Question

In a single slit diffraction experiment first minimum for lambda 1 is equal to 660 nm coincides with first maximum for wavelength lambda to calculate lambda 2

Answer 1

Answer:

final answer

Explanation:

first maxima approximately lies between first and second maxima. for wavelength lambda $2$ its position will be

                  $\begin{aligned}a \sin \theta_{2} &=\frac{3}{2} \lambda_{2} \\\sin \theta_{2} &=\frac{3 \lambda_{2}}{2 a}\end{aligned}$

Two will coincide if

$\begin{aligned}&\theta_{1}=\theta_{2} \text { or } \sin \theta_{1}=\sin \theta_{2} \\&\frac{\lambda_{1}}{\mathrm{a}}=\frac{3 \lambda_{2}}{2 \mathrm{a}} \\&\lambda_{2}=\frac{2}{3} \lambda_{1}=\frac{2}{3} \times 660 \mathrm{~nm}=440 \mathrm{~nm}\end{aligned}$