Question

A double-slit arrangement produces interference, in which angular fringe width is ${0.25}^{\circ }$, using the wavelength of light 600nm. If angular fringe width is decreased to ${0.2}^{\circ }$ by changing wavelength of the light used, then the new wavelength must be

• A. $480nm$
• B. $540nm$
• C. $720nm$
• D. $840nm$

Answer 1

The correct option is A $480nm$

The optical path difference is given by

$dsin\theta =n\lambda$

For small values of $\theta$, $sin\theta \simeq \theta$

Therefore, $d\theta =\lambda$, since $n=1$

$⟹d=\frac{\lambda }{\theta }$

If we consider, two angles then we have

$d=\frac{{\lambda }_1}{{\theta }_1}=\frac{{\lambda }_2}{{\theta }_2}$

Given ${\lambda }_1=600nm$, ${\theta }_1={0.25}^o$ and ${\theta }_2={0.2}^o$

$⟹\frac{600\times {10}^{-9}}{0.25}=\frac{{\lambda }_2}{0.2}$

$⟹{\lambda }_2=\frac{600\times {10}^{-9}\times 0.2}{0.25}$

$⟹{\lambda }_2=480nm$