Question

In a Young's double slit experiment, the intensity at a point where the path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of the light used) is $I$. If $I_{0}$ denotes the maximum intensity, $\frac{I}{I_{0}}$ is

• A. $\frac{1}{\sqrt{2}}$
• B. $\frac{\sqrt{3}}{2}$
• C. $\frac{1}{2}$
• D. $\frac{3}{4}$

Hint:

A path difference of $\lambda$ equals a $360^{\circ}$ phase change. So, a path difference of $\frac{\lambda}{6}$ corresponds to $60^{\circ}$ phase difference. Half of that angle is $30^{\circ}$, and $\cos^2(30^{\circ}) = 3/4$.

Answer 1

The Correct Option is D

Step 1: Intensity in interference.
In Young's experiment $I = I_{0}\cos^{2}\!\left(\dfrac{\phi}{2}\right)$, where $\phi$ is the phase difference and $I_{0}$ is the brightest value.

Step 2: Link path and phase.
The phase difference comes from the path difference: $\phi = \dfrac{2\pi}{\lambda}\,\Delta x$.

Step 3: Put in the path difference.
Given $\Delta x = \dfrac{\lambda}{6}$, \[ \phi = \frac{2\pi}{\lambda}\times\frac{\lambda}{6} = \frac{\pi}{3} \]
Step 4: Halve the phase.
\[ \frac{\phi}{2} = \frac{\pi}{6} = 30^{\circ} \]
Step 5: Square the cosine.
\[ \frac{I}{I_{0}} = \cos^{2}(30^{\circ}) = \left(\frac{\sqrt{3}}{2}\right)^{2} = \frac{3}{4} \]
Step 6: State the answer.
So $\dfrac{I}{I_{0}} = \dfrac{3}{4}$, which is option 4.
\[ \boxed{\frac{I}{I_{0}} = \frac{3}{4}} \]