Question

A double-slit arrangement produces interference fringes for sodium light of, wavelength $589\text{nm}$. For what wavelength, would the fringe width, be $10$% greater?

• A. $527\text{nm}$
• B. $648\text{nm}$
• C. $722\text{nm}$
• D. $449\text{nm}$

Answer 1

The correct option is B $648\text{nm}$

The fringe width is given by,
$\beta =\frac{\lambda D}{d}$

As $d$ and $D$ remain constant, $\beta \propto \lambda$

$\Rightarrow$ for fringe width to be $10$% greater, $\lambda$ should be $10$% greater.

Therefore, new wavelength is,
${\lambda }^{\prime }=(589+\frac{589\times 10}{100})\approx 648\text{nm})$

Hence, $\left(B\right)$ is the correct answer.