A pair of narrow slits is positioned 1.10 m from a screen and illuminated with l

Question

A pair of narrow slits is positioned 1.10 m from a screen and illuminated with light made up of two wavelengths 464.00 nm and 622.00 nm.

First a filter is used to block the 622.00 nm light from reaching the screen.  The resulting interference maxima are found to be separated by s = 0.90 mm.

a) What is in meters the separation of the slits?

We now remove that filter and place a filter that blocks the 464.00 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?

Explanation / Answer

In interfreence or diffraction pattern,

the needed equation is Y = mLR/d---------------1

and d sin theta = mL--------------------2

where L = wavelgnth,

m = order = 1,2,3,4, ......... for brigth bands

m = 1.5, 2.5, 3.5, 4.5, ......for dark bands

R is the distance from slit to screen,

Y = disatnce from central spot to nth order fringe

so here

part A:

apply Y = mLR/d

so

d = mLR/Y

d = (1 * 622 e-9 * 1.1)/(0.9 e-3)

d = 0.76 mm

----------------------

b. Y = 1 * 464 e-9 * 1.1/(0.76 e-3)

Y = 0.671 mm

---------------------------

c.. Y2-Y1 = 0.9-0.671 = 0.229 mm

---------------------------------

d.

Y4   = 4 * (622-464)*1 e-9 * 1.1/(0.76 e-3)

Y4 = 9.14 *10^-4 m

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