Question

In a Young's double slit experiment, $I_0$ is the maximum intensity and $\beta$ is the fringe width. Intensity at point P which is distance $x$ from central maxima is

• A. $I_0 \cos \dfrac{\pi x}{\beta}$
• B. $4I_0 \cos^2 \dfrac{\pi x}{\beta}$
• C. $I_0 \cos^2 \dfrac{\pi x}{\beta}$
• D. $\dfrac{I_0}{4} \cos^2 \dfrac{\pi x}{\beta}$

Answer 1

Answer: (C)

$I_0 \cos^2 \dfrac{\pi x}{\beta}$

Answer 2

Key idea: Intensity $I\propto$ (amplitude)$^2$.

1. Central fringe intensity:
   $I_0 = k(2A)^2.$

2. At point P with phase difference $\phi$, resultant amplitude
   $A_P = 2A\cos\frac{\phi}{2}.$

3. Hence intensity at P:
   $I = k\bigl(2A\cos\frac{\phi}{2}\bigr)^2 = I_0\cos^2\frac{\phi}{2}.$

4. Phase difference $\phi = \frac{2\pi}{\lambda}\Delta x = \frac{2\pi x}{\beta}$, so
   $\frac{\phi}{2} = \frac{\pi x}{\beta}.$

5. Therefore,
   $I = I_0\cos^2\frac{\pi x}{\beta}.$