Physics - Wave Optics Question with Solution | TestHub

For a light of wavelength $\lambda$, the fringe width in double slit experiment is given by

$\beta =\frac{\lambda D}{d} ...\left(1\right)$

For a light of wavelength $\lambda \text{'}$, the fringe width is given by

$\beta \text{'}=\frac{\lambda \text{'}D}{d} ...\left(2\right)$

Divide equation (2) by equation (1) and simplify to obtain the fringe width for the later case.

$\begin{array}{rcl}\frac{\beta \text{'}}{\beta }& =& \frac{\frac{\lambda \text{'}D}{d}}{\frac{\lambda D}{d}}\\ & =& \frac{\lambda \text{'}}{\lambda }\\ & \Rightarrow & \beta \text{'}= \frac{{\displaystyle \lambda \text{'}}}{{\displaystyle \lambda }}\beta ...\left(3\right)\end{array}$

Substitute the values of the known parameters into equation (3) to calculate the required fringe width.

$\begin{array}{rcl}\beta \text{'}& =& \frac{600 \mathrm{nm}}{400 \mathrm{nm}}\times 2 \mathrm{mm}\\ & =& 3 \mathrm{mm}\end{array}$