Question

Light of wavelength$550\mathrm{nm}$ illuminates a double slit, and the interference pattern is observed on a screen. At the position of the $\mathrm{m}=2$ bright fringe, how much farther is it to the more distant slit than to the nearer slit$?$

Short answer

Distance of distant slit than to the nearer slit is$1\mathrm{\mu m}$.

Step-by-step solution

Definition for path distance

The path difference between the two slits to the fringe is an integral number of light wavelengths, constructive interference occurs, resulting in vivid fringes.

When the route difference is a half-integral number of wavelengths, destructive interference and dark fringes occur.

Calculation for distance

To find path difference first we calculate phase difference.

Phase difference is,

$∆\mathrm{phase}=2\mathrm{m\pi }$

Where,

$\mathrm{m}$is number of bright fringe

So,

$∆\mathrm{phase}=2\times 2\times \mathrm{\pi }$

$=4\mathrm{\pi }$

path difference,

$∆\mathrm{D}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }}\times ∆\mathrm{phase}$

$∆\mathrm{D}=\frac{\mathrm{\lambda }}{2\mathrm{\pi }}\times 4\mathrm{\pi }$

$∆\mathrm{D}=2\times \mathrm{\lambda }$

$∆\mathrm{D}=2\times 5\times {10}^{-7}$

$∆\mathrm{D}=1\mathrm{\mu m}$