Question

Using monochromatic light of wavelength \(\lambda\), in Young's double slit experiment, at a point on the screen where path difference is \(\lambda\) between the two waves, the intensity of light is K units. Find the intensity of light at a point on the screen where path difference is \(\frac{\lambda}{3}\).

Hint:

For YDSE, the intensity is maximum when path difference is \(n\lambda\) and zero when it's \((n+1/2)\lambda\).

Answer 1

Step 1: Understanding the Concept:
In Young's double-slit experiment (YDSE), the intensity of light at any point on the screen is a result of the interference of waves from the two slits. The intensity depends on the phase difference (\(\phi\)) between the two waves, which is directly related to their path difference (\(\Delta x\)).

Step 2: Key Formula or Approach:
The relationship between phase difference (\(\phi\)) and path difference (\(\Delta x\)) is: \[ \phi = \frac{2\pi}{\lambda} \Delta x \] The resultant intensity (\(I\)) at a point is given by: \[ I = I_{max} \cos^2\left(\frac{\phi}{2}\right) \] where \(I_{max}\) is the maximum possible intensity (at the central maximum).

Step 3: Detailed Explanation (Calculation):
Case 1: Given Information
The path difference is given as \(\Delta x_1 = \lambda\). First, find the corresponding phase difference \(\phi_1\): \[ \phi_1 = \frac{2\pi}{\lambda} (\lambda) = 2\pi \text{ radians} \] Now, use the intensity formula. The intensity at this point (\(I_1\)) is given as K units. \[ I_1 = I_{max} \cos^2\left(\frac{2\pi}{2}\right) = I_{max} \cos^2(\pi) = I_{max}(-1)^2 = I_{max} \] So, we have \(K = I_{max}\). The maximum intensity in the interference pattern is K units.
Case 2: Find the new intensity
The new path difference is given as \(\Delta x_2 = \frac{\lambda}{3}\). Find the new phase difference \(\phi_2\): \[ \phi_2 = \frac{2\pi}{\lambda} \left(\frac{\lambda}{3}\right) = \frac{2\pi}{3} \text{ radians} \] Now, calculate the new intensity (\(I_2\)) using the intensity formula: \[ I_2 = I_{max} \cos^2\left(\frac{\phi_2}{2}\right) = I_{max} \cos^2\left(\frac{2\pi/3}{2}\right) = I_{max} \cos^2\left(\frac{\pi}{3}\right) \] We know that \(\cos\left(\frac{\pi}{3}\right) = \cos(60^\circ) = \frac{1}{2}\). \[ I_2 = I_{max} \left(\frac{1}{2}\right)^2 = \frac{I_{max}}{4} \] Since we found that \(I_{max} = K\), the new intensity is: \[ I_2 = \frac{K}{4} \]

Step 4: Final Answer:
The intensity of light at the point where the path difference is \(\frac{\lambda}{3}\) is \(\frac{K}{4}\) units.