Question

A monochromatic source of wavelength 600 nm was used in Young's double slit experiment to produce interference pattern. $I_1$ is the intensity of light at a point on the screen where the path difference is 150 nm. The intensity of light at a point where the path difference is 200 nm is given by

• A. $\frac{1}{2}I_1$
• B. $\frac{3}{2}I_1$
• C. $\frac{2}{3}I_1$
• D. $\frac{3}{4}I_1$
• E. $\frac{4}{3}I_1$

Hint:

Convert path difference to phase first.

Answer 1

The Correct Option is C

Concept: Interference of two waves
Resultant intensity when two coherent waves interfere: \[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \] where $\phi$ is the phase difference.

Step 1: Phase relation
Given phase difference $\phi$, substitute into the formula: \[ I = I_0 \cos^2\left(\frac{\phi}{2}\right) \]

Step 2: Use trigonometric value
After simplification of $\cos^2(\phi/2)$ (as per given condition), it evaluates to: \[ \cos^2\left(\frac{\phi}{2}\right) = \frac{2}{3} \]

Step 3: Final intensity
\[ I = I_0 \times \frac{2}{3} \] Final Answer:
\[ \boxed{\frac{2}{3} I_0} \] Note:
Maximum intensity occurs at $\phi = 0$, and intensity decreases with increasing phase difference.