) \]
where \( \phi \) is the phase difference between the two waves arriving at the point on the screen.

4. **Calculating Maximum Intensity**: For the maximum intensity when the waves are in phase (constructive interference), we have:
\[ I_{max} = \left(I + \frac{I}{2}\right)^2 + 2 \sqrt{I \cdot \frac{I}{2}} = \left(\frac{3I}{2}\right)^2 + 2 \sqrt{\frac{I^2}{2}} = \frac{9I^2}{4} + 2 \cdot \frac{I}{\sqrt{2}} = \frac{9I^2}{4} + \sqrt{2}I \]

5. **Calculating Minimum Intensity**: For the minimum intensity (destructive interference), we have:
\[ I_{min} = I_1 + I_2 - 2\sqrt{I_1 I_2} = I + \frac{I}{2} - 2\sqrt{I \cdot \frac{I}{2}} = \frac{3I}{2} - 2 \cdot \frac{I}{\sqrt{2}} = \frac{3I}{2} - \sqrt{2}I \]

6. **Fringe Shift**: The introduction of the glass slab will also cause a phase shift due to the difference in the optical path length. The light traveling through the glass slab will take longer, causing the fringe pattern to shift. The fringe pattern will shift towards the covered slit (S2).

7. **Brightness of Fringes**: The bright fringes will become less bright due to the reduced intensity from the covered slit, while the dark fringes will become brighter because the minimum intensity is no longer zero.

8. **Fringe Width**: The fringe width, given by \( \beta = \frac{\lambda D}{d} \) (where \( D \) is the distance from the slits to the screen and \( d \) is the distance between the slits), remains unchanged because it depends on the wavelength and the geometry of the setup, not on the intensity.

### Conclusion:
1. The fringe pattern will shift towards the covered slit.
2. The bright fringes will become less bright, and the dark fringes will become brighter.
3. The fringe width will remain unchanged.