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
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
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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 $\cos^2(30^{\circ}) = 3/4$.
Updated On: Jun 3, 2026
- $\frac{1}{\sqrt{2}}$
- $\frac{\sqrt{3}}{2}$
- $\frac{1}{2}$
- $\frac{3}{4}$
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The Correct Option is D
Solution and Explanation
Step 1: Concept
The resultant intensity $I$ at any point in an interference pattern depends on the phase difference ($\phi$) and is given by the formula $I = I_{0} \cos^2\left(\frac{\phi}{2}\right)$, where $I_{0}$ is the maximum possible intensity.
Step 2: Meaning
The phase difference ($\phi$) can be computed from the path difference ($\Delta x$) using the standard relationship: $\phi = \frac{2\pi}{\lambda} \cdot \Delta x$.
Step 3: Analysis
Given the path difference $\Delta x = \frac{\lambda}{6}$, the phase difference becomes: $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} \text{ rad}$ (or $60^{\circ}$). Now, substituting $\phi$ into the intensity equation gives: $\frac{I}{I_{0}} = \cos^2\left(\frac{\pi / 3}{2}\right) = \cos^2\left(\frac{\pi}{6}\right) = \cos^2(30^{\circ}) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$.
Step 4: Conclusion
Consequently, the ratio of the point intensity to the maximum intensity is $\frac{3}{4}$.
Final Answer: (D)
The resultant intensity $I$ at any point in an interference pattern depends on the phase difference ($\phi$) and is given by the formula $I = I_{0} \cos^2\left(\frac{\phi}{2}\right)$, where $I_{0}$ is the maximum possible intensity.
Step 2: Meaning
The phase difference ($\phi$) can be computed from the path difference ($\Delta x$) using the standard relationship: $\phi = \frac{2\pi}{\lambda} \cdot \Delta x$.
Step 3: Analysis
Given the path difference $\Delta x = \frac{\lambda}{6}$, the phase difference becomes: $\phi = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} = \frac{\pi}{3} \text{ rad}$ (or $60^{\circ}$). Now, substituting $\phi$ into the intensity equation gives: $\frac{I}{I_{0}} = \cos^2\left(\frac{\pi / 3}{2}\right) = \cos^2\left(\frac{\pi}{6}\right) = \cos^2(30^{\circ}) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4}$.
Step 4: Conclusion
Consequently, the ratio of the point intensity to the maximum intensity is $\frac{3}{4}$.
Final Answer: (D)
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