Question

In a single slit diffraction experiment first minimum of red light $\left(660 \, nm\right)$ coincides with first maximum of some other wavelength $\left(\right.in \, nm\left.\right)$ $\lambda $ . Calculate $\lambda $ .

Answer 1

Secondary minimum
$sin \theta =\frac{n\lambda }{a}$
$sin \theta =\frac{n\lambda _{r}}{d}$
$n=1$
$sin \theta =\frac{\left(\lambda \right)_{r}}{a}\ldots \ldots \left(\right.i\left.\right)$
Secondary maximum
$sin \theta =\frac{\left(\right. 2 n + 1 \left.\right) \lambda }{2 a}$
$n=1$
$sin \theta =\frac{3 \lambda }{2 d}\ldots \ldots \left(\right.ii\left.\right)$
$\left(i\right)=\left(ii\right)$
$\frac{\lambda _{r}}{d}=\frac{3 \lambda }{2 d}$
$\lambda =\frac{2 \lambda _{r}}{3}$
$=\frac{2 \times 660 \times 10^{- 9}}{3}$
$\lambda =440 \, nm$