Question 1

A researcher is investigating a cubic crystal with x-rays.He is looking at Bragg reflection from the planes parallel to the cube faces.He finds that when using x-rays of 0.165 nm a strong first maximum occurs when the beam makes an angle of 23.5° with the planes.What is the spacing of adjacent atoms in the crystal?

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Question 2

A 10-inch telescope (25.4 cm in diameter)is used to determine if what appears to be one star is actually two stars.Stars are so far away that they are essentially point sources.How close (in angle)can the two stars be and still be resolved by this telescope if it is focusing light of wavelength of 550 nm? (Consider only the limitation due to diffraction.) A) 4.2 × 10^-8 degree B) 2.6 × 10^-6 degree C) 3.0 × 10^-4 degree D) 1.5 × 10^-4 degree E) 6.6 × 10^-8 degree

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Question 3

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the aperture.The diameter of the central bright area is 5.4 cm.What is the diameter of the aperture?
A) 110 μm
B) 2.0 μm
C) 6600 μm
D) 960 μm

Accepted Answer

The diameter of the central bright area (first dark fringe) in a diffraction pattern is given by $D = 2L \tan(\theta)$, where $L$ is the distance to the screen, and $\theta$ is the angle to the first dark fringe. For small angles, $\tan(\theta) \approx \sin(\theta) = \frac{\lambda}{d}$, where $\lambda$ is the wavelength of light, and $d$ is the diameter of the aperture. Rearranging for $d$, we get $d = \frac{\lambda L}{D/2}$. Substituting $\lambda = 633 \times 10^{-9} m$, $L = 4.0 m$, and $D = 5.4 \times 10^{-2} m$, we find $d \approx 110 \times 10^{-6} m$ or 110 μm.

Question 4

A radio telescope 200 m in diameter is used to investigate sources emitting a 21-cm wavelength wave.What is the minimum angular separation of the sources that can be resolved by this system?
A) 0.073°
B) 0.030°
C) 0.0013°
D) 0.154°
E) 0.0026°

Accepted Answer

The minimum angular resolution of a telescope can be calculated using the formula $\theta = \frac{1.22 \lambda}{D}$, where $\theta$ is the angular resolution in radians, $\lambda$ is the wavelength of the light (or other electromagnetic radiation), and $D$ is the diameter of the telescope's aperture. Given $\lambda = 21$ cm = 0.21 m and $D = 200$ m, substituting these values gives $\theta \approx 1.22 \times \frac{0.21}{200} = 0.00127$ radians. Converting to degrees ($1 \text{ radian} = 57.2958^\circ$), $\theta \approx 0.00127 \times 57.2958 = 0.073^\circ$.

Question 5

A thin beam of laser light of wavelength 514 nm passes through a diffraction grating having 3952 lines/cm.The resulting pattern is viewed on a distant curved screen that can show all bright fringes up to and including ±90.0° from the central spot.What is the TOTAL number of bright fringes that will show up on the screen?
A) 4
B) 5
C) 8
D) 9
E) 10

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Question 6

What is the limiting angle of resolution for the eye if the pupil diameter of the eye is 4.0 mm,the wavelength of the light is 600 nm,and index of refraction of the liquid in the eye is 1.34?
A) 0.42 mrad
B) 0.21 mrad
C) 0.14 mrad
D) 0.11 mrad
E) 0.26 mrad

Accepted Answer

The limiting angle of resolution for the eye can be approximated using the Rayleigh criterion:

θ = 1.22 (λ / D)

where:
- θ is the limiting angle of resolution
- λ is the wavelength of light
- D is the diameter of the pupil

Substituting the given values:
θ = 1.22 (600 nm / 4.0 mm)
θ = 0.183 rad

However, since the question is asking for the limiting angle of resolution in milliradians (mrad), we need to convert radians to milliradians:

θ (mrad) = (θ * 1000) / π

Substituting the value of θ we obtained:
θ (mrad) = (0.183 * 1000) / π
θ (mrad) = 58.25 mrad

But this is the total angle, so we also need to divide this by 2 to get the angle for one eye:
θ (mrad) = 58.25 / 2
θ (mrad) = 29.12 mrad

Finally, we need to take into account the index of refraction of the liquid in the eye, which affects the wavelength of light inside the eye:

λ' = λ / n

where:
- λ' is the effective wavelength of light inside the eye
- n is the index of refraction of the liquid in the eye

Substituting the given values:
λ' = 600 nm / 1.34
λ' = 448.9 nm

Now we can use the same formula as before to calculate the limiting angle of resolution with the effective wavelength:

θ' = 1.22 (λ' / D)

Substituting the values:
θ' = 1.22 (448.9 nm / 4.0 mm)
θ' = 0.136 rad

Converting to milliradians:
θ' (mrad) = (θ' * 1000) / π
θ' (mrad) = 43.26 mrad

Again, we divide by 2 to get the angle for one eye:
θ' (mrad) = 43.26 / 2
θ' (mrad) = 21.63 mrad

Therefore, the limiting angle of resolution for the eye is approximately 21.63 mrad, which corresponds to choice C.

Question 7

If an x-ray beam of wavelength 1.4 × 10^-10 m makes an angle of 20° with a set of planes in a crystal causing first order constructive interference,at what angle will the second order line appear? A) 40° B) 20° C) 43° D) 4.0° E) 11°

Accepted Answer

The Bragg's law for constructive interference is given by:
2d sinθ = nλ
where d is the distance between two adjacent planes in the crystal, θ is the angle of incidence, n is the order of reflection and λ is the wavelength of the x-ray beam.

For first order reflection:
2d sin20° = λ
2d = λ/sin20° = 4.14 × 10^-10 m

For second order reflection:
2d sinθ = 2λ
sinθ = (2λ/2d) = λ/d = 2sin20°
θ = sin^-1(2sin20°) = 43.1°

Therefore, the second order line will appear at an angle of 43.1°. Hence, the correct answer choice is C.

Question 8

Treat each of your eyes as a circular aperture of diameter 3.5 mm.Light of wavelength 500 nm is used to view two point sources that are 894 m distant from you.How far apart must these two point sources be if they are to be just resolved by your eye? Assume that the resolution is diffraction limited and use Rayleigh's criterion.

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Question 9

Light of wavelength 500 nm illuminates a round 0.50-mm diameter hole.A screen is placed 6.3 m behind the slit.What is the diameter of the central bright area on the screen?
A) 15 mm
B) 270 μm
C) 7.7 mm
D) 3800 μm

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Question 10

A camera used for aerial surveillance has a lens with a 30-cm maximum aperture and a 42-cm focal length.Assume light of 550-nm wavelength is used and that the resolution of the camera is limited solely by diffraction.What is the angular resolution of the camera at maximum aperture?
A) 1.6 µrad
B) 2.2 µrad
C) 3.2 µrad
D) 4.5 µrad
E) 6.3 µrad

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Question 11

Certain planes of a crystal of halite have a spacing of 0.399 nm.The crystal is irradiated by a beam of x-rays.First order constructive interference occurs when the beam makes an angle of 20° with the planes.What is the wavelength of the x-rays?
A) 0.14 nm
B) 0.17 nm
C) 0.21 nm
D) 0.24 nm
E) 0.27 nm

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Question 12

A diffraction grating has 450 lines per mm.What is the highest order m that contains the entire visible spectrum from 400 nm to 700 nm?
A) m = 2
B) m = 3
C) m = 4
D) m = 5
E) m = 6

Accepted Answer

The highest order m that contains the entire visible spectrum is given by the formula nλ = d sinθ for bright spots, where n is the order, λ is the wavelength of light, d is the spacing between the grating lines, and θ is the angle of diffraction. Since we want the entire visible spectrum to diffract, we need to find the angle θ that will result in the longest wavelength (700 nm) being diffracted at the edge of the spectrum. Using the formula θ = sin^-1(nλ/d), we can solve for n when θ is equal to the inverse sine of 700 nm and d is equal to 1/450 mm. We get n = 3.03, which means that the third order (m=3) diffraction maxima will contain the entire visible spectrum.

Question 13

Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV.What is the maximum kinetic energy of the photoelectrons emitted from the metal? (c = 3.00 × 10⁸ m/s, H = 6.626 × 10^-34 J • s = 4.141 × 10^-15 ev • s,1 eV = 1.60 × 10^-19 J) A) 4.52 eV B) 3.11 eV C) 1.41 eV D) 2.82 eV E) 1.70 eV

Accepted Answer

The energy of the incident photon is calculated using E = hc/λ. Converting 400 nm to meters, we get 400 × 10^-9 m. Substituting the given values, E = (4.141 × 10^-15 eV • s × 3.00 × 10^8 m/s) / (400 × 10^-9 m) = 3.11 eV. The maximum kinetic energy of the photoelectrons is the photon energy minus the work function: 3.11 eV - 1.70 eV = 1.41 eV.

Question 14

If the headlights on a car are separated by 1.3 m,how far down the road can they be resolved if the angular resolution of the eye is 5.0 × 10^-4 rad and the person has excellent vision? A) 1.3 km B) 5.0 km C) 4.8 km D) 0.65 km E) 2.6 km

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Question 15

A metallic sheet has a large number of slits,5.0 mm wide and 20 cm apart,and is used as a diffraction grating for microwaves.A wide parallel beam of microwaves is incident normally on the sheet.What is the smallest microwave frequency for which only the central maximum occurs? (c = 3.00 × 10⁸ m/s) A) 0.50 GHz B) 0.70 GHz C) 1.0 GHz D) 1.5 GHz E) 2.0 GHz

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Question 16

A thin beam of laser light of wavelength 514 nm passes through a diffraction grating having
3952 lines/cm.The resulting pattern is viewed on a distant curved screen that can show all bright fringes up to and including ±90.0° from the central spot.If the experiment were performed with all of the apparatus under water (which has an index of refraction of 1.33),what would be the TOTAL number of bright spots that would show up on the screen?
A) 6
B) 7
C) 12
D) 13
E) 14

Accepted Answer

The number of bright spots in a diffraction grating can be calculated using the formula:

nλ = d sinθ

where n is the order of the bright spot (n=1,2,3...), λ is the wavelength of the light, d is the spacing between the lines in the grating, and θ is the angle between the central bright spot and the nth bright spot.

First, we need to calculate the spacing between the lines in the grating:

d = 1/3952 cm = 2.527 x 10^-4 cm

Next, we need to find the maximum angle at which we can see bright spots:

sinθ = 90°
θ = sin^-1(1) = 90°

Now we can calculate the maximum order of the bright spot that we can see:

nλ = d sinθ
n = d sinθ / λ
n = (2.527 x 10^-4 cm)(sin 90°) / (514 nm)
n = 0.000491

However, we need to take into account the refractive index of water. The wavelength of the light in water will be:

λwater = λ / nwater
λwater = 514 nm / 1.33
λwater = 386.5 nm

Now we can recalculate the maximum order of the bright spot:

nλwater = d sinθ
n = d sinθ / λwater
n = (2.527 x 10^-4 cm)(sin 90°) / (386.5 nm)
n = 0.000635

The maximum order of the bright spot that we can see is now 0.000635. We need to round up to the nearest whole number, so the maximum order is 1.

The total number of bright spots is given by:

# of bright spots = (2n + 1)^2
# of bright spots = (2(1) + 1)^2
# of bright spots = 9

However, we need to subtract the central bright spot, so the total number of bright spots that we will see on the screen is:

# of bright spots = 9 - 1
# of bright spots = 8

But we need to take into account that we can see bright fringes up to and including ±90.0° from the central spot. This means that we need to count both positive and negative orders of bright spots. So we double our result:

# of bright spots = 8 x 2
# of bright spots = 16

However, we need to subtract the central bright spot twice (once for positive orders and once for negative orders), so the final answer is:

# of bright spots = 16 - 2
# of bright spots = 14

Therefore, the answer is (D) 14.

Question 17

A diffraction grating is to be used to find the wavelength of the light in the emission spectrum of a gas.The grating spacing is not known,but a light having a known wavelength of 632.8 nm is deflected by 43.2° away from the central maximum in the second order by this grating.Light of the wavelength to be measured is deflected by 48.2° away from the central maximum in the second order.What is the wavelength of this light?

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Question 18

If an x-ray beam of wavelength 1.4 × 10^-10 m makes an angle of 30° with a set of planes in a crystal causing first order constructive interference,what is the plane spacing? A) 0.28 nm B) 0.070 nm C) 0.32 nm D) 0.16 nm E) 0.14 nm

Accepted Answer

Using Bragg's Law, nλ = 2d sin θ, for first order (n=1), d = λ / (2 sin θ). Substituting λ = 1.4 × 10^-10 m and θ = 30°, we get d = 0.14 nm.

Question 19

A certain astronomical telescope has a diameter of 5.60 m.Considering only the limitation due to diffraction,what is the minimum angle of resolution for this telescope at a wavelength of 620 nm?
A) 0.111 µrad
B) 0.311 µrad
C) 0.270 µrad
D) 0.135 µrad
E) 0.405 µrad

Accepted Answer

The minimum angle of resolution for a telescope is given by the Rayleigh criterion: $	heta_{min} = 1.22 \frac{\lambda}{D}$, where $\lambda$ is the wavelength of light and $D$ is the diameter of the telescope. Plugging in the given values, we get $	heta_{min} = 1.22 \frac{620 	imes 10^{-9}\ m}{5.60\ m} \approx 0.135\ \mu rad$. Therefore, the correct answer is D.

Question 20

When a certain metal is illuminated by light,photoelectrons are observed provided that the wavelength of the light is less than 669 nm.Which one of the following values is closest to the work function of this metal? (h = 6.626 × 10^-34J • s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19 J) A) 1.9 eV B) 2.0 eV C) 2.2 eV D) 2.3 eV

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Quiz 34: Wave and Particle Optics

Light of wavelength 633 nm from a He-Ne laser passes through a circular aperture and is observed on a screen 4.0 m behind the aperture.The diameter of the central bright area is 5.4 cm.What is the diameter of the aperture?

A radio telescope 200 m in diameter is used to investigate sources emitting a 21-cm wavelength wave.What is the minimum angular separation of the sources that can be resolved by this system?

What is the limiting angle of resolution for the eye if the pupil diameter of the eye is 4.0 mm,the wavelength of the light is 600 nm,and index of refraction of the liquid in the eye is 1.34?

If an x-ray beam of wavelength 1.4 × 10-10 m makes an angle of 20° with a set of planes in a crystal causing first order constructive interference,at what angle will the second order line appear?

Light of wavelength 500 nm illuminates a round 0.50-mm diameter hole.A screen is placed 6.3 m behind the slit.What is the diameter of the central bright area on the screen?

A camera used for aerial surveillance has a lens with a 30-cm maximum aperture and a 42-cm focal length.Assume light of 550-nm wavelength is used and that the resolution of the camera is limited solely by diffraction.What is the angular resolution of the camera at maximum aperture?

Certain planes of a crystal of halite have a spacing of 0.399 nm.The crystal is irradiated by a beam of x-rays.First order constructive interference occurs when the beam makes an angle of 20° with the planes.What is the wavelength of the x-rays?

A diffraction grating has 450 lines per mm.What is the highest order m that contains the entire visible spectrum from 400 nm to 700 nm?

Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV.What is the maximum kinetic energy of the photoelectrons emitted from the metal? (c = 3.00 × 108 m/s, H = 6.626 × 10-34 J • s = 4.141 × 10-15 ev • s,1 eV = 1.60 × 10-19 J)

If the headlights on a car are separated by 1.3 m,how far down the road can they be resolved if the angular resolution of the eye is 5.0 × 10-4 rad and the person has excellent vision?

A metallic sheet has a large number of slits,5.0 mm wide and 20 cm apart,and is used as a diffraction grating for microwaves.A wide parallel beam of microwaves is incident normally on the sheet.What is the smallest microwave frequency for which only the central maximum occurs? (c = 3.00 × 108 m/s)

If an x-ray beam of wavelength 1.4 × 10-10 m makes an angle of 30° with a set of planes in a crystal causing first order constructive interference,what is the plane spacing?

A certain astronomical telescope has a diameter of 5.60 m.Considering only the limitation due to diffraction,what is the minimum angle of resolution for this telescope at a wavelength of 620 nm?

When a certain metal is illuminated by light,photoelectrons are observed provided that the wavelength of the light is less than 669 nm.Which one of the following values is closest to the work function of this metal? (h = 6.626 × 10-34 J • s,c = 3.00 × 108 m/s,1 eV = 1.60 × 10-19 J)

If an x-ray beam of wavelength 1.4 × 10^-10 m makes an angle of 20° with a set of planes in a crystal causing first order constructive interference,at what angle will the second order line appear?

Light of wavelength 400 nm falls on a metal surface having a work function 1.70 eV.What is the maximum kinetic energy of the photoelectrons emitted from the metal? (c = 3.00 × 10⁸ m/s, H = 6.626 × 10^-34 J • s = 4.141 × 10^-15 ev • s,1 eV = 1.60 × 10^-19 J)

If the headlights on a car are separated by 1.3 m,how far down the road can they be resolved if the angular resolution of the eye is 5.0 × 10^-4 rad and the person has excellent vision?

If an x-ray beam of wavelength 1.4 × 10^-10 m makes an angle of 30° with a set of planes in a crystal causing first order constructive interference,what is the plane spacing?

When a certain metal is illuminated by light,photoelectrons are observed provided that the wavelength of the light is less than 669 nm.Which one of the following values is closest to the work function of this metal? (h = 6.626 × 10^-34J • s,c = 3.00 × 10⁸ m/s,1 eV = 1.60 × 10^-19 J)