Question

In Young's double slit experiment, the distance between the slits is $1.9mm$. The distance between the slit and the screen is $1m$. If the bandwidth is $0.35mm$, calculate the wavelength of light used.

Answer 1

Given,
$d=1.9mm=1.9\times {10}^{-3}m$;
$D=1m$;
$\beta =0.35mm=0.35\times {10}^{-3}m$;
Now, we know that:
$\beta =\frac{D\lambda }{d}$, $\lambda =\frac{\beta d}{D}$
$\lambda =\frac{0.35\times {10}^{-3}\times 1.9\times {10}^{-3}}{1}$
$\lambda =6.65\times {10}^{-7}m=6650\mathring{A}$