Interference fringes from sodium light (1 = 5890 Å) in a double slit experiment have an angular width 0.20°. To increase the fringe width by 10%, wavelength of light used should be
Answer & explanation
Correct answer: option 4
$\beta=\frac{\lambda D}{d}$ and angular fringe width, $\theta=\frac{\beta}{D}=\frac{\lambda}{d}$
$\theta_1=\lambda_1 / d, \theta_2=\lambda_2 / d$
∴ $\frac{\theta_1}{\theta_2}=\frac{\lambda_1}{\lambda_2}$ or $\lambda_2=\lambda_1 . \frac{\theta_2}{\theta_1}$
∴ $5890 \times \frac{0.22}{0.20}$ = 6479 Å