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Step-by-Step Solution
Step 1: Determine the phase difference when the path difference is Ī»
The phase difference $ \phi $ is related to the path difference by the formula
$ \phi = \frac{2\pi}{\lambda} \times (\text{path difference}). $
When the path difference is $ \lambda $:
$ \phi = \frac{2\pi}{\lambda} \times \lambda = 2\pi.
$
Step 2: Express the intensity for the path difference Ī»
In Young's double slit experiment, the resultant intensity $ I $ can be written as
$ I = 4 I_0 \cos^2 \left(\frac{\phi}{2}\right). $
Substituting $ \phi = 2\pi $:
$ I = 4 I_0 \cos^2\left(\frac{2\pi}{2}\right)
= 4 I_0 \cos^2(\pi).
$
Since $ \cos(\pi) = -1 $, $ \cos^2(\pi) = 1 $, so
$ I = 4 I_0.
$
Given that this intensity is $ K $, we have
$ 4 I_0 = K. $
Step 3: Determine the phase difference when the path difference is Ī»/4
When the path difference is $ \frac{\lambda}{4} $,
$ \phi = \frac{2\pi}{\lambda} \times \left(\frac{\lambda}{4}\right) = \frac{\pi}{2}.
$
Step 4: Express the intensity for the path difference Ī»/4
Using the formula
$ I = 4 I_0 \cos^2\left(\frac{\phi}{2}\right),
$
substitute $ \phi = \frac{\pi}{2} $:
$ I = 4 I_0 \cos^2\left(\frac{\pi}{4}\right).
$
We know $ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, $ so
$ \cos^2\left(\frac{\pi}{4}\right) = \left(\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}.
$
Thus,
$ I = 4 I_0 \times \frac{1}{2} = 2 I_0.
$
Step 5: Relate 2Iā to K
From Step 2, we found $ 4 I_0 = K, $ hence $ I_0 = \frac{K}{4}. $
So,
$ 2 I_0 = 2 \times \frac{K}{4} = \frac{K}{2}.
$
Therefore, the intensity at the point where the path difference is $ \frac{\lambda}{4} $ is
$ \frac{K}{2}.
Final Answer
The intensity at a point where the path difference is $ \frac{\lambda}{4} $ is $ \frac{K}{2}.
