In Young's double slit experiment, the interference bright fringes are of:

Options

A. equal widths and unequal intensities
B. unequal width and equal intensities
C. equal widths and equal intensities
D. unequals widths and unequal intensities

Answer

To solve the question regarding the characteristics of bright fringes in Young's double slit experiment, we can break it down into the following steps:

### Step-by-Step Solution:

1. **Understanding Young's Double Slit Experiment**:
- In this experiment, light from a coherent source passes through two closely spaced slits (S1 and S2) and creates an interference pattern on a screen. The overlapping light waves from the two slits can interfere constructively or destructively.

2. **Constructive Interference**:
- Bright fringes (maxima) occur where the light waves from the two slits arrive in phase, which means the path difference between the two waves is an integer multiple of the wavelength (Nλ, where N is an integer).

3. **Mathematical Representation**:
- The position of the bright fringes on the screen can be calculated using the formula:
\[ y_n = \frac{n \lambda D}{d} \]
where:
- \( y_n \) is the position of the nth bright fringe,
- \( n \) is the order of the fringe (0, 1, 2, ...),
- \( \lambda \) is the wavelength of light,
- \( D \) is the distance from the slits to the screen,
- \( d \) is the distance between the two slits.

4. **Calculating Fringe Width**:
- The width of the bright fringe (fringe width) can be determined by finding the distance between two consecutive bright fringes:
\[ \text{Fringe Width} = y_{n+1} - y_n = \frac{(n+1) \lambda D}{d} - \frac{n \lambda D}{d} = \frac{\lambda D}{d} \]
- This shows that the fringe width is constant for all bright fringes.

5. **Intensity of Bright Fringes**:
- The intensity of the bright fringes can be derived from the superposition of the two waves. The intensity at a bright fringe is given by:
\[ I = (A_1 + A_2)^2 \]
where \( A_1 \) and \( A_2 \) are the amplitudes of the waves from the two slits. Since the amplitudes are constant, the intensity remains the same for all bright fringes.

6. **Conclusion**:
- From the analysis, we conclude that in Young's double slit experiment, the bright fringes are of equal width and equal intensity.

### Final Answer:
The interference bright fringes in Young's double slit experiment are of **equal width and equal intensity**.