1© Manhattan Press (H.K.) Ltd. Young’s double slit experiment Young’s double slit experiment 9.10 Interference of light waves Relationship between x,,

© Manhattan Press (H.K.) Ltd. 1

• • Young’s double slit experimentYoung’s double slit experiment

9.10 Interference of 9.10 Interference of light waveslight waves

• • Relationship between Relationship between xx, , , , DD and and aa

• • Changes in the interference patternChanges in the interference pattern

• • Measurement of the wavelength of Measurement of the wavelength of lightlight

• • Energy distributionEnergy distribution


Interference of light waves

9.10 Interference of light waves (SB p. 76)

Conditions required for interference of light:1. Light waves from coherent sources2. Same or almost the same amplitudes3. Small separation between two coherent sources4. Small path difference


Young’s double slit experiment


By Huygen’s principle,- C sends out circular wavefronts- Points at A and B emit wavelets in phase- This method is called division of wavefront


Young’s double slit experiment


Superposition of waves from A and B- interference- constructive interference (bright fringes)- destructive interference (dark fringes)


PB – PA = a sin = n (constructive interference at P)and sin = xn /D

Relationship between x, , D and a


1. Bright fringe

aDnx,

aDnx nn 11

aDx separation Fringe


a sin = (n - ) (destructive interference at P)and sin = xn /D

Relationship between x, , D and a


2. Dark fringe

aDnx,

aDnx nn

2

1121

1

aDx separation Fringe

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Example 11Example 11

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Changes in the interference pattern


1. Change in the space between S and double slit

Move slit S and light source to double slit- bright fringes are brighter (intensity of light through double slit is greater)- fringe separation x unchanged

aDx




2. Change in a

a increases, x decreases

aDx

Larger aSmaller a




3. Change in source of light

Use white light (different wavelengths)

aDx

white

multi-coloured at

edges

Blue on edge nearest central maximumRed on outer edge




4. Change in slits width

wider S, A and B - no interference pattern (wide wavefronts from S, A and B are no more coherent)




5. Change in medium

Transparent medium between double slit and screen

aDx

ooo

1

oo

1

separation fringe New

xx

aD

x

x

o – wavelength in vacuum – refractive indexxo – fringe separation in vacuum




6. One of the slits covered with a film

aDx

From A to O, optical path length = AOFrom B to O, optical path length = BO + ( - 1)d = AO + ( - 1)d Optical path difference = [AO - ( - 1)d] – AO = ( - 1)d 0

Central maximum shifts to O’

If ( - 1)d = n (the nth bright fringe at O’)If ( - 1)d = (n - ) (the nth dark fringe at O’)

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7. Slits covered with polaroids

Light are plane polarized- No interference pattern (lights are mutually perpendicular)- or visible interference pattern (planes of polarization are parallel)




8. Changes in source of light

Use two separate sources of light- No interference pattern sources not coherent)


Measurement of the wavelength of light


Dxa,

aDx

x

x

From20

20 o1

o1 1 – o = length of 20

dark fringes

wavelength can be calculated

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Energy distribution


Amplitude at bright fringe = 2 x amplitude of single slit (a)Intensity at bright fringe = 4 x intensity of single slit (4Io) [since I a2]

Note: Wave energy can neither be created nor destroyed, but they can be redistributed for the constructive and destructive interference.


End


Q: Q: (a) What is meant by interference?(b) State the conditions necessary for the production of an interference pattern.(c) State the conditions necessary to produce an interference pattern where the fringes are of the same separation and of about the same intensity in a Young’s double slit experiment.(d) In a Young’s double slit experiment, light of wavelength 644 nm was used. Initially, the slit separation is 0.20 mm and the distance between the screen and the double slit is 1.00 m. Throughout the experiment, the interference pattern was viewed from the position of the double slit.(i) What is the angular separation between two neighbouring bright fringes formed on the screen?(ii) If the screen is then shifted further away from the double slit, what changes, if any, will occur to the interference pattern?

Solution



Solution:Solution:

(a) Interference is the effect produced by the superposition of waves from two coherent sources moving through the same region.

(b) The conditions necessary for the production of an interference pattern are:1. The waves must be from two coherent sources.2. The amplitudes of the waves must be the same or almost the same.3. The two coherent sources must be close to each other.



Solution (cont’d):Solution (cont’d):

(c)

The conditions necessary to produce evenly spaced fringes are:1. The use of a monochromatic source.2. The single slit S must be narrow and lie on the perpendicular divider of AB (see figure above).3. The slit separation a must be small.4. The distance of the screen from the double slit, D, must be large.




(d) (i) Use the equation x = ............................. (1)

From the figure above, the angular separation between two neighbouring bright fringes,

θ= From equation (1): = θ=

=

= 3.22 × 103 rad

aD

Dx

Dx

a

a

3

9

10 20010 446

.




(ii) When the screen is further away from the double slit,

1. the angular separation ( θ= ) remains unchanged.

2. the linear separation x = increases since D is larger.

3. the intensity of the bright fringes decreases.

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a

aD



Q: Q: (a) What is meant by Huygens’ principle?(b) With the aid of a diagram, use Huygens’ principle to explain how the interference pattern is produced in a Young’s double slit experiment.(c) In a Young’s double slit experiment, the slit separation is 0.05 cm and the distance between the double slit andscreen is 200 cm. When blue light is used, the distance of the first bright fringe from the centre of the interferencepattern is 0.13 cm.(i) Calculate the wavelength of the blue light used in the experiment.(ii) Calculate the distance of the fourth dark fringe from the centre of the interference pattern.

Solution



Solution:Solution:

(a) Huygens’ principle states that every point on a wavelength acts as a point source emitting secondary wavelets in phase with one other.(b)




Spherical wavefronts emerge from the slit S. Since slits A and B are of an equal distance from S, the same wavefront from S arrives at A and B simultaneously. According to Huygens’ Principle, the points on the wavefront at A and B emit wavelets in phase with one other. Hence, the waves from A and B are coherent.

At the points where the optical path difference of waves from A and B is nλ ( n = 0, 1, 2...), constructive interference occurs and a bright fringe is obtained. If the optical path difference is ( n – )λ, then destructive interference occurs and a dark fringe is obtained.

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(c) (i) Slit separation, a = 0.05 cm = 5.0 × 10–4 m Distance, D = 200 cm = 2.00 m Fringe separation, x = 0.13 cm = 1.3 × 10–3 m

(ii) Using x =

the wavelength of the blue light, =

=

= 3.25 × 10–7 m

aD

Dxa

00210 10 .31 43

. ) )(5.0(




From the figure, distance of the fourth dark fringe from the centre= = 3.5 × (1.3 ×10–3) = 4.55 × 10–3 m

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x213



Q:Q: A Young’s double slit system was set up at one end of an empty rectangular glass tank, and a sodium lamp with a narrow slit placed outside the tank. Interference fringes were observed on the opposite side of the tank. When the tank was filled with oil, the fringe separation changes by 35%. Explain why this happens and discuss whether the fringe separation increases or decreases. Calculate the refractive index of the oil. Solution



Solution:Solution:Fringe separation, x = ……………….. (1)

In oil, the wavelength, 1 = where = refractive index

Fringe separation, x1 = ……………….. (2) = ……………….. (from (1))

Also x1 = 65% or 0.65x i.e. fringe separation decreases

Hence = 0.65x

Refractive index, = = 1.54

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Dxa

aD

x

x

6501.



Q:Q: In a Young’s double slit experiment, an interference pattern was obtained using monochromatic light of wavelength 500 nm. When a thin film of transparent material with refractive index 1.5 was placed in front of one of the slits, the central maximum was shifted to the original position of the 9th bright fringe. What is the thickness of the film?

Solution



Solution:Solution:

When the central maximum was shifted to O’, the original position of the 9th bright fringe above O, all the other fringes were also shifted upwards. Therefore, at the point O on the axis of the double slit system, we will find the 9th bright fringe which was originally below O.




If d = thickness of the film, for light originating from A and B and arriving at O,

the optical path difference = nλ (since there is constructive interference) [AO +( μ – 1) d] – BO = 9λ ( n = 9) (μ – 1)d = 9λ ( AO = BO)

Thickness of the film (∴ d) = = 9.0 × 10–6 m

1) (1.5)10 (500 9 9

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Q:Q: In a Young’s double slit arrangement, the distance

between the centres of the slits is 0.25 mm and light of wavelength 6.0 × 10–4 mm is used. Calculate the angle θ subtended at the double slit by the neighbouring maxima of the interference pattern (refer to figure below).



Q: Q: Describe and explain what happens to the

interference fringes if(a) the slit A is covered with a thin piece of transparent material of high refractive index,(b) the intensity of light from A is reduced to half of that from B,(c) A and B are covered with thin films of polaroid, and one of the films is rotated slowly,(d) the distance between A and B is increased slowly.

Solution



Solution:Solution:Using the equation x = ,

and since the angle is small, = tan =

=

=

= 2.4 ×103 rad

aD

Dx

a

3

7

10 25010 .06

.

aDx



Solution (cont’d):Solution (cont’d):(a) When the slit A is covered with a thin piece of transparent material of high refractive index, the central maximum which was originally at O is shifted upwards.When A is covered with the transparent material, the optical path for light travelling from A to O is AO + (μ – 1)dwhere μ is the refractive index of the material and d its thickness.Hence the optical path difference between light travelling from A to O and from B to O is optical path difference = [AO + (μ – 1)d] – OB = (μ – 1) d (AO = OB) ≠ 0Hence O is no more the position of the central maximum.If (μ – 1)d = nλ where n is an integerthen the nth bright fringe from central maximum will be formed at O.If (μ – 1)d = (n – )λthen the nth dark fringe from the central maximum is found at O.

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Solution (cont’d):Solution (cont’d):(b) Initially, the intensity of the light across the interference pattern is as shown in Fig. (a). When the intensity of the light from A is halved, the intensity of the light across the interference pattern is as shown in Fig. (b).

Fig. (a)

Fig. (b)The intensity of the maxima has decreased but the intensity of the minima has increased. Hence the difference between the intensities of the maxima and minima is reduced. There is no change in the fringe separation.




(c) When the axis of polarization of the polaroids are parallel, the bright fringes can be differentiated clearly for the dark fringes. When one of the polaroids is rotated, the difference in intensities of the maxima and minima decreases until the axes of the polaroids are mutually perpendicular. Then no interference pattern will be seen.

(d) Using the equation x = ,

when the slit separation a is slowly increased the slit separation x decreases. The fringes become closer and closer until the neighbouring bright fringes cannot be resolved by the eye.

aD

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Q:Q: In a Young’s double slit experiment, a source of light which emits light of wavelengths 7 × 10–7 m (red) and 5.6 × 10–7 m (green) is used. Draw a sketch to show the interference fringes of the two colours formed. Show that the fringes ofthe two colours first exactly overlap while moving out from the centre of the pattern when

What is the significance of n? What is the colour of the fringe at that point?

Solution

451) (

nn



Solution:Solution: For red light, the fringe separation

= (7 × 10–7)

For green light, the fringe separation = (5.6 × 10–7) x1

When a red fringe and a green fringe overlap for the first time, the nth red fringe overlap with the (n + 1)th green fringe; and their distances from the central maximum are equal.i.e. nx1 = (n + 1) x2

n(7 × 10–7) = (n + 1)(5.6 × 10–7)

n = 4

aD

x 11 a

D

aD

x 22 a

D

aD a

D

45

5.67 1

nn



Solution (cont’d):Solution (cont’d):n represents the fourth red fringe which coincides with the fifth green fringe. The colour of the fringe produced is

Note that for n = 0 and n = 4, a red and green fringe coincide exactly to produce a yellow fringe.


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1© Manhattan Press (H.K.) Ltd. Young’s double slit experiment Young’s double slit experiment 9.10 Interference of light waves Relationship between x,,