Question

A double-slit arrangement produces interference fringes that are ${0.25}^0$ apart using wavelength$00nm$.if angular separation is increased to ${0.3}^0$ by changing the length of the light used, then the new wavelength must be :

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

Answer 1

Answer: (C)

$720mm$

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 ${\theta }_2={0.3}^o$

$⟹\frac{600\times {10}^{-9}}{0.25}=\frac{{\lambda }_2}{0.3}$

$⟹{\lambda }_2=\frac{600\times {10}^{-9}\times 0.3}{0.25}$

$⟹{\lambda }_2=720nm$