Question

A pair of narrow slits is positioned 1.90 m from a screen and illuminated with light made up of two wavelengths 556.00 nm and 507.60 nm.

First a filter is used to block the 507.60 nm light from reaching the screen.  The resulting interference maxima are found to be separated by s = 1.60 mm.

a) What is in meters the separation of the slits?
We now remove that filter and place a filter that blocks the 556.00 nm light.

b) What is in mm the obtained fringe separation?

We remove both filters.

c) What is in meters the separation on the screen of first bright maxima due to both wavelengths?

d) What is in meters the separation on the screen of fourth bright maxima due to both wavelengths?

Answer 1

Solution:

Given:

a) wavelength 1 = 556 nm =556x10^-9 m

fringe width = y1 =1.6 mm= 1.6 x10^-3 m

screen -slit distance = D =1.9 m

slit separation = d = m 1 D /d = 18556 x10^-9 x 1.9 /(1.6 x10^-3) = 660.25 m

b) when 556 nm is blocked the wavelength incident is 507.6 nm

slit separation = d=660.25 x10^-6 m

D=1.9m

fringe separation = y2 = 2 D/d =(507.6x10^-9)(1.9)/(660.25x10^-6)

                                                 = 1.46 mm

c) Difference due to both wavelengths = 1.6-1.46 = 0.14 mm

d) y1 =m1D/d= (4 x 556 x1.9/1.6 ) m =2641 m

y2 = m2D/d =4 x 507.6 x 1.9 /1.6 m = 2411.1 m

y1-y2 = 2641 - 2411.1 = 230 m = The separation on the screen of fourth bright maxima due to both wavelengths