Question

In Young's double slit experiment, the wavelength of light used is \(λ\). The intensity on the screen at a point for path difference '\(λ\)' is 'X'. The intensity at the point for path difference \((\frac{λ}{6})\) is (\(cos180^{\circ} = -1\), \(cos30^{\circ} = \frac{\sqrt{3}}{2}\))

• A. \(\frac{X}{6}\)
• B. \(\frac{X}{2}\)
• C. \(\frac{3X}{4}\)
• D. \(\frac{4X}{3}\)

Hint:

Intensity is I = 4 I0 cos^2(phi/2), with phase = 2 pi / lambda times the path difference.

Answer 1

The Correct Option is C

Step 1: Ratio:
$\dfrac I{I_{max}}=\cos^2\dfrac\phi2$. At path difference $\lambda$ we have a maximum ($\phi=2\pi$), so $I_{max}=X$.

Step 2: Phase for lambda/6:
$\phi=60^{\circ}$, $\cos^2(30^{\circ})=\dfrac34$.

Step 3: Answer:
$I=\dfrac34X$. Option (C).

Final Answer:
Option (C). \[ \boxed{\text{(C) } \frac{3X}{4}} \]