Physics - Wave Optics Question with Solution | TestHub

The fringe width in Young's double slit experiment is given by $\beta =\frac{\lambda D}{d}$.

Let the total width of the field of view on the screen be $W$. The number of fringes $n$ observed in this width is given by $n=\frac{W}{\beta }$.

Therefore, $n\propto \frac{1}{\lambda }$, which implies $n_1{\lambda }_1=n_2{\lambda }_2$.

Given $n_1=18$, ${\lambda }_1=600$ nm, and ${\lambda }_2=400$ nm.

Substituting the values:

$18\times 600=n_2\times 400$

$n_2=\frac{18\times 600}{400}=27$

The number of fringes observed on the screen is $27$.

Answer: $27$