Question

In a single slit diffraction experiment, the first minimum for $\lambda _{1}=600nm$ coincides with the first maximum for wavelength $\lambda _{2}$ . Calculate $\lambda _{2}$ .

• A. $400nm$
• B. $440nm$
• C. $600nm$
• D. $660nm$

Answer 1

Answer: (A)

In the case of a single-slit experiment, for minima,
$asin\theta =n\lambda $ ,
and for maxima,
$asin\theta =\frac{\left(2 n + 1\right)}{2}\lambda $ .
So, for the first minima and maxima,
$\frac{a sin \theta }{a sin \theta }=\frac{1 \left(\lambda \right)_{1}}{\left(\left(3 \left(\lambda \right)_{2}\right)/2\right)}\Rightarrow \frac{3 \left(\lambda \right)_{2}}{2}=600nm$
$\Rightarrow \lambda _{2}=400nm$ .