Question

Coherent light that contains two wavelengths, 660 nm (red) and 470 nm (blue), passes through two narrow slits that are separated by 0.300 mm. Their interference pattern is observed on a screen 4.00 m from the slits. What is the distance on the screen between the first-order bright fringes for the two wavelengths?

Answer 1

Answer:

0.00254 m or 2.54 mm

Explanation:

y = Fringe distance

d= Distance between slits = 0.3 mm

L = Screen distance = 4 m

$\lambda$ = Wavelength

In the case of constructive interference

$\frac{y}{L}=\frac{m\lambda}{d}$

For first order wavelength

$\frac{y_1}{4}=\frac{1\times 660\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y_1=4\times \frac{1\times 660\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y_1=0.0088\ m$

$\frac{y_2}{4}=\frac{1\times 470\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y_2=4\times \frac{1\times 470\times 10^{-9}}{0.3\times 10^{-3}}\\\Rightarrow y_2=0.00626\ m$

The distance between the fringes is given by

$y_1-y_2=0.0088-0.00626=0.00254\ m$

The distance on the screen between the first-order bright fringes for the two wavelengths is 0.00254 m or 2.54 mm