Physics - Wave Optics Question with Solution | TestHub

Position of the $n$-th bright fringe from the central maximum is given by $y_n=\frac{n\lambda D}{d}$

Position of the $m$-th dark fringe from the central maximum is given by $y_m^{\prime }=\frac{(2m-1)\lambda D}{2d}$

For the 10th bright fringe, $y_{10}=\frac{10\lambda D}{d}$

For the 3rd dark fringe, $y_3^{\prime }=\frac{5\lambda D}{2d}=2.5\frac{\lambda D}{d}$

The distance between them on the same side is $\Delta y=y_{10}-y_3^{\prime }=7.5\frac{\lambda D}{d}$

Given $\Delta y=8.85\text{ mm}=8.85\times 10^{-3}\text{ m}$, $D=1.2\text{ m}$, and $d=0.6\text{ mm}=0.6\times 10^{-3}\text{ m}$

Substituting the values:

$7.5\times \frac{\lambda \times 1.2}{0.6\times 10^{-3}}=8.85\times 10^{-3}$

$7.5\times \lambda \times 2000=8.85\times 10^{-3}$

$15000\lambda =8.85\times 10^{-3}$

$\lambda =\frac{8.85\times 10^{-3}}{15000}=0.59\times 10^{-6}\text{ m}$

$\lambda =5900\times 10^{-10}\text{ m}=5900\text{ A˚}$

Answer: $5900\mathring{A}$