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: Understanding the Question:
The problem asks for the ratio of the light intensity $I'$ at a specific observation point where the path difference is $\frac{\lambda}{6}$ relative to the maximum possible constructive interference fringe intensity $I_0$ in a Young's Double Slit Experiment (YDSE).

Step 2: Key Formula or Approach:
The resultant intensity $I$ at any point in an interference pattern where the two slits have equal initial intensities is given by:
$$I = I_0 \cos^2\left(\frac{\phi}{2}\right)$$ where $\phi$ represents the phase difference between the two overlapping wave fronts.
The fundamental relationship connecting phase difference $\phi$ to spatial path difference $\Delta x$ is:
$$\phi = \frac{2\pi}{\lambda} \cdot \Delta x$$

Step 3: Detailed Explanation:
First, let's compute the phase difference $\phi$ corresponding to the given path difference $\Delta x = \frac{\lambda}{6}$:
$$\phi = \frac{2\pi}{\lambda} \cdot \left(\frac{\lambda}{6}\right) = \frac{2\pi}{6} = \frac{\pi}{3} \text{ rad} = 60^\circ$$ Now, substitute this calculated value of $\phi$ into the intensity equation to find $I'$:
$$I' = I_0 \cos^2\left(\frac{\pi / 3}{2}\right) = I_0 \cos^2\left(\frac{\pi}{6}\right) = I_0 \cos^2(30^\circ)$$ We know from standard trigonometry that $\cos(30^\circ) = \frac{\sqrt{3}}{2}$. Squaring this value yields:
$$I' = I_0 \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = I_0 \cdot \frac{3}{4}$$ To isolate the requested ratio, divide both sides by $I_0$:
$$\frac{I'}{I_0} = \frac{3}{4}$$

Step 4: Final Answer:
The intensity ratio $\frac{I'}{I_0}$ equals $\frac{3}{4}$, which corresponds to option (A).