Question

In Young's double slit experiment, waves from slits ' \(\mathrm{S}_1\) ' and ' \(\mathrm{S}_2\) ' have a path difference of \(\frac{\lambda}{4}\) and \(\frac{\lambda}{6}\) respectively at points \(x\) and \(y\) on the screen. The ratio of intensities at points \(y\) to that at \(x\) is \(\left(\cos 45^{\circ}=\frac{1}{\sqrt{2}}, \cos 30^{\circ}=\sqrt{3} / 2,\right)\)

• A. \(1:3\)
• B. \(3:2\)
• C. \(3:4\)
• D. \(2:3\)

Hint:

Convert path difference to phase difference and use I proportional to cosine squared of half the phase difference.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
Phase difference \(\phi = \dfrac{2\pi}{\lambda}\Delta x\) and intensity \(I = I_0\cos^2\dfrac{\phi}{2}\).

Step 2: Point x
\(\Delta x = \dfrac\lambda4\Rightarrow\phi = \dfrac\pi2\), so \(I_x = I_0\cos^2 45^{\circ} = \dfrac{I_0}{2}\).

Step 3: Point y
\(\Delta x = \dfrac\lambda6\Rightarrow\phi = \dfrac\pi3\), so \(I_y = I_0\cos^2 30^{\circ} = \dfrac{3I_0}{4}\).

Step 4: Ratio
\(\dfrac{I_y}{I_x} = \dfrac{3/4}{1/2} = \dfrac32\), option B.

Final Answer:
The ratio of intensities at y to x is 3 to 2. \[ \boxed{\text{(B) }3:2} \]