Question

Interference fringes from sodium light $(\lambda =5890\mathring{A})$ in a double slit experiment have an angular width $0.20^{\circ }$ To increase the fringe width by $10\%$, new wavelength should be

• A. $5896\mathring{A}$
• B. $7321\mathring{A}$
• C. $6300\mathring{A}$
• D. $6479\mathring{A}$

Answer 1

Answer: (D)

${\theta }_1=\frac{{\lambda }_1}{d},{\theta }_2=\frac{{\lambda }_2}{d}$
$\therefore \frac{{\theta }_1}{{\theta }_2}=\frac{{\lambda }_1}{{\lambda }_2}$
$\therefore {\lambda }_2={\lambda }_1\times \frac{{\theta }_2}{{\theta }_1}$
or ${\lambda }_2=5890\times \frac{0.22}{0.20}$
$=6479\mathring{A}$