Question

Ramon has a coherent light source with wavelength 521 nm. He wishes to send light through a double slit with slit separation of 1.50 mm to a screen 90.0 cm away. What is the minimum width of the screen if Ramon wants to display three interference maxima?

Answer 1

the width of the screen has to be:

$w = 2y = \frac{2m\lambda L}{d}$

for m = 3 and using given data we have:

$w =1.8756\times 10^{-3} m$

Answer 2

lambda = 521 nm
d= 1.5 mm
D = 90 cm

three maximas => 2 maximas on either side of the central maxima.
distance of the first maxima from the central maxima (y) = lambda*D/d
= 521x10^(-9) * 90x10^(-2)/1.5x10^(-3) = 3.13 x10^-4 m = 0.313 mm
hence the screen should be 2*.313 mm = .626 mm to get 3 maximas