Question

Single-Slit Diffraction?

Single-Slit Diffraction

1. You have been asked to measure the width of a slit in a piece of paper. You mount the paper 80 centimeters from a screen and illuminate it from behind with laser light of wavelength 633 nanometers (in air). You mark two of the intensity minima as shown in the figure, and measure the distance between them to be 17.9 millimeters

Express your answer in micrometers, to three significant figures.

2.If the entire apparatus were submerged in water, would the width of the central peak change?

a. The width would increase.

b.The width would decrease.

c.The width would not change.

here's the link to the page :D

Answer 1

(1) In single slit diffraction ,for position of minima ,path difference d sine theta =n*wavelength, where n is order of minima.In the figure shown 17. 9 mm is total distance between 3 rd minima on either side.

therefore 2y=17.9

y=8.95 mm

sine theta=tan theta=y/screen distance'D'

d sine theta= d * y/ screen distance 'D

d * y/ screen distance 'D= 3 wavelength

d =3wavelength*D / y

=3*633*E-9*0.8/8.95*E-3

=169.743 micro meter.

(2)width in water= width in air / refractive index of water

hence width in water will decrease ( choice b correct)

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Answer 2

Have you considered that the thing the slit is cut out of is made of matter? And that matter, being made of protons, electrons, and neutrons, will influence the shooting protons through the strong nuclear force, the weak nuclear force, electromagnetism, and gravity? And that the position of a proton as it goes through the slit will not be known, but actually be a "cloud" of quantum probability, which will result in lots of protons being affected by the matter in the slit slightly differently, resulting in basically an enlarged "cloud" of quantum probability at the output end, which is exactly what a Fraunhofer diffraction pattern is (though he didn't know it at the time he came up with the idea)? This one really isn't at all difficult to understand. Peace.

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Answer 3

Let denote a the width of the slit.

In single slit experiment amplitude of electric field of

light in direction α is calculated as superposition

of amplitudes of dE of small elements dx of the slit.

dE = Eo exp(2πi r(x) / λ) dx =

= Eo exp(2πi [x sin α] / λ) dx

E(α) = integral [-a/2 .. +a/2] of Eo exp(2πi [x sin α] / λ) dx =

= int [-a/2 .. +a/2] of 1/(2πi) λ/sinα Eo exp(2πi [x sin α] / λ) d [2πisinα/λ x]] =

= int [-πiasinα/λ .. +πiasinα/λ] of 1/(2πi) λ/sinα Eo exp(z) dz =

-1/π λ/sinα sin(π a sinα / λ) Eo

The intensity A(α) of light is proportional to square of electic

field E(α):

A(α) = 1/π² (λ/sinα)² sin²(πa sinα / λ) Eo² = Ao (λ/sinα)² sin²(πa sinα / λ)

Minima of intensity occur when

πa sinα / λ = n π, n = 1,2,3...

Answer:

a = n λ/sinα =

3 x 633 nm / (0.5 x 17.9 mm / 80 cm) = 170 μm

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