Question

In Young's double slit experiment, the intensities at two points, for the path difference \(\frac{\lambda}{4}\) and \(\frac{\lambda}{3}\) are \(\mathrm{I}_1\) and \(\mathrm{I}_2\) respectively. If \(\mathrm{I}_0\) denotes the intensity produced by each one of the individual slits then the ratio \(\left(\mathrm{I}_1+\mathrm{I}_2\right): \mathrm{I}_0\) is \(\left(\cos 45^{\circ}=1 / \sqrt{2}\right)\left(\cos 60^{\circ}=0.5\right)\)

• A. 2
• B. 3
• C. 4
• D. 1

Hint:

Use \(I=4I_0\cos^2(\phi/2)\) with phase difference \(\phi=2\pi\Delta/\lambda\) at each point.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
For two identical slits, the resultant intensity is \(I=4I_0\cos^2\dfrac{\phi}{2}\), where \(\phi=\dfrac{2\pi}{\lambda}\Delta\).

Step 2: First point
\(\Delta=\dfrac{\lambda}{4}\Rightarrow\phi=\dfrac{\pi}{2}\).\n\(I_1=4I_0\cos^2 45^{\circ}=4I_0\cdot\dfrac12=2I_0\).

Step 3: Second point
\(\Delta=\dfrac{\lambda}{3}\Rightarrow\phi=\dfrac{2\pi}{3}\).\n\(I_2=4I_0\cos^2 60^{\circ}=4I_0\cdot\dfrac14=I_0\).

Step 4: Ratio
\(\dfrac{I_1+I_2}{I_0}=\dfrac{2I_0+I_0}{I_0}=3\).

Final Answer:
The required ratio is 3. \[ \boxed{\text{(B) }3} \]