Question

In Young's double slit experiment, the light of wavelength '$\lambda$' is used. The intensity at a point on the screen is 'I' where the path difference is $\lambda/4$. If '$I_0$' denotes the maximum intensity then the ratio of '$I_0$' to 'I' is ($\cos 45^\circ = 1/\sqrt{2}$)

• A. 2 : 1
• B. 4 : 1
• C. 8 : 1
• D. 12 : 1

Hint:

At a path difference of $\lambda/4$, the phase difference is $90^\circ$, leading to exactly half the maximum intensity.

Answer 1

The Correct Option is A

Step 1: Calculate Phase Difference
Path difference $\Delta x = \lambda/4$.
Phase difference $\phi = \frac{2\pi}{\lambda} \cdot \Delta x = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{4} = \frac{\pi}{2} (90^\circ)$.
Step 2: Intensity Formula
$I = I_0 \cos^2(\phi/2)$.
Step 3: Calculation
$I = I_0 \cos^2(\frac{\pi/2}{2}) = I_0 \cos^2(\frac{\pi}{4})$
$I = I_0 \cdot (\frac{1}{\sqrt{2}})^2 = I_0 \cdot \frac{1}{2} = \frac{I_0}{2}$.
Step 4: Ratio
$I_0/I = I_0 / (I_0/2) = 2/1$.
Final Answer:(A)