A flat screen is located 0.55 m away from a single slit. Light with a wavelength of 544 nm (in vacuum) shines through the slit and produces a diffraction pattern. The width of the central bright fringe on the screen is 0.052 m. What is the width of the slit?

Explanation:

The diffraction pattern is given as

Sinθ = mλ/ω

Where m=1,2,3,4

Now, when m=1

Sinθ = λ/ω

Then,

ω = λ/Sinθ

The width of the central bright fringe is given as

y=2Ltanθ. From trigonometric

Then,

θ=arctan(y/2L)

Given that,

y=0.052m

L=0.55m

θ=arctan(0.052/2×0.55)

θ=arctan(0.0473)

θ=2.71°

Substituting this into

ω = λ/Sinθ

Since λ=544nm=544×10^-9m

Then,

ω = 544×10^-9/Si.2.71

ω = 1.15×10^-5m.

which three statements are true about all alloys?

the components of a vector are the parts of the vectors that are to each other.

b perpendicular