Question

Coherent light from a sodium-vapor lamp is passed through a filter that blocks everything except for light of a single wavelength. It then falls on two slits separated by 0.460 mm. In the resulting interference pattern on a screen 2.20 m away, adjacent bright fringes are separated by 2.82 mm. What is the wavelength of the light that falls on the slits

Answer 1

Answer:

$5.896\times 10^{-7}\ \text{m}$

Explanation:

D = Distance of the screen from the light source = 2.2 m

d = Distance between slits = 0.46 mm

m = Order

Distance between adjacent bright fringes is 2.82 m

$y_{m+1}-y_m=2.82\ \text{mm}\\\Rightarrow \dfrac{D(m+1)\lambda}{d}-\dfrac{Dm\lambda}{d}=2.82\times 10^{-3}\\\Rightarrow \dfrac{D\lambda}{d}(m+1-m)=2.82\times 10^{-3}\\\Rightarrow \lambda=\dfrac{d}{D}2.82\times 10^{-3}\\\Rightarrow \lambda=\dfrac{0.46\times 10^{-3}\times 2.82\times 10^{-3}}{2.2}\\\Rightarrow \lambda=5.896\times 10^{-7}\ \text{m}$

The wavelength of the light that falls on the slits is $5.896\times 10^{-7}\ \text{m}$.