In Young's interference experiment, if the slit are of unequal width, then

Options

A. fringes will not be formed
B. the positions of minimum intensity will not be completely dark
C. bright fringe will not be formed at the centre of the screen
D. distance between two consecutive bright fringes will not be equal to the distance between two consecutive dark fringes.

Answer

### Step-by-Step Solution:

1. **Understanding Young's Experiment**: In Young's double-slit experiment, light waves from two slits interfere with each other to produce a pattern of bright and dark fringes on a screen.

2. **Effect of Unequal Slit Widths**: When the slits have unequal widths, the intensity of light coming from each slit will differ. The slit with the larger width will emit more light (higher intensity) compared to the slit with the smaller width.

3. **Intensity and Amplitude Relationship**: The intensity (I) of light is proportional to the square of the amplitude (A) of the wave:
\[ I \propto A^2 \]
Therefore, if one slit has a greater amplitude due to its larger width, it will contribute more to the overall intensity of the light.

4. **Formation of Maxima and Minima**: In the case of equal slit widths, complete destructive interference can occur at certain points, leading to completely dark fringes (minima). However, with unequal slit widths, the amplitudes are different, which affects the condition for minima.

5. **Resulting Interference Pattern**: At points where destructive interference would normally occur (minima), the resulting intensity will not be zero because the amplitudes from the two slits are not equal. Thus, while there will still be dark fringes, they will not be completely dark.

6. **Conclusion**: The presence of unequal slit widths means that the minima will not be completely dark. Instead, they will be partially illuminated due to the unequal contributions of the two slits.

### Final Answer:
The position of minima will not be completely dark.