Question

In a single slit diffraction experiment, first minimum for red light $\left(660\,nm\right)$ coincides with first maximum of some other wavelength $\lambda ^{'}.$ The value of $\lambda ^{'}$ is

• A. $4400\,\mathring{A}$ .
• B. $6600\,\mathring{A}$
• C. $2000\,\mathring{A}$
• D. $3500\,\mathring{A}$

Answer 1

Answer: (A)

In a single slit diffraction experiment, position of minima is given by $dsin\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
$dsin\theta ^{'}=\frac{\lambda ^{'} + 2 \lambda ^{'}}{2}\Rightarrow sin\theta ^{'}=\frac{3 \lambda ^{'}}{2 d}.$
According to given condition $sin\theta =sin\theta '$
$\Rightarrow \lambda ^{'}=\frac{2}{3}\lambda _{R}$
So, $\lambda ^{'}=\frac{2}{3}\times 6600=440nm=4400\,\mathring{A}$ .