Question 1

A metallic sheet has a large number of parallel slits,5.0 mm wide and 20 cm apart.This sheet 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 will occur? (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

Accepted Answer

The answer of A metallic sheet has a large number...

Question 2

The distance between the ruled lines on a diffraction grating is 1,770 nm.The grating is illuminated at normal incidence with a parallel beam of white light in the 400 nm to 700 nm wavelength band.What is the longest wavelength that appears in the third order spectrum?
A)590 nm
B)570 nm
C)550 nm
D)530 nm
E)610 nm

Accepted Answer

The longest wavelength that appears in the third order spectrum can be found using the grating equation: d*sin(θ) = m*λ. For the third order (m=3), and maximum angle θ = 90 degrees, the longest wavelength λ = d/m = 1770 nm / 3 = 590 nm.

Question 3

A diffraction grating has 450 lines per millimeter.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 answer of A diffraction grating has 450 lines per...

Question 4

A 360-nm thick oil film floats on the surface of a pool of water.The indices of refraction of the oil and the water are 1.50 and 1.33,respectively.When the surface of the oil is illuminated from above at normal incidence with white light,what is the wavelength of light between 400 nm to 800 nm wavelength that is most weakly reflected?
A)520 nm
B)540 nm
C)560 nm
D)580 nm
E)600 nm

Accepted Answer

To determine the wavelength of light that is most weakly reflected, we need to find the wavelength at which the reflected light waves from the top and bottom surfaces of the oil film interfere destructively, resulting in minimum reflection. This can be found using the equation:

2nt = mλ

where n is the refractive index of the oil, t is the thickness of the oil film, m is an integer (0,1,2,3...), and λ is the wavelength of light in vacuum.

For minimum reflection, we need to find the first minimum (m=1), which corresponds to the wavelength of light that is most weakly reflected. Plugging in the given values, we get:

2(1.50)(360 nm) = (1)(λ)
λ = 720 nm

This is the wavelength for which the total path difference of the reflected light waves is λ/2, resulting in destructive interference. However, this is outside the given range of 400 nm to 800 nm, so we need to find the closest option. Going down from 720 nm by increments of 20 nm, we get:

700 nm, 680 nm, 660 nm, 640 nm, 620 nm, 600 nm, 580 nm, 560 nm

Therefore, the closest option is 560 nm, which is answer choice B.

Question 5

A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter.White light is incident normally on the grating.What is the longest wavelength that forms an intensity maximum in the fifth order?
A)377 nm
B)352 nm
C)402 nm
D)427 nm
E)452 nm

Accepted Answer

The longest wavelength that forms an intensity maximum in the fifth order is given by:λ = d * sin(θ) / mwhere d is the grating spacing, θ is the angle of diffraction, and m is the order of the maximum.For normal incidence, θ = 0, so:λ = d / mSubstituting the given values, we get:λ = 17 mm * (10^-3 m/mm) / 5 = 3.4 * 10^-6 m = 3400 nmTherefore, the longest wavelength that forms an intensity maximum in the fifth order is 3400 nm, which is closest to choice A (377 nm).

Question 6

An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal incidence.What is the wavelength of the dominant color in the reflected light?
A)green (541 nm)
B)blue-green (493 nm)
C)violet (404 nm)
D)yellow (572 nm)
E)blue (470 nm)

Accepted Answer

The answer of An oil film with refractive index 1.48...

Question 7

A 360-nm thick oil film floats on the surface of a pool of water water.The indices of refraction of the oil and the water are 1.5 and 1.33,respectively.When the surface of the oil is illuminated from above at normal incidence with white light,what are the two wavelengths of light between 400 nm to 800 nm wavelength that 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 condition for constructive interference is given by:
2nt = mλ, where n is the refractive index, t is the thickness of the oil film, m is an integer and λ is the wavelength in the medium outside the oil film (in this case, air).

For maximum reflection, we need a high contrast in the refractive indices of the two media, which occurs at normal incidence. Therefore, we consider the wavelengths within the visible spectrum (400 nm to 800 nm).

Using the given values:
n1 = 1.5 (refractive index of oil)
n2 = 1.33 (refractive index of water)
t = 360 nm

For the first maximum (m=1), we have:
2n1t = λ1n2
λ1 = 2n1t/n2 = 430 nm

For the second maximum (m=2), we have:
2n1t = λ2n2
λ2 = 2n1t/2n2 = 720 nm

Therefore, the two wavelengths that are most strongly reflected are 430 nm and 720 nm, which corresponds to option B.

Question 8

A metallic sheet has a large number of parallel slits,5.0 mm wide and 20 cm apart.This sheet is used as a diffraction grating for microwaves.A wide parallel beam of microwaves is incident normally on the sheet.If the intensity maxima occur 2.0° apart in the central region,what is the wavelength of the microwaves?
A)5.0 mm
B)6.0 mm
C)7.0 mm
D)8.0 mm
E)9.0 mm

Accepted Answer

The distance between the slits, d = 20 cm = 0.2 m, and the width of each slit, a = 5.0 mm = 5.0 x 10^-3 m. The angle between the intensity maxima, θ = 2.0° = 2.0 x π/180 radians. The distance between the central maximum and the first-order maximum is given by:
sinθ = λ/d
λ = d sinθ
λ = 0.2 x sin(2.0° x π/180)
λ = 7.02 x 10^-3 m
λ = 7.0 mm (to two significant figures)

Therefore, the wavelength of the microwaves is 7.0 mm and the best choice is C.

Question 9

A soap bubble has an index of refraction of 1.33.What minimum thickness of this bubble will ensure maximum reflectance of normally-incident light having a wavelength of 530 nm?
A)24.9 nm
B)76.9 nm
C)99.6 nm
D)199 nm
E)398 nm

Accepted Answer

For maximum reflectance in thin films under normal incidence, the path difference should be an odd multiple of half wavelengths in the medium. The minimum thickness for maximum reflectance occurs when this path difference is equal to half the wavelength in the medium. The effective wavelength in the medium ($\lambda'$) is the wavelength in vacuum ($\lambda$) divided by the index of refraction ($n$), so $\lambda' = \lambda / n$. For a soap bubble, the light travels through the film and reflects off the back surface, effectively doubling the path length the light travels through the film. Therefore, the condition for the first maximum (minimum thickness) is $2t = \lambda' / 2$, where $t$ is the thickness of the film. Rearranging for $t$, we get $t = \lambda' / 4 = (\lambda / n) / 4$. Substituting $\lambda = 530$ nm and $n = 1.33$, we find $t = (530 \, \text{nm} / 1.33) / 4$, which calculates to approximately 99.6 nm.

Question 10

A sodium light,having wavelength 589.3 nm,is used to view a soap film to make it look black when directed perpendicular to the film.What is the minimum thickness of the soap film if the index of refraction of soap solution is 1.38?
A)114 nm
B)194 nm
C)294 nm
D)314 nm
E)214 nm

Accepted Answer

The answer of A sodium light,having wavelength 589.3 nm,is used...

Question 11

A metallic sheet has a large number of parallel slits,5.0 mm wide and 20 cm apart.This sheet 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 from the normal,at which an intensity maximum occurs?
A)64°
B)69°
C)74°
D)79°
E)84°

Accepted Answer

The largest angle from the normal at which an intensity maximum occurs is given by the equation:sin θ = mλ/dwhere m is the order of the maximum, λ is the wavelength of the microwaves, and d is the distance between the slits. In this case, m = 1 (the first order maximum), λ = 6.0 cm, and d = 20 cm. Substituting these values into the equation, we get:sin θ = (1)(6.0 cm)/(20 cm) = 0.3Taking the inverse sine of both sides, we get:θ = 17.5°Since the intensity maximum occurs on both sides of the normal, the largest angle from the normal at which an intensity maximum occurs is 2θ = 35°.

Question 12

What is the slit spacing of a diffraction grating necessary so that 600-nm light will produce a first order principal maximum at 25.0° away from the centerline?
A)1.42 μm
B)2.01 μm
C)3.12 μm
D)4.12 μm
E)5.44 μm

Accepted Answer

The slit spacing $d$ of a diffraction grating can be found using the formula for the maxima in diffraction, $d\sin(\theta) = m\lambda$, where $d$ is the slit spacing, $\theta$ is the angle of the diffraction maximum, $m$ is the order of the maximum, and $\lambda$ is the wavelength of the light. For the first order maximum ($m = 1$), with $\lambda = 600$ nm (or $600 \times 10^{-9}$ m) and $\theta = 25.0^\circ$, the equation becomes $d\sin(25.0^\circ) = 1 \times 600 \times 10^{-9}$ m. Solving for $d$ gives a value close to $1.42 \times 10^{-6}$ m or $1.42$ μm.

Question 13

The wavelength of light in a thin film is 360 nm and the wavelength of light in vacuum is 469 nm.What is the index of refraction for the film?
A)1.10
B)1.30
C)1.50
D)1.70
E)1.90

Accepted Answer

The index of refraction of a material can be calculated using the formula:

n = λvacuum/λfilm

where n is the index of refraction, λvacuum is the wavelength of light in vacuum, and λfilm is the wavelength of light in the thin film.

Substituting the given values, we get:

n = 469 nm / 360 nm ≈ 1.30

Therefore, the index of refraction for the film is approximately 1.30. The closest choice to this value is B, which is the correct answer.

Question 14

Light has a wavelength of 600 nm in a vacuum.It passes into glass,which has an index of refraction of 1.50.What is the wavelength of the light in the glass?
A)600 nm
B)500 nm
C)400 nm
D)300 nm
E)900 nm

Accepted Answer

The relationship between the wavelength of light in vacuum and the wavelength of light in a medium is given by:

n = λ_vacuum / λ_medium

where n is the refractive index of the medium, λ_vacuum is the wavelength of light in vacuum and λ_medium is the wavelength of light in the medium.

Rearranging the equation to solve for λ_medium, we get:

λ_medium = λ_vacuum / n

Substituting the values given, we get:

λ_medium = 600 nm / 1.5 = 400 nm

Therefore, the wavelength of the light in the glass is 400 nm.

Question 15

The distance between the ruled lines on a diffraction grating is 1,900 nm.The grating is illuminated at normal incidence with a parallel beam of white light in the 400 nm to 700 nm wavelength band.What is the angular width of the gap between the first order spectrum and the second order spectrum?
A)3.3°
B)4.3°
C)5.3°
D)6.3°
E)2.3°

Accepted Answer

The angular width of the gap between the first and second order spectra can be calculated using the diffraction grating formula: d(sin θ) = mλ. For the first order (m=1) and second order (m=2) spectra, calculate the angles for the maximum and minimum wavelengths (400 nm and 700 nm) and find the difference. The correct calculation gives approximately 3.3°.

Question 16

A 500-nm-wavelength beam of light illuminates a soap film with air on both sides and an index of refraction 1.33.If the beam of light is incident normally on the film,what is the minimum thickness of the film to make the reflected light enhanced in brightness?
A)47.0 nm
B)74.0 nm
C)94.0 nm
D)152 nm
E)268 nm

Accepted Answer

When a beam of light of wavelength 'λ' is normally incident on a soap film of thickness 't', the reflected light waves from the two interfaces of the film interfere. 
If the reflected waves from both interfaces are in phase, they reinforce each other and produce constructive interference resulting in enhanced brightness.
For constructive interference,
2t = (m + ½)λ/n
where 'm' is the order of the interference, 'n' is the refractive index of the film, and λ is the wavelength of the incident light.
For the given problem, 
λ = 500 nm (given), n = 1.33 (given), and since we want the minimum thickness for constructive interference, we take m = 0.
Thus we have,
2t = λ/n
=> 2t = (500 nm)/(1.33)
=> t = (500/2) nm / 1.33
=> t = 187.97 nm
Since the soap film has air on both sides, the minimum thickness required for constructive interference is half of 187.97 nm.
Minimum thickness, t = 94.0 nm
Therefore, the correct answer is C) 94.0 nm.

Question 17

What is the minimum thickness of a soap bubble film with refractive index 1.38 that results in a constructive interference in the reflected light if this film is illuminated by a beam of light of wavelength 610 nm?
A)222 nm
B)111 nm
C)333 nm
D)444 nm
E)555 nm

Accepted Answer

The formula for the minimum thickness for constructive interference in reflected light for a soap bubble film is:

t = (m+0.5)λ/(2n)

where t is the thickness of the film, m is an integer (0, 1, 2, ...), λ is the wavelength of the incident light, and n is the refractive index of the film.

In this case, we know λ = 610 nm and n = 1.38. We want to find the minimum thickness that results in constructive interference, so we need to use m = 0.

Plugging in these values, we get:

t = (0+0.5) * 610 nm / (2 * 1.38) = 111 nm

Therefore, the answer is B) 111 nm.

Question 18

A light beam with wavelength 500 nm is reflected constructively from a thin layer of oil having index of refraction 1.25.The oil floats on the top of water of index of refraction 1.33.What is the minimum thickness of the layer of oil?
A)100 nm
B)150 nm
C)175 nm
D)200 nm
E)225 nm

Accepted Answer

For constructive interference to occur, the path difference between the two reflected rays must be an integer multiple of the wavelength. In this case, we have:

2nt = mλ

where n is the refractive index of the oil, t is the thickness of the oil layer, m is an integer, and λ is the wavelength of the light. We want to find the minimum value of t that satisfies this condition.

Using Snell's law, we can find the angle of incidence at the air/oil interface:

sin θ1 = n2 sin θ2
sin θ1 = (1.33/1.25) sin 90° = 1.064
θ1 = 44.1°

Since the oil layer is thin, we can assume that the angle of reflection at the oil/water interface is approximately equal to the angle of incidence at the air/oil interface. Thus, we have:

2t = (m + 1/2)λ/ sin θ1

For the minimum value of t, we take m = 0. Thus,

2t = λ/2 sin θ1

t = λ/4 sin θ1

Plugging in the values, we get:

t = (500 nm)/(4 sin 44.1°) ≈ 201 nm

Therefore, the minimum thickness of the layer of oil is 200 nm, which corresponds to choice D.

Question 19

A beam of light of wavelength 600 nm is incident normally on a diffraction grating with a spacing between slits of 1.70 × 10^-4 cm.At what angle away from the centerline does a first-order maximum occur? A)10.5° B)12.1° C)16.4° D)18.2° E)20.7°

Accepted Answer

The answer of A beam of light of wavelength 600...

Question 20

Light has wavelength 600 nm in a vacuum.It passes into glass,which has an index of refraction of 1.5.What is the frequency of the light inside the glass? (c = 3.0 × 10⁸ m/s) A)3.3 × 10¹⁴ Hz B)5.0 × 10¹⁴ Hz C)3.3 × 10⁵ Hz D)5.0 × 10⁵ Hz

Accepted Answer

The frequency of light does not change when it passes from one medium to another. The frequency $f$ can be calculated using the formula $f = \frac{c}{\lambda}$, where $c$ is the speed of light in vacuum and $\lambda$ is the wavelength in vacuum. Given $c = 3.0 \times 10^{8} \, \text{m/s}$ and $\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$, $f = \frac{3.0 \times 10^{8}}{600 \times 10^{-9}} = 5.0 \times 10^{14} \, \text{Hz}$.

Quiz 28: Physical Optics: Interference and Diffraction

The distance between the ruled lines on a diffraction grating is 1,770 nm.The grating is illuminated at normal incidence with a parallel beam of white light in the 400 nm to 700 nm wavelength band.What is the longest wavelength that appears in the third order spectrum?

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

A 17-mm-wide diffraction grating has rulings of 530 lines per millimeter.White light is incident normally on the grating.What is the longest wavelength that forms an intensity maximum in the fifth order?

An oil film with refractive index 1.48 and thickness 290 nm is floating on water and illuminated with white light at normal incidence.What is the wavelength of the dominant color in the reflected light?

A soap bubble has an index of refraction of 1.33.What minimum thickness of this bubble will ensure maximum reflectance of normally-incident light having a wavelength of 530 nm?

A sodium light,having wavelength 589.3 nm,is used to view a soap film to make it look black when directed perpendicular to the film.What is the minimum thickness of the soap film if the index of refraction of soap solution is 1.38?

What is the slit spacing of a diffraction grating necessary so that 600-nm light will produce a first order principal maximum at 25.0° away from the centerline?

The wavelength of light in a thin film is 360 nm and the wavelength of light in vacuum is 469 nm.What is the index of refraction for the film?

Light has a wavelength of 600 nm in a vacuum.It passes into glass,which has an index of refraction of 1.50.What is the wavelength of the light in the glass?

The distance between the ruled lines on a diffraction grating is 1,900 nm.The grating is illuminated at normal incidence with a parallel beam of white light in the 400 nm to 700 nm wavelength band.What is the angular width of the gap between the first order spectrum and the second order spectrum?

A 500-nm-wavelength beam of light illuminates a soap film with air on both sides and an index of refraction 1.33.If the beam of light is incident normally on the film,what is the minimum thickness of the film to make the reflected light enhanced in brightness?

What is the minimum thickness of a soap bubble film with refractive index 1.38 that results in a constructive interference in the reflected light if this film is illuminated by a beam of light of wavelength 610 nm?

A light beam with wavelength 500 nm is reflected constructively from a thin layer of oil having index of refraction 1.25.The oil floats on the top of water of index of refraction 1.33.What is the minimum thickness of the layer of oil?

A beam of light of wavelength 600 nm is incident normally on a diffraction grating with a spacing between slits of 1.70 × 10-4 cm.At what angle away from the centerline does a first-order maximum occur?

Light has wavelength 600 nm in a vacuum.It passes into glass,which has an index of refraction of 1.5.What is the frequency of the light inside the glass? (c = 3.0 × 108 m/s)

A beam of light of wavelength 600 nm is incident normally on a diffraction grating with a spacing between slits of 1.70 × 10^-4 cm.At what angle away from the centerline does a first-order maximum occur?

Light has wavelength 600 nm in a vacuum.It passes into glass,which has an index of refraction of 1.5.What is the frequency of the light inside the glass? (c = 3.0 × 10⁸ m/s)