In a single slit diffraction experiment first minimum for red lig… - Vidyadip

MCQ

In a single slit diffraction experiment first minimum for red light ($ \gamma = 660\, nm$) coincides with first maximum of some other wavelength $\lambda$'. The value of $\lambda$' is......$\mathop A\limits^o $

✓
$4400 $
B
$6600 $
C
$2000$
D
$3500$

✓

Answer

Correct option: A.

$4400 $

a
(a)In a single slit diffraction experiment, position of minima is given by $d\sin \theta = n\lambda $
So for first minima of red $\sin \theta = 1 \times \left( {\frac{{{\lambda _R}}}{d}} \right)$
and as first maxima is midway between first and second minima, for wavelength $\lambda '$,
its position will be
$d\sin \theta ' = \frac{{\lambda ' + 2\lambda '}}{2} \Rightarrow \sin \theta ' = \frac{{3\lambda '}}{{2d}}$
According to given condition $\sin \theta = \sin \theta '$
$ \Rightarrow \lambda ' = \frac{2}{3}{\lambda _R}$ so $\lambda ' = \frac{2}{3} \times 6600 = 440\,nm$$ = 4400\,{Å}$

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