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: Intensity Formula:
\(I=4I_0\cos^2\dfrac\phi2\), where \(\phi=\dfrac{2\pi}\lambda\Delta\).

Step 2: Path Difference lambda:
\(\phi=2\pi\), so \(I=4I_0\cos^2\pi=4I_0\). This is given as \(X\), so \(4I_0=X\).

Step 3: Path Difference lambda/6:
\(\phi=\dfrac{2\pi}\lambda\cdot\dfrac\lambda6=\dfrac\pi3\). Then
\[ I=4I_0\cos^2\frac\pi6=4I_0\times\frac34=3I_0 \]

Step 4: Express in X:
\(3I_0=\dfrac34(4I_0)=\dfrac{3X}4\). So (C) is correct. Option (A) \(X/6\) and (B) \(X/2\) correspond to \(\cos^2\) values of \(1/6\) and \(1/2\), and (D) \(4X/3\) exceeds the maximum intensity \(X\).

Final Answer:
The intensity is \(\dfrac{3X}4\), option (C). \[ \boxed{\text{(C) } \frac{3X}{4}} \]