The single slit diffraction pattern shown below was made with light with λ = 630 nm. The screen was 2.8 m from the slit. If the slit had a width 7 mm, what is the width w of the central maximum in cm?

Question

anonymous · Answer

is it 0.0252 cms????

anonymous · Answer

Use this formula and get the answer....
The distance on the screen from the center to either edge of the central maximum is
 
y = L tan(theta)=Lsin(theta)=L(lambda/a). 
Hope it helps! :D

anonymous · Answer

so would it be 280(cm) *(630E-9/.7cm)? @aryandecoolest

anonymous · Answer

which I get 2.52E-4 which keeps telling me its wrong

anonymous · Answer

Oh is that the case. let me work out :)

anonymous · Answer

So here it goes...!! sorry it took tie

The width of the central maximum can be determined as the distance between first minima: 

d = 2L*sin(alpha) 

where L=2.8 m is the distance between the slit and screen and alpha is the angle for direction to the first minimum. Because 
sin(alpha) = lambda/a, 

d = 2L*lambda/a = 2*2.8*630e-9/0.007 = 0.00504 m = 0.504 cm.

Hope it helps :D