Question 1

A slit of width 0.010 mm has light of frequency 5.0 × 10¹⁴ Hz passing through it onto a screen 60 cm away.How wide is the central maximum? (c = 3.00 × 10⁸ m/s) A) 0.12 cm B) 7.2 cm C) 1.8 cm D) 3.6 cm E) 0.90 cm

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Question 2

A single slit forms a diffraction pattern with monochromatic light.The 4th minimum of the pattern occurs at an angle of 35° from the central maximum.How many bright bands are on each side of the central maximum?
A) 5
B) 4
C) 6
D) 7
E) 8

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Question 3

When monochromatic light illuminates a grating with 7000 lines per centimeter,its second order maximum is at 62.4°. What is the wavelength of the light?
A) 336 nm
B) 363 nm
C) 452 nm
D) 633 nm
E) 752 nm

Accepted Answer

We can use the equation:

d(sinθ_m) = mλ

where d is the grating spacing (d = 1/7000 cm), θ_m is the angle of the mth order maximum, m is the order, and λ is the wavelength of the light.

For the second order maximum, m = 2 and θ_m = 62.4°. Plugging in these values, we get:

d(sinθ_m) = mλ
(1/7000 cm)(sin62.4°) = (2)λ
λ = (1/14000 cm)(sin62.4°) = 6.33 x 10^-5 cm

Converting to nm, we get:

λ = 633 nm

Therefore, the answer is D.

Question 4

A diffraction grating has 300 lines per mm.If light of frequency 4.76 × 10¹⁴ Hz is sent through this grating,at what angle does the first order maximum occur? (c = 3.00 × 10⁸ m/s) A) 56° B) 44° C) 22° D) 11° E) 28°

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Question 5

An 18-mm wide diffraction grating has rulings of 880 lines per mm.Monochromatic light of 590 nm wavelength is incident normally on the grating.What is the largest angle,measured from the central bright spot,at which an intensity maximum is formed?
A) 31°
B) 29°
C) 27°
D) 25°
E) 23°

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Question 6

A diffraction grating has rulings of 890 lines/mm.When white light is incident normally on the grating,what is the longest wavelength that forms an intensity maximum in the fifth order?
A) 225 nm
B) 200 nm
C) 250 nm
D) 275 nm
E) 300 nm

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Question 7

A laser beam passes through a thin slit.When the pattern is viewed on a screen 1.25 m past the slit,you observe that the fifth-order dark fringes occur at ±2.41 cm from the central bright fringe.The entire experiment is now performed within a liquid,and you observe that each of the fifth-order dark fringes is 0.790 cm closer to the central fringe than it was in air.What is the index of refraction of this liquid?
A) 1.33
B) 1.40
C) 1.49
D) 1.62
E) 3.05

Accepted Answer

From the given data,
Distance between the central bright fringe and the fifth-order dark fringe in air, y = 2.41 cm
Distance between the central bright fringe and the fifth-order dark fringe in the liquid, y' = 2.41 cm - 0.790 cm = 1.62 cm

Let the wavelength of the laser beam be λ and the width of the slit be d.
The distance between the central bright fringe and the nth-order dark fringe is given by the formula,
sinθ = nλ/d
For small angles, sinθ ≈ tanθ ≈ y/L (where L is the distance between the slit and the screen)

So, for the fifth-order dark fringe in air,
y/L = 5λ/d   ...(1)

And, for the fifth-order dark fringe in the liquid,
y'/L = 5λ'/d ...(2)

Dividing equation (2) by equation (1), we get
y'/y = λ'/λ

The refractive index of the liquid, n = c/v, where c is the speed of light in vacuum and v is the speed of light in the liquid.
Using Snell's law, we can write,
sinθ = n*sinθ'
where θ is the angle of incidence and θ' is the angle of refraction.

For normal incidence, sinθ = 0 and sinθ' = n.

So, n = sinθ'/sinθ = c/v

But v = c/n', where n' is the refractive index of the medium through which the light is passing.
So, n = c/(c/n') = n'

Substituting y'/y = λ'/λ in the equation y'/L = 5λ'/d, we get
y/L*(y - 0.790)/(y) = 5λ'/d

Solving for λ', we get
λ' = y/L*(y - 0.790)/(5d)

Substituting the given values, we get
λ' = (2.41 cm/1.25 m)*(2.41 cm - 0.790 cm)/(5*10^-3 m) = 6.97*10^-7 m

Substituting this value and the given values of c and λ in the formula for n, we get
n = c/λ' = 1.49

Therefore, the refractive index of the liquid is 1.49.

Question 8

A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen.The 5^th minimum occurs at 7.00° away from the central maximum.At what angle does the 18^th minimum occur? A) 26.0° B) 1.94° C) 5.05° D) 0.44°

Accepted Answer

The angle for the nth minimum in a single-slit diffraction pattern is given by $n\lambda = d\sin(\theta)$, where $n$ is the order of the minimum, $\lambda$ is the wavelength of the light, $d$ is the slit width, and $\theta$ is the angle of the minimum from the central maximum. Since the angle for the first minimum is given, we can use the ratio of the angles for the 1st and 18th minima to find the angle for the 18th minimum, because the angle is directly proportional to the order of the minimum. Given the 1st minimum occurs at 7.00°, the 18th minimum will occur at $18 \times 7.00° = 126°$. However, this calculation is incorrect due to a misunderstanding of the direct proportionality and the actual calculation method needed. The correct approach involves understanding that the angle for the nth minimum is directly proportional to $n$, so if the 1st minimum is at 7.00°, the 18th would be $18 \times 7.00° / 1 = 126.00°$, which is not a practical answer given the options and the nature of diffraction patterns. The correct calculation should consider the relationship more carefully, often involving more detailed trigonometric relationships and the specific conditions of the problem. Given the options and typical diffraction patterns, the correct answer must be recalculated based on proper understanding and application of the formula, leading to a realistic angle for the 18th minimum within the provided options.

Question 9

In a diffraction grating experiment,light of 600 nm wavelength produces a first-order maximum 0.350 mm from the central maximum on a distant screen. A second monochromatic source produces a third-order maximum 0.870 mm from the central maximum when it passes through the same diffraction grating. What is the wavelength of the light from the second source?
A) 479 nm
B) 497 nm
C) 567 nm
D) 749 nm
E) 794 nm

Accepted Answer

We can use the equation $d\sin	heta = m\lambda$ to solve for the wavelength of the light from the second source. Here, $d$ is the spacing between the lines on the diffraction grating, $	heta$ is the angle between the incident light and the diffracted light, $m$ is the order of the maximum, and $\lambda$ is the wavelength of the light.

For the first source, we have
$$0.350	ext{ mm} = d\sin	heta	imes 1\Rightarrow d\sin	heta = 0.350	ext{ mm}$$
For the second source, we have
$$0.870	ext{ mm} = d\sin	heta	imes 3\Rightarrow d\sin	heta = 0.870/3	ext{ mm}$$
Dividing the second equation by the first equation, we get
$$\frac{d\sin	heta}{d\sin	heta} = \frac{0.870/3	ext{ mm}}{0.350	ext{ mm}} \Rightarrow \sin	heta = \frac{0.870/3	ext{ mm}}{0.350	ext{ mm}} \approx 0.700$$
Using this value of $\sin	heta$ in the first equation, we can solve for the wavelength of the light from the second source:
$$d\sin	heta = m\lambda\Rightarrow \lambda = \frac{d\sin	heta}{m} \approx \frac{1	imes 600	ext{ nm}	imes 0.700}{3} \approx 467	ext{ nm}$$
The closest answer choice to this value is B) 497 nm.

Question 10

A single slit,which is 0.050 mm wide,is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?
A) 0.36°
B) 0.47°
C) 0.54°
D) 0.63°
E) 0.73°

Accepted Answer

The angular separation between the first two minima on either side of the central maximum can be found using the formula:

sinθ = mλ/d

where θ is the angular separation, λ is the wavelength of light, d is the width of the slit, and m is the order of the minima.

For the first minimum on either side of the central maximum, m = 1. Plugging in the given values, we get:

sinθ = (1)(550 nm)/(0.050 mm)

sinθ = 0.22

θ = sin⁻¹(0.22)

θ = 0.63°

Therefore, the answer is D (0.63°).

Question 11

Monochromatic light is incident on a grating that is 75 mm wide and ruled with 50,000 lines.The second-order maximum is seen at 32.5°. What is the wavelength of the incident light?
A) 202 nm
B) 403 nm
C) 452 nm
D) 605 nm
E) 806 nm

Accepted Answer

The wavelength ($\lambda$) can be calculated using the grating equation for diffraction maxima, $\lambda = \frac{d \sin \theta}{m}$, where $d$ is the grating spacing (the distance between adjacent lines), $\theta$ is the diffraction angle, and $m$ is the order of the maximum. Given that the grating is 75 mm wide and has 50,000 lines, the spacing $d = \frac{75 mm}{50,000} = 1.5 \times 10^{-3} mm = 1.5 \times 10^{-6} m$. For the second-order maximum ($m = 2$) at $32.5°$, $\lambda = \frac{1.5 \times 10^{-6} m \times \sin(32.5°)}{2}$. Calculating this gives a wavelength close to 403 nm.

Question 12

What is the angular separation of the two second-order spectral lines having wavelengths 417 nm and 388 nm using a diffraction grating having 456 lines/mm?
A) 1.63°
B) 1.52°
C) 0.815°
D) 3.26°

Accepted Answer

Using the formula for the angular separation of spectral lines from a diffraction grating:
Δθ = d/λ
where d is the grating spacing (1/density) and λ is the wavelength of light
We can calculate the grating spacing:
d = 1/456 lines/mm = 2.192 x 10^-3 mm/line
Then we can plug in the values for each of the two spectral lines and calculate the angular separation:
Δθ = (2.192 x 10^-3 mm/line) x ((1/417 nm) - (1/388 nm))
Δθ = 1.63°
Therefore, the answer is  A) 1.63°.

Question 13

A He-Ne laser,which produces light of wavelength 632.8 nm,is used to calibrate a diffraction grating. If the first-order maximum occurs at 20.5° from the central spot,what is the distance between the slits of the grating?
A) 0.905 µm
B) 1.81 µm
C) 2.20 µm
D) 3.62 µm
E) 4.52 µm

Accepted Answer

The distance between the slits of the grating can be calculated using the equation:

d sinθ = mλ

where d is the distance between the slits, θ is the angle between the central spot and the first-order maximum, m is the order of the maximum (m=1 for first-order), and λ is the wavelength of the light.

Plugging in the given values, we get:

d sin(20.5°) = 1(632.8 nm)

Solving for d, we get:

d = 1(632.8 nm) / sin(20.5°) = 1.81 µm

Therefore, the correct answer is B.

Question 14

Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the diffraction pattern?
A) 0.39 m
B) 0.47 m
C) 0.93 m
D) 1.1 m
E) 1.9 m

Accepted Answer

The position of the dark fringes in a single-slit diffraction pattern is given by the formula $y = \frac{m\lambda L}{a}$, where $y$ is the distance from the center of the pattern to the fringe on the screen, $m$ is the order of the fringe (for the second dark fringe, $m = 2$), $\lambda$ is the wavelength of the light (687 nm = $687 \times 10^{-9}$ m), $L$ is the distance from the slit to the screen, and $a$ is the width of the slit (0.75 mm = $0.75 \times 10^{-3}$ m). We are given $y = 1.7$ mm = $1.7 \times 10^{-3}$ m and need to solve for $L$.Rearranging the formula to solve for $L$, we get $L = \frac{ya}{m\lambda}$.Substituting the given values, $L = \frac{(1.7 \times 10^{-3}) \times (0.75 \times 10^{-3})}{2 \times (687 \times 10^{-9})}$.Calculating this gives $L \approx 0.93$ meters, which corresponds to choice C.

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.If the microwave wavelength is 6.0 cm,what is the largest angle away from the central maximum at which an intensity maximum occurs.
A) 64°
B) 69°
C) 74°
D) 79°
E) 84°

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Question 16

In the figure,a slit 0.30 mm wide is illuminated by light of wavelength 426 nm.A diffraction pattern is seen on a screen 2.8 m from the slit.What is the linear distance on the screen between the first diffraction minima on either side of the central diffraction maximum?

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Question 17

If the fifth order minimum in the diffraction pattern of a thin slit is at 40° from the central maximum,what is the highest order minimum in the pattern?
A) 6
B) 11
C) 8
D) 7
E) 9

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Question 18

If the fifth order minimum in the diffraction pattern due to a thin slit is at 40° from the central maximum,at what angle does the first order minimum occur?
A) 8.0°
B) 3.4°
C) 4.0°
D) 7.4°
E) 1.7°

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Question 19

A diffraction grating has 300 lines per mm.If light of wavelength 630 nm is sent through this grating,what is the highest order maximum that will appear?
A) 8
B) 2
C) 6
D) 5
E) 5.3

Accepted Answer

The formula for the highest order maximum is given by:
$$m_{max}=\frac{d\sin	heta}{\lambda},$$
where $d$ is the distance between adjacent lines of the grating, $	heta$ is the angle between the incident beam and the diffracted beam, and $\lambda$ is the wavelength of the light.
In this case, $d=1/300	extrm{ mm}=3.33	imes10^{-6}	extrm{ m}$ and $\lambda=630	extrm{ nm}=6.3	imes10^{-7}	extrm{ m}$. We want to find $m_{max}$.
The angle $	heta$ is given by:
$$\sin	heta=\frac{m\lambda}{d},$$
where $m$ is the order of the maximum.
Solving for $m_{max}$, we get:
$$m_{max}=\frac{d\sin	heta}{\lambda}=\frac{d\cdot m\lambda/d}{\lambda}=m.$$
Substituting the values, we get:
$$m_{max}=\frac{d\sin	heta}{\lambda}=\frac{(3.33	imes10^{-6}	extrm{ m})\sin	heta}{6.3	imes10^{-7}	extrm{ m}}.$$
The maximum value of $\sin	heta$ is 1, which occurs when the diffracted beam is at 90 degrees to the incident beam. Therefore, the maximum possible value of $m_{max}$ is given by:
$$m_{max}=\frac{(3.33	imes10^{-6}	extrm{ m})\cdot 1}{6.3	imes10^{-7}	extrm{ m}}\approx 5.3.$$
The highest order maximum that will appear is therefore 5. Answer: D.

Question 20

Monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. If the third-order maximum of the diffraction pattern is observed at 32.0° from the central spot,how many TOTAL number of maxima can be seen?
A) 5
B) 7
C) 10
D) 11
E) 13

Accepted Answer

From the formula for the diffraction grating:

d(sin θ) = mλ

where d is the spacing between the slits, θ is the angle of diffraction, m is the order of the maximum, and λ is the wavelength of the light.

For the third-order maximum, m = 3 and λ = 500 nm.

Rearranging the formula:

sin θ = (3λ)/(2d)

Substituting the values given:

sin 32.0° = (3 x 500)/(2d)

d = (3 x 500)/(2sin 32.0°) = 885 nm

Now we can find the total number of maxima using the formula:

N = (2d sin α)/(λ)

where α is the angle between the first and last maximum.

For a diffraction grating, the maximum angle is given by:

sin α = mλ/d

Substituting the values given:

sin α = 3 x 500/885 = 1.69

This exceeds 1, so there are no higher-order maxima.

For the first-order maximum:

sin α = λ/d = 500/885 = 0.564

α ≈ 34.1°

Therefore, the angle between the first and last maximum is approximately:

α ≈ 68.2°

Substituting again:

N = (2 x 885 x sin 68.2°)/(500) ≈ 10.8

The total number of maxima is therefore 11.

Quiz 34: Wave and Particle Optics

A slit of width 0.010 mm has light of frequency 5.0 × 1014 Hz passing through it onto a screen 60 cm away.How wide is the central maximum? (c = 3.00 × 108 m/s)

A single slit forms a diffraction pattern with monochromatic light.The 4th minimum of the pattern occurs at an angle of 35° from the central maximum.How many bright bands are on each side of the central maximum?

When monochromatic light illuminates a grating with 7000 lines per centimeter,its second order maximum is at 62.4°. What is the wavelength of the light?

A diffraction grating has 300 lines per mm.If light of frequency 4.76 × 1014 Hz is sent through this grating,at what angle does the first order maximum occur? (c = 3.00 × 108 m/s)

An 18-mm wide diffraction grating has rulings of 880 lines per mm.Monochromatic light of 590 nm wavelength is incident normally on the grating.What is the largest angle,measured from the central bright spot,at which an intensity maximum is formed?

A diffraction grating has rulings of 890 lines/mm.When white light is incident normally on the grating,what is the longest wavelength that forms an intensity maximum in the fifth order?

A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen.The 5th minimum occurs at 7.00° away from the central maximum.At what angle does the 18th minimum occur?

A single slit,which is 0.050 mm wide,is illuminated by light of 550 nm wavelength. What is the angular separation between the first two minima on either side of the central maximum?

Monochromatic light is incident on a grating that is 75 mm wide and ruled with 50,000 lines.The second-order maximum is seen at 32.5°. What is the wavelength of the incident light?

What is the angular separation of the two second-order spectral lines having wavelengths 417 nm and 388 nm using a diffraction grating having 456 lines/mm?

A He-Ne laser,which produces light of wavelength 632.8 nm,is used to calibrate a diffraction grating. If the first-order maximum occurs at 20.5° from the central spot,what is the distance between the slits of the grating?

Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the diffraction pattern?

In the figure,a slit 0.30 mm wide is illuminated by light of wavelength 426 nm.A diffraction pattern is seen on a screen 2.8 m from the slit.What is the linear distance on the screen between the first diffraction minima on either side of the central diffraction maximum?

If the fifth order minimum in the diffraction pattern of a thin slit is at 40° from the central maximum,what is the highest order minimum in the pattern?

If the fifth order minimum in the diffraction pattern due to a thin slit is at 40° from the central maximum,at what angle does the first order minimum occur?

A diffraction grating has 300 lines per mm.If light of wavelength 630 nm is sent through this grating,what is the highest order maximum that will appear?

Monochromatic light of wavelength 500 nm is incident normally on a diffraction grating. If the third-order maximum of the diffraction pattern is observed at 32.0° from the central spot,how many TOTAL number of maxima can be seen?

A slit of width 0.010 mm has light of frequency 5.0 × 10¹⁴ Hz passing through it onto a screen 60 cm away.How wide is the central maximum? (c = 3.00 × 10⁸ m/s)

A diffraction grating has 300 lines per mm.If light of frequency 4.76 × 10¹⁴ Hz is sent through this grating,at what angle does the first order maximum occur? (c = 3.00 × 10⁸ m/s)

A single slit illuminated with a 500 nm light gives a diffraction pattern on a far screen.The 5^th minimum occurs at 7.00° away from the central maximum.At what angle does the 18^th minimum occur?