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प्रश्न
A narrow slit S transmitting light of wavelength λ is placed a distance d above a large plane mirror, as shown in the following figure. The light coming directly from the slit and that coming after the reflection interfere at a screen ∑ placed at a distance D from the slit. (a) What will be the intensity at a point just above the mirror, i.e. just above O? (b) At what distance from O does the first maximum occur?
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उत्तर
(a) The phase of a light wave reflecting from a surface differs by \[\pi\] from the light directly coming from the source.
Thus, the wave fronts reaching just above the mirror directly from the source and after reflecting from the mirror have a phase difference of \[\pi,\] which is the condition of distractive interference. So, the intensity at a point just above the mirror is zero.
(b) Here, separation between two slits is 2d.
Wavelength of the light is `lambda.`
Distance of the screen from the slit is `D.`
Consider that the bright fringe is formed at position y. Then,
path difference, \[∆ x = \frac{y \times 2d}{D} = n\lambda.\]
After reflection from the mirror, path difference between two waves is \[\frac{\lambda}{2}.\]
\[\Rightarrow \frac{y \times 2d}{D} = \frac{\lambda}{2} + n\lambda\]
For first order, put n = 0
\[ \Rightarrow y = \frac{\lambda D}{4d}\]
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