Question

In Young's double slit experiment, the intensity at a point where path difference is $\frac{\lambda}{6}$ ($\lambda$ being the wavelength of light used) is $I'$. If '$I_0$' denotes the maximum intensity, then $\frac{I'}{I_0}$ is equal to ($\cos 0^\circ = 1, \cos 60^\circ = \frac{1}{2}$)

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

Hint:

Always remember that a full wave path difference of $\lambda$ corresponds to a complete phase rotation of $2\pi$ ($360^\circ$). Therefore, a path shift of $\frac{\lambda}{6}$ is a $60^\circ$ phase shift. Halving this inside the cosine gives $\cos(30^\circ)$, which squares directly to $\frac{3}{4}$.

Answer 1

The Correct Option is A

Step 1: Read the problem.
In a double-slit experiment the path difference at a point is $\dfrac{\lambda}{6}$, giving intensity $I'$. With $I_0$ the maximum intensity, we want $\dfrac{I'}{I_0}$.
Step 2: Recall the intensity formula.
For two equal sources, $I = I_0 \cos^2\!\left(\dfrac{\phi}{2}\right)$, where $\phi$ is the phase difference.
Step 3: Convert path difference to phase.
Phase difference $\phi = \dfrac{2\pi}{\lambda}\,\Delta x$. With $\Delta x = \dfrac{\lambda}{6}$, $\phi = \dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{6} = \dfrac{\pi}{3} = 60^\circ$.
Step 4: Take half the phase.
$\dfrac{\phi}{2} = 30^\circ$, so we need $\cos^2 30^\circ$.
Step 5: Evaluate.
$\cos 30^\circ = \dfrac{\sqrt{3}}{2}$, so $\cos^2 30^\circ = \dfrac{3}{4}$. Thus $I' = I_0 \cdot \dfrac{3}{4}$.
Step 6: Form the ratio.
$\dfrac{I'}{I_0} = \dfrac{3}{4}$, option (1). Since the point lies between a maximum and a minimum, a value of $3/4$ (between $1$ and $0$) is reasonable.
\[ \boxed{\dfrac{I'}{I_0} = \dfrac{3}{4}} \]