Question

A diffraction pattern is formed on a screen 120 cm away from a 0.4 mm wide slit.

Answer 1

Answer:

Solution

First we find where we are. The angle to the side is small so

sinθ≈tanθ=

L

y

=

1.20m

4.10×10

−3

m

=3.417×10

−3

The parameter controlling the intensity is

λ

πasinθ

=

546.1×10

−9

m

π(4.00×10

−4

m)(3.417×10

−3

)

=7.862rad

This is between 2π and 3π, so the point analyzed is off in the second side fringe. The fractional intensity is

I

max

I

=[

πasinθ/λ

sin(πasinθ/λ)

]

2

=[

7.862rad

sin(7.862rad)

]

2

=[1.62×10

−2

Solve