Question

In a Young's double slit experiment, the intensities at two points, for the path difference \(\frac{λ}{4}\) and \(\frac{λ}{3}\) (\(λ\) being the wavelength of light used) are \(I_1\) and \(I_2\) respectively. If \(I_0\) denotes the intensity produced by each one of the individual slits, then \(\frac{I_1+I_2}{I_0} =\)
\((cos45^{\circ} = \frac{1}{\sqrt{2}},cos60 = \frac{1}{2})\)

• A. \(2\)
• B. \(3\)
• C. \(4\)
• D. \(5\)

Hint:

Use $I=4I_0\cos^2\frac\phi2$ with $\phi=\frac{2\pi}{\lambda}\Delta$.

Answer 1

The Correct Option is B

Step 1: First point
\(\Delta=\frac\lambda4\) gives \(\phi=\frac\pi2\). \(I_1=4I_0\cos^245^{\circ}=4I_0\times\frac12=2I_0\).

Step 2: Second point
\(\Delta=\frac\lambda3\) gives \(\phi=\frac{2\pi}{3}\). \(I_2=4I_0\cos^260^{\circ}=4I_0\times\frac14=I_0\).

Step 3: Sum
\(\frac{I_1+I_2}{I_0}=2+1=3\). Option (B).

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