Question

The interference fringes for sodium light of wavelength 5980angstrom in a double slit experiment have an angular width of 0.20degree . For what wavelength will this width be 10% greater.

Answer 1

Dear Student,


Angular width of fringes in an interference pattern is given by

$$\begin{gathered} \theta =\frac{\lambda }{d} \\ \Rightarrow \theta \propto \lambda \\ \mathrm{New}\mathrm{angular}\mathrm{width}\theta \text{'}=\theta +\frac{10}{100}\theta =1.1\theta \\ \therefore \frac{\theta \text{'}}{\theta }=\frac{\lambda \text{'}}{\lambda } \\ \Rightarrow \lambda \text{'}=1.1\times \lambda =1.1\times 5980\dot{\mathrm{A}} \\ \Rightarrow \lambda \text{'}=6578\dot{\mathrm{A}} \end{gathered}$$