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 $\lambda_{2}=5890 \times \frac{0.22}{0.20}$
$=6479 \, \mathring{A}$