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: Use the general intensity formula
$I=I_0+I_0+2I_0\cos\phi=2I_0(1+\cos\phi)$.
At $\phi=90^{\circ}$: $2I_0$. At $\phi=120^{\circ}$: $2I_0\times\frac12=I_0$. Sum $=3I_0$.

Final Answer:
Option (B). \[ \boxed{\text{(B)}} \]