Question

Coherent light that contains two wavelengths, 660 nmnm (red) and 470 nmnm (blue), passes through two narrow slits that are separated by 0.490 mmmm. Their interference pattern is observed on a screen 4.50 mm from the slits.Required:What is the distance on the screen between the first order bright fringes for the two wavelengths?

Answer 1

Answer:

$1.94\times10^{-3}$ m

Explanation:

Condition for constructive interference is

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

y= width of the first bright fringe

λ= wavelength of the incident light

d= distance between the slits

D= distance of the screen from the slit

for first order 1st wavelength

$y_1 =\frac{1\times660\times10^{-9}}{0.49\times10^{-3}} 5$

$y_1=6.73\times10^{-3} m$

Now, for first order 2nd wavelength

$y_2 =\frac{1\times470\times10^{-9}}{0.49\times10^{-3}} 5$

$y_2=4.79\times10^{-3} m$

The distance between the first bright fringe for each wavelength

$d=y_1-y_2\\=(6.73-4.79)\times10^{-3} m\\=1.94\times10^{-3} m$