In Young's double-slit experiment (Y.D.S.E), light of intesity `I_0`, is used at both slits. Calculate the intensity on screen at a point for which the phase difference is `60^@`.

Answer

To solve the problem of finding the intensity on the screen at a point where the phase difference is \(60^\circ\) in Young's double-slit experiment, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Intensity Formula**: In Young's double-slit experiment, the intensity \(I\) at a point on the screen is given by the formula:
\[ I = I_0 \left(1 + \cos \Delta \phi\right) \]
where \(I_0\) is the intensity from each slit, and \(\Delta \phi\) is the phase difference between the two waves arriving at that point.

2. **Convert Phase Difference to Radians**: The phase difference given is \(60^\circ\). We need to convert this to radians for calculations:
\[ \Delta \phi = 60^\circ = \frac{\pi}{3} \text{ radians} \]

3. **Calculate \(\cos \Delta \phi\)**: Now, we calculate \(\cos 60^\circ\):
\[ \cos 60^\circ = \frac{1}{2} \]

4. **Substitute into the Intensity Formula**: Now, substitute \(\cos 60^\circ\) into the intensity formula:
\[ I = I_0 \left(1 + \frac{1}{2}\right) = I_0 \left(1 + 0.5\right) = I_0 \cdot 1.5 \]

5. **Final Calculation**: Thus, the intensity \(I\) at the point where the phase difference is \(60^\circ\) is:
\[ I = 1.5 I_0 \]

### Conclusion:
The intensity on the screen at a point for which the phase difference is \(60^\circ\) is:
\[ I = 1.5 I_0 \]