Question

The distance between the coherent source is $0.3mm$ and the screen is $90cm$ from the source. The second dark band is $0.3cm$ away from central bright fringe. Find the wavelength and the distance of the fourth bring fringe from central bright fringe.

Answer 1

Given,

Distance between two coherent sources, $d=0.03cm$.

Distance between sources from the screen, $D=90cm$

Second dark fringe form center bright fringe, $y_1=0.3cm$

For Dark fringe $y_1=(n+\frac{1}{2})\frac{\lambda D}{d}$ , $(n=0,1,2,\mathrm{3......})$,

for second dark band $(n=1)$, $\lambda =\frac{2y_{1}d}{D}=\frac{2\times 0.03\times 0.3}{90}=2\times {10}^{-4}cm$

For Bright fringe $y_2=\frac{n\lambda D}{d}$ $(n=0,1,2,\mathrm{3......})$

Distance of fourth bright fringe from center $(n=3)$ , $y_2=\frac{n\lambda D}{d}=\frac{3\times (2\times {10}^{-4})\times 90}{0.03}=1.8cm$

Wavelength $\lambda =2\times {10}^{-4}cm=2\mu m$

& distance of bright fringe from center bright fringe is $1.8cm$