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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?

User Gnarf
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1 Answer

5 votes

Answer:


1.94*10^(-3) m

Step-by-step explanation:

Condition for constructive interference is


y =(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 =(1*660*10^(-9))/(0.49*10^(-3)) 5


y_1=6.73*10^(-3) m

Now, for first order 2nd wavelength


y_2 =(1*470*10^(-9))/(0.49*10^(-3)) 5


y_2=4.79*10^(-3) m

The distance between the first bright fringe for each wavelength


d=y_1-y_2\\=(6.73-4.79)*10^(-3) m\\=1.94*10^(-3) m

User Amr Magdy
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