Review  

Single/  Double  Slit,  Diffrac6on  Gra6ngs  

Young’s Double Slit Experiment •  Constructive Interference •  Order number refers to

BRIGHT fringe locations.

•  Destructive Interference

Single Slit Diffraction

•  Order number relates to the minima

where m = 1, 2, 3, ….

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Dsinθ = mλ

DyL

⎛

⎝ ⎜

⎞

⎠ ⎟ = mλ

Variation in slit width

Question

In a single slit diffraction experiment, as the width of the slit is made smaller, the width of the central maximum of the diffraction pattern becomes

A)  Smaller B)  Larger C)  Remains the same

Question

In a single slit diffraction experiment, as the width of the slit is made smaller, the width of the central maximum of the diffraction pattern becomes

A)  Smaller B)  Larger C)  Remains the same

Diffraction Grating

•  Order number refers to maxima locations

•  Many slits

•  Cannot approximate

Spectroscope

Mercury Vapor

Helium Vapor

White Light

Example #1

A helium neon laser is used with a diffraction grating containing 6000 lines/cm. The first order fringe is located at an angle of 22.31° from central.

A) What is the wavelength of the laser? B) How far from central is the first order if

the screen is 1.2 m away from the grating?

C) What is the third order angle?

Lloyd’s Mirror

•  An arrangement for producing an interference pattern with a single light source

•  Waves reach point P either by a direct path or by reflection

•  The reflected ray can be treated as a ray from the source S’ behind the mirror

•  So, while the picture shows what looks like a double slit experiment, we might expect the same interference pattern. But that’s not what we get if one of the “sources” is the result of reflection.

•  There must be something else going on.

The Explanation:

Reflected light is sometimes inverted, resulting in destructive interference that is not because of path difference alone (as in Young’s double slit experiment).

Phase Changes Due To Reflection

•  An electromagnetic wave undergoes a phase change of 180° (ie π, or ½ of a cycle) upon reflection from a medium of higher index of refraction than the one in which it was traveling –  Analogous to a pulse on

a string reflected from a rigid support

A phase shift (or inversion) occurs: No phase shift occurs:

Interference by Thin Films   Thin film interference patterns seen in

Thin film of soapy water

A thin layer of oil on the Water of a street puddle

Seashell

Parallel-sided Thin Film   Consider a film of soap with uniform thickness

in air When a beam of light is incident on to the surface of the film, part of incident light is reflected on the top surface and part of that transmitted is reflected on the lower surface.

If the film is not too thick, the two reflected beams produces an interference effect.

t air

Soap film

Parallel-sided Thin Film   If light travelling in a less dense medium

is reflected by a dense medium, the reflected wave is phase-shifted by π.

  If light travelling in a dense medium is reflected by a less dense medium, the reflected wave does not experience any phase shift.

Parallel sided Thin Film   However, the part reflected at the lower

surface must travel the extra distance of 2 t, where t is the thickness of the film.

  That is, 2t is the path difference between the two reflected beams.

  If 2t = (m +½) λ then constructive interference occurs (since one phase shift due to reflection is cancelled by a second phase shift due to path difference, 2t).

  If 2t = mλ then destructive interference occurs.   When t is large, several values of λ satisfy the

equation. The film will appear to be generally illuminated.

Parallel sided Thin Film   Constructive interference occurs if the path

difference, 2t, between the two reflected light beams is

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(m +12)λn Where m = 0, 1, 2, …

€

mλn Where m = 0, 1, 2, …

  Destructive interference occurs if the path difference, 2t, between the two reflected light beams is

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λn =λon

  And if the film has a refractive index n then we know

Interference in Thin Films •  (Assume the light rays are traveling in

air nearly normal to the two surfaces of the film)

•  Ray 1 undergoes a phase change of 180° with respect to the incident ray

•  Ray 2, which is reflected from the lower surface, undergoes NO phase change with respect to the incident wave

•  For constructive interference

2t = (m + ½)λn (m = 0, 1, 2 …)

•  For destructive interference

2t = mλn (m = 0, 1, 2 …)

Interference in Thin Films •  For constructive interference

2t = (m + ½)λn (m = 0, 1, 2 …)

And since

2nt = (m + ½) λo

•  For destructive interference 2t = mλn (m = 0, 1, 2 …)

so 2nt = mλo

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λn =λon

must travel an extra distance of 2t before the waves recombine in the air above the surface.

The bottom layer does not have to be air for these equations to apply. Here, the bottom layer is water. And since 1.33 is still less than 1.5, there is only one phase shift from reflection, so same equations.

Wha What if…

n = 1.45

If both surfaces cause phase shift, then

for destructive interference:

constructive interference:

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2nt = (m + 12)λ

Equations for Thin Film Interference •  If n1 > n2, then no phase change. •  If n1 < n2, then a phase shift of 180° (half of a wave cycle)

If one surface causes phase shift , then destructive interference constructive interference

If both surfaces cause phase shift, then destructive interference

constructive interference €

2nt = (m + 12)λ

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2nt = (m + 12)λ

Example #2 A) Calculate the minimum thickness of soap

bubble film that will result in constructive interference in the reflected light if the film is illuminated by light with a wavelength in free space of 602 nm.

B)  What other film thickness will produce constructive interference?

Semiconductors such a silicon are used to build solar cells. They are coated with a transparent thin film, whose index of refraction is 1.45, in order to minimize reflected light. If the index of refraction of silicon is 3.5, what is the minimum width of the coating that will produce the least reflection at a wavelength of 552nm?

n=1.45

Semiconductors such a silicon are used to build solar cells. They are coated with a transparent thin film, whose index of refraction is 1.45, in order to minimize reflected light. If the index of refraction of silicon is 3.5, what is the minimum width of the coating that will produce the least reflection at a wavelength of 552nm?

Both rays undergo 180 phase changes at reflection, therefore for destructive interference (no reflection), the distance travelled (twice the thickness) should be equal to half a wavelength in the coating n=1.45

Semiconductors such a silicon are used to build solar cells. They are coated with a transparent thin film, whose index of refraction is 1.45, in order to minimize reflected light. If the index of refraction of silicon is 3.5, what is the minimum width of the coating that will produce the least reflection at a wavelength of 552nm?

Both rays undergo 180 phase changes at reflection, therefore for destructive interference (no reflection), the distance travelled (twice the thickness) should be equal to half a wavelength in the coating n=1.45

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2nt =λ2

Semiconductors such a silicon are used to build solar cells. They are coated with a transparent thin film, whose index of refraction is 1.45, in order to minimize reflected light. If the index of refraction of silicon is 3.5, what is the minimum width of the coating that will produce the least reflection at a wavelength of 552nm?

Both rays undergo 180 phase changes at reflection, therefore for destructive interference (no reflection), the distance travelled (twice the thickness) should be equal to half a wavelength in the coating n=1.45

€

2nt =λ2

€

t =λ4n

Semiconductors such a silicon are used to build solar cells. They are coated with a transparent thin film, whose index of refraction is 1.45, in order to minimize reflected light. If the index of refraction of silicon is 3.5, what is the minimum width of the coating that will produce the least reflection at a wavelength of 552nm?

Both rays undergo 180 phase changes at reflection, therefore for destructive interference (no reflection), the distance travelled (twice the thickness) should be equal to half a wavelength in the coating n=1.45

€

2nt =λ2

€

t =λ4n = 95.2 nm