Physics - Wave Optics Question with Solution | TestHub

Let the length of the segment on the screen be $y$.

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

The number of fringes $n$ in the segment $y$ is $n=\frac{y}{\beta }=\frac{yd}{\lambda D}$.

This implies that $n\lambda =\text{constant}$, or $n_1{\lambda }_1=n_2{\lambda }_2$.

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

Substituting the values, we get:

$18\times 400=n_2\times 600$

$n_2=\frac{18\times 400}{600}=12$

Thus, 12 fringes will be observed in the same segment.

Answer: $12$