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Question
Find the intensity at a point on a screen in Young's double slit experiment where the interfering waves have a path difference of (i) λ/6, and (ii) λ/2.
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Solution
Resultant intensity of the interfering waves can be given by
I' = I1 + I2 +2√I1I2 cosδ
where I1 and I2 are the intensities of light emitted by first and second source respectively and δ is the phase difference between two waves emitted by light sources.
In Young's double slit experiment since a single light source is used. Therefore I1 = I2
This implies, I' = 2I + 2Icosδ
(i) If path difference is λ/6
\[\text { Then phase difference }, \delta = ( \frac{2\pi}{\lambda} ) \times \text { path difference } \]
\[ = \frac{2\pi}{\lambda} \times \frac{\lambda}{6} \]
\[ = \frac{\pi}{3}\]
∴ I' = 2I +2I cos(π/3)
= 2I + 2I(1/2)
= 2I + I
= 3I
(ii) If path difference is λ/2
\[\text { Then phase difference }, \delta = ( \frac{2\pi}{\lambda} ) \times \text { path difference } \]
\[ = \frac{2\pi}{\lambda} \times \frac{\lambda}{2} \]
\[ = \pi\]
I' = 2I +2I cos(π)
= 2I + 2I (-1)
= 2I - 2I
= 0
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