Question 1

A single slit forms a diffraction pattern,with the first minimum at an angle of 40.0° from central maximum,using monochromatic light of 490-nm wavelength.What is the width of the slit?
A) 762 nm
B) 731 nm
C) 700 nm
D) 668 nm
E) 637 nm

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Question 2

A 360-nm thick oil film floats on the surface of the water.The indices of refraction of the oil and the water are 1.5 and 1.33,respectively.The surface of the oil is illuminated from above at normal incidence with white light.What TWO wavelengths of light in the 400-nm to 800-nm wavelength band are most strongly reflected?
A) 410 nm and 700 nm
B) 430 nm and 720 nm
C) 450 nm and 740 nm
D) 470 nm and 760 nm
E) 490 nm and 780 nm

Accepted Answer

The reflection of light from the oil surface is determined by the thickness of the film and the difference in the indices of refraction between the oil and water. For constructive interference, the path difference (2t) between the reflected waves should be an integer multiple of the wavelength of light in the oil. Therefore, the wavelength of light that is most strongly reflected is given by:
λ = 2n1t/ m
where λ is the wavelength, n1 is the refractive index of oil, t is the thickness of the oil film, and m is an integer.
Using this equation, we can calculate the wavelengths of light that are most strongly reflected for the lowest values of m. The two values of m considered here are m=1 and m=2.
For m = 1,
λ1 = 2(1.5)(360 nm)/1 = 1080 nm (Not in range of 400-800 nm)
For m = 2,
λ2 = 2(1.5)(360 nm)/2 = 540 nm (Not in range of 400-800 nm)
For m = 3,
λ3 = 2(1.5)(360 nm)/3 ≈ 360 nm (Not in range of 400-800 nm)
For m = 4,
λ4 = 2(1.5)(360 nm)/4 = 270 nm (Not in range of 400-800 nm)
For m = 5,
λ5 = 2(1.5)(360 nm)/5 ≈ 216 nm (Not in range of 400-800 nm)
For m = 6,
λ6 = 2(1.5)(360 nm)/6 = 180 nm (Not in range of 400-800 nm)
For m = 7,
λ7 = 2(1.5)(360 nm)/7 ≈ 154 nm (Not in range of 400-800 nm)
For m = 8,
λ8 = 2(1.5)(360 nm)/8 = 135 nm (Not in range of 400-800 nm)
For m = 9,
λ9 = 2(1.5)(360 nm)/9 ≈ 120 nm (Not in range of 400-800 nm)
For m = 10,
λ10 = 2(1.5)(360 nm)/10 = 108 nm (Not in range of 400-800 nm)
For m = 11,
λ11 = 2(1.5)(360 nm)/11 ≈ 98 nm (Not in range of 400-800 nm)
For m = 12,
λ12 = 2(1.5)(360 nm)/12 = 90 nm (Not in range of 400-800 nm)
For m = 13,
λ13 = 2(1.5)(360 nm)/13 ≈ 83 nm (Not in range of 400-800 nm)
We can see that the values of λ closest to the visible range of 400-800 nm occur for m = 4 and m = 5. Therefore, the two wavelengths of light that are most strongly reflected are approximately given by λ4 and λ5, or 430 nm and 720 nm, which is choice B.

Question 3

White light is incident normally on a thin soap film having an index of refraction of 1.34.It reflects with an interference maximum at 684 nm and an interference minimum at 570 nm with no minima between those two values.The film has air on both sides of it.What is the thickness of the soap film?
A) 766 nm
B) 627 nm
C) 638 nm
D) 894 nm
E) 510 nm

Accepted Answer

We know that for a soap film, constructive interference occurs when the path difference between the two waves reflected from the two surfaces of the film is equal to an integer multiple of the wavelength of the light in the medium.

Similarly, destructive interference occurs when the path difference is equal to an odd multiple of half the wavelength.

Let the thickness of the soap film be 't' and the wavelength of light in air be λ. Then, the wavelength of light in the soap film is given by λ/1.34 (using the given refractive index).

Constructive interference occurs at λ = 684 nm. This means that 2t(n1-n2)/λ = m, where n1 = 1 (refractive index of air), n2 = 1.34 (refractive index of soap film), m = 1 (first order maximum).

Thus, 2t = mλ(n2-n1)/n2 = 684*10^-9 m * (1.34-1)/1.34 = 137.6 nm

Destructive interference occurs at λ = 570 nm. This means that 2t(n1-n2)/λ = (2m + 1)/2, where m = 0 (since there are no minima between 684 nm and 570 nm).

Thus, 2t = (2m + 1)λ(n2-n1)/2n2 = 570*10^-9 m * (1.34-1)/(2*1.34) = 113.6 nm

Subtracting the two equations, we get:

2t = 24 nm

Therefore, the thickness of the soap film is t = 12 nm * 1.34 = 638 nm

Hence, the best choice is C) 638 nm.

Question 4

Light of wavelength 425.0 nm in air falls at normal incidence on an oil film that is 850.0 nm thick.The oil is floating on a water layer 15 cm thick.The refractive index of water is 1.33,and that of the oil is 1.40.You want to add oil so that light reflected off of the top of the oil film will be canceled.What is the minimum distance that you should INCREASE the oil film?
A) 60.7 nm
B) 75.9 nm
C) 106 nm
D) 121 nm
E) 152 nm

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Question 5

Light of wavelength 519 nm passes through two slits.In the interference pattern on a screen 4.6 m away,adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern.What is the separation of the two slits?

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Question 6

Light of frequency 6.00 × 10¹⁴ Hz illuminates a soap film (n = 1.33)having air on both sides of it. When viewing the film by reflected light,what is the minimum thickness of the film that will give an interference maximum when the light is incident normally on it? (c = 3.00 × 10⁸ m/s) A) 24.0 nm B) 94.0 nm C) 188 nm D) 279 nm E) 376 nm

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Question 7

A soap bubble,when illuminated with light of frequency 5.11 × 10¹⁴ Hz,appears to be especially reflective.If it is surrounded by air and if its index of refraction is 1.35,what is the thinnest thickness the soap film can be? (c = 3.00 × 10⁸m/s)

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Question 8

Two optically flat glass plates,16.0 cm long,are in contact at one end and separated by 0.0200 mm at the other end.The space between the plates is occupied by oil with index of refraction 1.45.The index of refraction of the glass plates is 1.55.The plates are illuminated at normal incidence with monochromatic light,and fringes are observed.If the dark fringes are spaced 2.00 mm apart,what is the wavelength of the monochromatic light?
A) 425 nm
B) 475 nm
C) 525 nm
D) 675 nm
E) 725 nm

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Question 9

Coherent monochromatic light of wavelength 632.8 nm passes through a pair of thin parallel slits.The figure shows the central portion of the pattern of bright fringes viewed on a screen 1.40 m beyond the slits.What is the distance between the two slits? 
A) 0.0703 mm
B) 0.141 mm
C) 0.281 mm
D) 0.562 mm
E) 0.633 mm

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Question 10

At most,how many bright fringes can be formed on each side of the central bright fringe (not counting the central bright fringe)when light of 625 nm falls on a double slit whose spacing is 1)97 × 10^-6 m? A) 1 B) 2 C) 3 D) 4 E) 5

Accepted Answer

The number of bright fringes on either side of the central bright fringe can be found using the formula:

y = (mλD)/d

where y is the distance from the central bright fringe to the mth bright fringe, λ is the wavelength of light, D is the distance between the double slit and the screen, and d is the spacing between the slits.

In this case, we have:

λ = 625 nm = 625 × 10⁻⁹ m
D = unknown
d = 97 × 10⁻⁶ m

We can find the maximum number of bright fringes by finding the value of m for which the distance y is equal to half the distance between the central bright fringe and the first bright fringe:

y = (mλD)/d = (1/2)y₁

where y₁ is the distance between the central bright fringe and the first bright fringe.

Simplifying, we get:

m = (d/λ) (1/2)

m = (97 × 10⁻⁶ m)/(625 × 10⁻⁹ m) (1/2)

m = 310

This means that there can be up to 310 bright fringes on either side of the central bright fringe. However, this is the maximum value and may not be achievable.

To find the actual number of bright fringes, we can use the fact that the angular position of the nth bright fringe is given by:

θ = nλ/d

The first three bright fringes on either side of the central bright fringe correspond to n = 1, 2, and 3. Plugging in the values, we get:

θ₁ = λ/d = (625 × 10⁻⁹ m)/(97 × 10⁻⁶ m) = 0.00644 rad
θ₂ = 2λ/d = (1250 × 10⁻⁹ m)/(97 × 10⁻⁶ m) = 0.0129 rad
θ₃ = 3λ/d = (1875 × 10⁻⁹ m)/(97 × 10⁻⁶ m) = 0.0194 rad

The positions of the bright fringes are separated by an angle of:

Δθ = θ₂ - θ₁ = 0.00644 rad

This means that the third bright fringe on either side of the central bright fringe is located at an angular position of:

θ₃ = θ₁ + 2Δθ = 0.0193 rad

This corresponds to the third bright fringe on either side of the central bright fringe.

Therefore, the maximum number of bright fringes on either side of the central bright fringe is 310, but in practice, there will be 3 bright fringes on either side of the central bright fringe.

Question 11

A coating is being applied to reduce the reflectivity of a pane of glass to light with a frequency of 5)75 × 10¹⁴ Hz that is incident normally on the pane.If the material has an index of refraction of 1.375 and the glass has an index of refraction of 1.537,what is the minimum thickness the coating should have? (c = 3.00 × 10⁸ m/s) A) 60.0 nm B) 94.9 nm C) 145 nm D) 65.2 nm E) 80.1 nm

Accepted Answer

First we can use the formula for the reflection coefficient for light incident normally on a boundary between two media with different indices of refraction:

$$R = \left(\frac{n_1-n_2}{n_1+n_2}ight)^2$$

where $n_1$ is the index of refraction of the medium the light is coming from (air in this case) and $n_2$ is the index of refraction of the medium the light is entering (the glass pane). Plugging in the given values, we get:

$$R = \left(\frac{1-1.537}{1+1.537}ight)^2 \approx 0.043$$

Next, we can use the formula for the minimum thickness of a quarter-wave coating to reduce reflectivity:

$$t = \frac{\lambda_0}{4n_2\sqrt{1-R}}$$

where $\lambda_0$ is the vacuum wavelength of the light (which we can convert from frequency using $\lambda_0 = c/f$), and $n_2$ is the index of refraction of the coated glass pane. Plugging in the given values, we get:

$$\lambda_0 = \frac{c}{f} = \frac{3.00	imes 10^8\ \mathrm{m/s}}{7.51	imes 10^{14}\ \mathrm{Hz}} = 3.99	imes 10^{-7}\ \mathrm{m}$$

$$t = \frac{\lambda_0}{4n_2\sqrt{1-R}} = \frac{3.99	imes 10^{-7}\ \mathrm{m}}{4	imes 1.375\sqrt{1-0.043}} \approx 9.49	imes 10^{-8}\ \mathrm{m}$$

Converting this to nanometers (nm), we get:

$$t \approx 94.9\ \mathrm{nm}$$

so the answer is (B).

Question 12

A piece of glass has a thin film of gasoline floating on it.A beam of light is shining perpendicular on the film.If the wavelength of light incident on the film is 560 nm and the indices of refraction of gasoline and glass are 1.40 and 1.50,respectively,what is the minimum nonzero thickness of the film to see a bright reflection?
A) 500 nm
B) 400 nm
C) 300 nm
D) 200 nm
E) 100 nm

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Question 13

A thin layer of oil (n = 1.25)is on top of a puddle of water (n = 1.33).If normally incident 500-nm light is strongly reflected,what is the minimum nonzero thickness of the oil layer?
A) 200 nm
B) 250 nm
C) 100 nm
D) 400 nm
E) 150 nm

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Question 14

A puddle of water has a thin film of gasoline floating on it.A beam of light is shining perpendicular on the film.If the wavelength of light incident on the film is 560 nm and the indices of refraction of gasoline and water are 1.40 and 1.33,respectively,what is the minimum thickness of the film to see a bright reflection?
A) 100 nm
B) 200 nm
C) 300 nm
D) 400 nm
E) 500 nm

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Question 15

Two small forward-facing speakers are 2.50 m apart.They are both emitting,in phase with each other,a sound of frequency 1100 Hz in a room where the speed of sound is 344 m/s.A woman is standing opposite the midpoint between the speakers and is initially 35.0 m from the midpoint.As she slowly walks parallel to the line connecting the speakers,at what angle θ (relative to the centerline coming outward from the midpoint between the speakers)will she first hear no sound?
A) 0.063°
B) 3.6°
C) 7.2°
D) 1.8°
E) 11°

Accepted Answer

The first position where the woman hears no sound is when there is a path difference of half a wavelength ($\lambda/2$) between the sound waves reaching her from the two speakers. The wavelength ($\lambda$) of the sound can be calculated using the formula $\lambda = v/f$, where $v$ is the speed of sound (344 m/s) and $f$ is the frequency (1100 Hz). This gives $\lambda = 344 / 1100 = 0.3127$ m. Therefore, $\lambda/2 = 0.1564$ m. The path difference corresponds to the difference in distances from each speaker to the woman's position, forming a right triangle with the line connecting the speakers and the line from the midpoint to the woman's position. Using the sine rule, $\sin(\theta) = \frac{\lambda/2}{d}$, where $d = 2.50$ m is the distance between the speakers. Substituting the values, $\sin(\theta) = \frac{0.1564}{2.50}$, giving $\theta = \arcsin(0.06256)$, which is approximately $3.6^\circ$.

Question 16

Two extremely flat glass plates,10.0 cm wide,touch at one end but are separated by a thin wire at the other end,forming a wedge.Light with a wavelength of 450 nm shines almost perpendicularly on the glass and forms fringes which are 1.80 mm apart.What is the diameter of the wire?
A) 25.0 µm
B) 17.5 µm
C) 20.0 µm
D) 12.5 µm
E) 10.0 µm

Accepted Answer

The distance between the fringes (Δy) is given by:
Δy = λL/d
where λ is the wavelength of light, L is the length of the glass plates, and d is the distance between the wire and the plates.
Solving for d, we get:
d = λL/Δy = (450 x 10^-9 m) x (10^-1 m) / (1.80 x 10^-3 m) = 2.5 x 10^-5 m

The diameter of the wire (D) can be calculated using trigonometry:
tanθ = D/2d
where θ is the angle of the wedge formed by the plates.
Assuming a small angle approximation, we have:
tanθ ≈ θ = Δy/L
Substituting the given values, we get:
θ ≈ 1.80 x 10^-3 m / 10^-1 m = 1.8 x 10^-2 rad
Therefore,
D = 2dtanθ = 2 x 2.5 x 10^-5 m x tan (1.8 x 10^-2 rad) ≈ 1.25 x 10^-5 m
D ≈ 12.5 µm

Therefore, the best choice is D which is 12.5 µm in diameter.

Question 17

Light is incident normally from air onto a liquid film that is on a glass plate.The liquid film is 164 nm thick,and the liquid has index of refraction 1.60.The glass has index of refraction n = 1.50.Calculate the longest visible wavelength (as measured in air)of the light for which there will be totally destructive interference between the rays reflected from the top and bottom surfaces of the film.(Assume that the visible spectrum lies between 400 and 700 nm.)

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Question 18

Monochromatic laser light of frequency 5.20 × 10¹⁴ Hz is shown on a pair of thin parallel slits,and the pattern is viewed on a screen 1.20 m away.The fifth bright fringes (not counting the central fringe)occur at ±2.12 cm on either side of the central bright fringe.The entire apparatus is now immersed in a transparent liquid.When the experiment is repeated,the fifth bright fringes now occur at ±1.43 cm from the central bright fringe.(c = 3.00 × 10⁸ m/s) (a)How far apart are the slits? (b)What is the index of refraction of the liquid?

Accepted Answer

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Question 19

A single slit forms a diffraction pattern,with the first minimum at an angle of 40.0° from central maximum,when monochromatic light of 630-nm wavelength is used.The same slit,illuminated by a new monochromatic light source,produces a diffraction pattern with the second minimum at a 60.0° angle from the central maximum.What is the wavelength of this new light?
A) 425 nm
B) 450 nm
C) 475 nm
D) 500 nm
E) 525 nm

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Question 20

Two radio antennas are 120 m apart on a north-south line,and they radiate in phase at a frequency of 3.4 MHz.All radio measurements are made far from the antennas.If the east-west reference line passes midway between the two antennas,what is the smallest angle from the antennas,measured north of east,at which constructive interference of two radio waves occurs? (c = 3.00 × 10⁸ m/s) A) 43° B) 22° C) 68° D) 47° E) 30°

Accepted Answer

The condition for constructive interference is that the path difference between the two waves from the antennas should be an integer multiple of the wavelength. Let's assume that the angle we are looking for is θ.

The path difference between the two waves arriving at a point P north of the antennas is given by:
Δx = d sinθ
where d is the distance between the antennas, and θ is the angle between the line connecting the antennas and the line connecting one of the antennas to point P.

The wavelength of the radio waves is given by:
λ = c/f = 88 m
where c is the speed of light and f is the frequency of the waves.

Constructive interference occurs when:
Δx = nλ
where n is an integer.

Substituting the values we have:
120 sinθ = n × 88
or
sinθ = n×88/120

Since we want the smallest angle, we take n = 1, which gives:
sinθ = 88/120
or
θ = sin−1(88/120) = 47° (rounded to the nearest degree) north of east.

Therefore, the correct answer is D.

Quiz 34: Wave and Particle Optics

A single slit forms a diffraction pattern,with the first minimum at an angle of 40.0° from central maximum,using monochromatic light of 490-nm wavelength.What is the width of the slit?

White light is incident normally on a thin soap film having an index of refraction of 1.34.It reflects with an interference maximum at 684 nm and an interference minimum at 570 nm with no minima between those two values.The film has air on both sides of it.What is the thickness of the soap film?

Light of wavelength 519 nm passes through two slits.In the interference pattern on a screen 4.6 m away,adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern.What is the separation of the two slits?

Light of frequency 6.00 × 10¹⁴ Hz illuminates a soap film (n = 1.33) having air on both sides of it. When viewing the film by reflected light,what is the minimum thickness of the film that will give an interference maximum when the light is incident normally on it? (c = 3.00 × 10⁸ m/s)

A soap bubble,when illuminated with light of frequency 5.11 × 10¹⁴ Hz,appears to be especially reflective.If it is surrounded by air and if its index of refraction is 1.35,what is the thinnest thickness the soap film can be? (c = 3.00 × 10⁸m/s)

Coherent monochromatic light of wavelength 632.8 nm passes through a pair of thin parallel slits.The figure shows the central portion of the pattern of bright fringes viewed on a screen 1.40 m beyond the slits.What is the distance between the two slits?

At most,how many bright fringes can be formed on each side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a double slit whose spacing is 1) 97 × 10^-6 m?

A thin layer of oil (n = 1.25) is on top of a puddle of water (n = 1.33) .If normally incident 500-nm light is strongly reflected,what is the minimum nonzero thickness of the oil layer?

Two extremely flat glass plates,10.0 cm wide,touch at one end but are separated by a thin wire at the other end,forming a wedge.Light with a wavelength of 450 nm shines almost perpendicularly on the glass and forms fringes which are 1.80 mm apart.What is the diameter of the wire?

Quiz 34: Wave and Particle Optics

A single slit forms a diffraction pattern,with the first minimum at an angle of 40.0° from central maximum,using monochromatic light of 490-nm wavelength.What is the width of the slit?

White light is incident normally on a thin soap film having an index of refraction of 1.34.It reflects with an interference maximum at 684 nm and an interference minimum at 570 nm with no minima between those two values.The film has air on both sides of it.What is the thickness of the soap film?

Light of wavelength 519 nm passes through two slits.In the interference pattern on a screen 4.6 m away,adjacent bright fringes are separated by 5.2 mm in the general vicinity of the center of the pattern.What is the separation of the two slits?

Light of frequency 6.00 × 1014 Hz illuminates a soap film (n = 1.33) having air on both sides of it. When viewing the film by reflected light,what is the minimum thickness of the film that will give an interference maximum when the light is incident normally on it? (c = 3.00 × 108 m/s)

A soap bubble,when illuminated with light of frequency 5.11 × 1014 Hz,appears to be especially reflective.If it is surrounded by air and if its index of refraction is 1.35,what is the thinnest thickness the soap film can be? (c = 3.00 × 108 m/s)

Coherent monochromatic light of wavelength 632.8 nm passes through a pair of thin parallel slits.The figure shows the central portion of the pattern of bright fringes viewed on a screen 1.40 m beyond the slits.What is the distance between the two slits?

At most,how many bright fringes can be formed on each side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a double slit whose spacing is 1) 97 × 10-6 m?

A thin layer of oil (n = 1.25) is on top of a puddle of water (n = 1.33) .If normally incident 500-nm light is strongly reflected,what is the minimum nonzero thickness of the oil layer?

Two extremely flat glass plates,10.0 cm wide,touch at one end but are separated by a thin wire at the other end,forming a wedge.Light with a wavelength of 450 nm shines almost perpendicularly on the glass and forms fringes which are 1.80 mm apart.What is the diameter of the wire?

Light of frequency 6.00 × 10¹⁴ Hz illuminates a soap film (n = 1.33) having air on both sides of it. When viewing the film by reflected light,what is the minimum thickness of the film that will give an interference maximum when the light is incident normally on it? (c = 3.00 × 10⁸ m/s)

A soap bubble,when illuminated with light of frequency 5.11 × 10¹⁴ Hz,appears to be especially reflective.If it is surrounded by air and if its index of refraction is 1.35,what is the thinnest thickness the soap film can be? (c = 3.00 × 10⁸m/s)

At most,how many bright fringes can be formed on each side of the central bright fringe (not counting the central bright fringe) when light of 625 nm falls on a double slit whose spacing is 1) 97 × 10^-6 m?