Physics - Wave Optics Question with Solution | TestHub

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

For the $3^{\text{rd}}$ order bright fringe of wavelength ${\lambda }_1$:
$y_1=\frac{3{\lambda }_{1}D}{d}$

For the $3^{\text{rd}}$ order bright fringe of wavelength ${\lambda }_2$:
$y_2=\frac{3{\lambda }_{2}D}{d}$

The separation between the two fringes on the screen is:
$\Delta y=y_2-y_1=\frac{3D}{d}({\lambda }_2-{\lambda }_1)$

Given ${\lambda }_2=1.5{\lambda }_1$, $D=1\text{ m}$, and $d=2\text{ mm}=2\times 10^{-3}\text{ m}$, substituting these values we get:
$\Delta y=\frac{3\times 1}{2\times 10^{-3}}(1.5{\lambda }_1-{\lambda }_1)$

$\Delta y=\frac{3}{2\times 10^{-3}}(0.5{\lambda }_1)$

$\Delta y=\frac{1.5}{2}\times 10^3{\lambda }_1=\left(0.75\times 10^{+3}\right){\lambda }_1$

Answer: $\left(0.75\times 10^{+3}\right){\lambda }_1$