Question

In Young’s double slit experiment, using monochromatic light of wavelength $\lambda$, the intensity of light at a point on the screen where the path difference is $\lambda/3$ is K units. The intensity of light at a point where the path difference is $\lambda/2$ will be: ____.

• A. K/2
• B. 2K
• C. K/4
• D. K

Hint:

Path difference $\lambda/2, 3\lambda/2, 5\lambda/2...$ always corresponds to destructive interference (zero intensity), regardless of the intensity at other points.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
The intensity of light in an interference pattern depends on the phase difference ($\phi$) between the two waves, which is related to the path difference ($\Delta x$).

Step 2: Key Formula or Approach:
1. Phase difference $\phi = \frac{2\pi}{\lambda} \cdot \Delta x$ 2. Resultant Intensity $I = I_{max} \cos^2\left(\frac{\phi}{2}\right)$

Step 3: Detailed Explanation:
1. Case 1: Path difference $\Delta x = \lambda/3$. - $\phi_1 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{3} = \frac{2\pi}{3} = 120^\circ$ - $I_1 = K = I_{max} \cos^2(60^\circ) = I_{max} \cdot \left(\frac{1}{2}\right)^2 = \frac{I_{max}}{4}$ - This means $I_{max} = 4K$. 2. Case 2: Path difference $\Delta x = \lambda/2$. - $\phi_2 = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{2} = \pi = 180^\circ$ - $I_2 = I_{max} \cos^2(90^\circ) = I_{max} \cdot 0 = 0$. (Note: If the question implies $I_2$ relative to $K$ and we assume standard options, $0$ is the physical answer. If $K$ was the max intensity, the answer would change; however, based on the calculation, the result is zero.)

Step 4: Final Answer:
The intensity at path difference $\lambda/2$ is zero (Destructive interference).