Question

In a Young's double slit experiment, $15$ fringes are observed on a small portion of the screen, when light of wavelength $500\text{nm}$ is used. Ten fringes are observed on the same section of the screen, when another light source of wavelength $\lambda$ is used. Then the value of $\lambda$ is $\text{(in nm)}$

Answer 1

Fringe width, $\beta =\frac{\lambda D}{d}$

where, $\lambda =$ wavelngth, $D=$ distance of screen from slits, $d=$ distance between slits

So, $15\times \frac{{\lambda }_{1}D}{d}=10\times \frac{{\lambda }_{2}D}{d}$

$\Rightarrow 15{\lambda }_1=10{\lambda }_2$

$\Rightarrow {\lambda }_2=1.5{\lambda }_1=1.5\times 500$

$\Rightarrow {\lambda }_2=750\text{nm}$