Question:
In a Young's double slit experiment 15 fringes are observed on a small portion of the screen when light of wavelength $500 \mathrm{~nm}$ 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 (in $\mathrm{nm}$ )
Solution:
(750) Fringe width, $\beta=\frac{\lambda D}{d}$ where, $\lambda=$ wavelength,
$D=$ distance of screen from slits, $d=$ distance
between slits
ATQ
$15 \times----=10 \times \frac{\lambda_{2} D}{d}$
$\Rightarrow 15 \lambda_{1}=10 \lambda_{2}$
$\Rightarrow \lambda_{2}=1.5 \lambda_{1} 15 \lambda_{1}=1.5 \times 500 \mathrm{~nm}$
$\Rightarrow \lambda_{2}=750 \mathrm{~nm}$