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Light Waves — Physics STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 SciencePhysicsLight Waves3 Marks
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.
  1. What will be the intensity at a point just above the mirror, i.e., just above $O?$
  2. At what distance from $0$ does the first maximum occur?
✓

Answer

  1. Since, there is a phase difference of $\pi$ between direct light and reflecting light, the intensity just above the mirror will be zero.
  2. 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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