A narrow slit S transmitting light of wavelength is placed a dist… - Vidyadip

Question

A narrow slit S transmitting light of wavelength $\lambda$ is placed a distance d above a large plane mirror as shown in figure (17-E1). The light coming directly from the slit and that coming after the reflection interfere at a screen $\sum$ placed at a distance D from the slit.

What will be the intensity at a point just above the mirror, i.e., just above O?
At what distance from 0 does the first maximum occur?

✓

Answer

Since, there is a phase difference of $\pi$ between direct light and reflecting light, the intensity just above the mirror will be zero.
Here, 2d = equivalent slit separation

D = Distance between slit and screen.

We know for bright fringe, $\Delta\text{x}=\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda$

But as there is a phase reversal of $\frac{\lambda}{2}.$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}+\frac{\lambda}{2}=\text{n}\lambda$

$\Rightarrow\frac{\text{y}\times2\text{d}}{\text{D}}=\text{n}\lambda-\frac{\lambda}{2}\Rightarrow\text{y}=\frac{\lambda\text{D}}{4\text{d}}$

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