Physics - Wave Optics Question with Solution | TestHub

Given:

$d=3\text{ mm}=3\times 10^{-3}\text{ m}$

$D=2\text{ m}$

${\lambda }_1=480\text{ nm}=480\times 10^{-9}\text{ m}$

${\lambda }_2=600\text{ nm}=600\times 10^{-9}\text{ m}$

The position of the $n^{\text{th}}$ order bright fringe is given by $y_n=\frac{n\lambda D}{d}$.

The separation between the $5^{\text{th}}$ order bright fringes of the two patterns is:

$\Delta y=y_{5,2}-y_{5,1}=\frac{5D}{d}({\lambda }_2-{\lambda }_1)$

Substituting the given values:

$\Delta y=\frac{5\times 2}{3\times 10^{-3}}(600-480)\times 10^{-9}$

$\Delta y=\frac{10}{3\times 10^{-3}}\times 120\times 10^{-9}$

$\Delta y=400\times 10^{-6}\text{ m}$

$\Delta y=4\times 10^{-4}\text{ m}$

Answer: $4\times 10^{-4}\text{m}$