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In a Young's double slit experiment the ...

In a Young's double slit experiment the intensity at a point where tha path difference is `(lamda)/(6)` (`lamda` being the wavelength of light used) is I. If `I_0` denotes the maximum intensity, `(I)/(I_0)` is equal to

A

`(3)/(4)`

B

`(1)/(sqrt2)`

C

`(sqrt3)/(2)`

D

`(1)/(2)`

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To solve the problem, we need to find the ratio \( \frac{I}{I_0} \) where \( I \) is the intensity at a point with a path difference of \( \frac{\lambda}{6} \) and \( I_0 \) is the maximum intensity in a Young's double slit experiment. ### Step-by-Step Solution: 1. **Understanding the Intensity Formula**: In a Young's double slit experiment, the intensity \( I \) at any point on the screen can be expressed as: \[ I = I_0 \cos^2\left(\frac{\delta \phi}{2}\right) \] where \( \delta \phi \) is the phase difference between the two waves arriving at that point. 2. **Calculating the Phase Difference**: The phase difference \( \delta \phi \) is related to the path difference \( \Delta x \) by the formula: \[ \delta \phi = \frac{2\pi}{\lambda} \Delta x \] Given that the path difference \( \Delta x = \frac{\lambda}{6} \), we can substitute this into the equation: \[ \delta \phi = \frac{2\pi}{\lambda} \cdot \frac{\lambda}{6} = \frac{2\pi}{6} = \frac{\pi}{3} \] 3. **Substituting Phase Difference into the Intensity Formula**: Now, we substitute \( \delta \phi \) back into the intensity formula: \[ I = I_0 \cos^2\left(\frac{\delta \phi}{2}\right) = I_0 \cos^2\left(\frac{\pi}{3 \cdot 2}\right) = I_0 \cos^2\left(\frac{\pi}{6}\right) \] 4. **Calculating \( \cos^2\left(\frac{\pi}{6}\right) \)**: We know that: \[ \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} \] Therefore: \[ \cos^2\left(\frac{\pi}{6}\right) = \left(\frac{\sqrt{3}}{2}\right)^2 = \frac{3}{4} \] 5. **Final Expression for Intensity**: Now we can express \( I \) in terms of \( I_0 \): \[ I = I_0 \cdot \frac{3}{4} \] 6. **Finding the Ratio \( \frac{I}{I_0} \)**: To find the ratio \( \frac{I}{I_0} \): \[ \frac{I}{I_0} = \frac{3}{4} \] ### Conclusion: Thus, the ratio \( \frac{I}{I_0} \) is: \[ \frac{I}{I_0} = \frac{3}{4} \]

To solve the problem, we need to find the ratio \( \frac{I}{I_0} \) where \( I \) is the intensity at a point with a path difference of \( \frac{\lambda}{6} \) and \( I_0 \) is the maximum intensity in a Young's double slit experiment. ### Step-by-Step Solution: 1. **Understanding the Intensity Formula**: In a Young's double slit experiment, the intensity \( I \) at any point on the screen can be expressed as: \[ I = I_0 \cos^2\left(\frac{\delta \phi}{2}\right) ...
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