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Question
The line-width of a bright fringe is sometimes defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum. Find the line-width of a bright fringe in a Young's double slit experiment in terms of \[\lambda,\] d and D where the symbols have their usual meanings.
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Solution
Given:-
Separation between two slits = d
Wavelength of the light = \[\lambda\]
Distance of the screen = D
Let Imax be the maximum intensity and I be half the maximum intensity at a point at a distance y from the central point.
So, \[I = a^2 + a^2 + 2 a^2 \cos\phi\]
Here, \[\phi\] is the phase difference in the waves coming from the two slits.
So, \[I = 4 a^2 \cos^2 \left( \frac{\phi}{2} \right)\]
\[\Rightarrow \frac{I}{I_\max} = \frac{1}{2}\]
\[ \Rightarrow \frac{4 a^2 \cos^2 \left( \frac{\phi}{2} \right)}{4 a^2} = \frac{1}{2}\]
\[ \Rightarrow \cos^2 \left( \frac{\phi}{2} \right) = \frac{1}{2}\]
\[ \Rightarrow \cos\left( \frac{\phi}{2} \right) = \frac{1}{\sqrt{2}}\]
\[ \Rightarrow \frac{\phi}{2} = \frac{\pi}{4}\]
\[ \Rightarrow \phi = \frac{\pi}{2}\]
Corrosponding path difference, \[∆ x = \frac{1}{4}\]
\[ \Rightarrow y = \frac{∆ xD}{d} = \frac{\lambda D}{4d}\]
The line-width of a bright fringe is defined as the separation between the points on the two sides of the central line where the intensity falls to half the maximum.
So, line-width = 2y
\[= 2\frac{D\lambda}{4d} = \frac{D\lambda}{2d}\]
Thus, the required line width of the bright fringe is \[\frac{D\lambda}{2d}.\]
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