To solve the given question, we will analyze the assertion and reason provided in the context of Young's Double Slit Experiment (YDSE).
### Step-by-Step Solution:
1. **Understanding the Assertion**:
- The assertion states that in YDSE, bright and dark fringes are equally spaced.
- In YDSE, the bright fringes occur at positions where the path difference between the two waves is an integer multiple of the wavelength (nλ, where n = 0, 1, 2,...), and dark fringes occur where the path difference is an odd multiple of half the wavelength ((2n-1)λ/2).
- The positions of the 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 fringe, \(D\) is the distance from the slits to the screen, \(d\) is the distance between the slits, and \(\lambda\) is the wavelength of light used.
- From this formula, we can see that the distance between consecutive bright or dark fringes is constant, confirming that they are equally spaced.
2. **Understanding the Reason**:
- The reason states that the spacing of the fringes only depends on the phase difference.
- The phase difference is related to the path difference, which is given by:
\[
\Delta x = d \sin \theta
\]
where \(\Delta x\) is the path difference, \(d\) is the slit separation, and \(\theta\) is the angle of the fringe from the central maximum.
- However, the spacing of the fringes also depends on the wavelength \(\lambda\) and the geometry of the setup (i.e., the distances \(D\) and \(d\)). Therefore, the statement that it only depends on phase difference is incorrect.
3. **Conclusion**:
- The assertion is true: bright and dark fringes are equally spaced.
- The reason is false: the spacing does not depend solely on the phase difference; it also depends on the wavelength and the geometry of the setup.
### Final Answer:
- The assertion is true, and the reason is false. Therefore, the correct answer is: **C** (Assertion is true, Reason is false).
