Question 1

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?

Accepted Answer

The wavelength ($\lambda$) can be calculated using the grating equation $\lambda = \frac{d \sin \theta}{m}$, where $d$ is the grating spacing (the inverse of the number of lines per meter), $\theta$ is the angle of the maximum, and $m$ is the order of the maximum. Given that the grating is 75 mm wide with 50,000 lines, the spacing $d = \frac{75 \times 10^{-3}}{50000} = 1.5 \times 10^{-6}$ m. For the second-order maximum ($m = 2$) at 32.5°, $\lambda = \frac{1.5 \times 10^{-6} \sin(32.5°)}{2}$, which calculates to approximately 403 nm.

Question 2

A soap bubble,when illuminated at normal incidence with light of 463 nm,appears to be especially reflective.If the index of refraction of the film is 1.35,what is the minimum thickness the soap film can be if it is surrounded by air?

Accepted Answer

The answer of A soap bubble,when illuminated at normal incidence...

Question 3

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?

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 4

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)

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 (3.0 × $10^{8}$ m/s) and $\lambda$ is the wavelength in vacuum (600 nm or 600 × $10^{-9}$ m). Thus, $f = \frac{3.0 \times 10^{8}}{600 \times 10^{-9}} = 5.0 \times 10^{14}$ Hz.

Question 5

A beam of light of wavelength 600 nm is incident normally on a 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?

Accepted Answer

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

Question 6

The distance between the ruled lines on a grating is 1770 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?

Accepted Answer

The longest wavelength in the third order spectrum is found using the grating equation: d*sin(θ) = m*λ. For maximum wavelength, sin(θ) = 1, so λ = d/m. Here, d = 1770 nm and m = 3, giving λ = 1770 nm / 3 = 590 nm.

Question 7

A 17-mm-wide 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?

Accepted Answer

The longest wavelength that forms an intensity maximum in the fifth order is given by:$$\lambda = \frac{d \sin 	heta}{m}$$where 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:$$\lambda = \frac{d}{m}$$Substituting the given values, we get:$$\lambda = \frac{17 	ext{ mm} 	imes 10^{-3} 	ext{ m/mm}}{5} = 3.4 	imes 10^{-6} 	ext{ m} = 3400 	ext{ nm}$$This is the longest wavelength that forms an intensity maximum in any order.

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 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?

Accepted Answer

The slit spacing is given as 20 cm which is equivalent to 200 mm.

We know that for a grating, the condition for maxima is given as:

d sinθ = mλ

where d is the distance between the slits, θ is the angle between the incident beam and the normal to the grating, m is the order of the maxima and λ is the wavelength of the light.

In this case, the incident beam is normal to the grating, so sinθ = 0.

The condition for maxima reduces to:

d = mλ

We are told that the intensity maxima occur 2.0° apart in the central region. This means that the angle between successive maxima is:

θ = 2.0° = 0.035 radians

Since the incident beam is normal to the grating, this angle is also the angle between the diffracted beams.

For the first order maxima (m = 1), the slit spacing should be equal to the wavelength of the microwaves.

Thus,

λ = d/m

Substituting d = 200 mm and m = 1, we get:

λ = 200 mm/1 = 200 mm = 20 cm = 7.0 mm

Therefore, the wavelength of the microwaves is 7.0 mm which corresponds to choice C.

Question 9

A metallic sheet has a large number of parallel slits,5.0 mm wide and 20 cm apart.This sheet is used as a 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)

Accepted Answer

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

Question 10

A grating has 6.00 × 10³ lines per centimeter ruled on it.What is the angular separation between the second and the third orders on the same side of the central order when the grating is illuminated with a beam of light of wavelength 550 nm?

Accepted Answer

The answer of A grating has 6.00 × 10³ lines...

Question 11

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

Accepted Answer

The highest order m that contains the entire visible spectrum can be found using the grating equation for the first order (m=1) at the shortest wavelength (400 nm), since this will be the limiting factor. The grating equation is d*sin(θ) = m*λ, where d is the grating spacing, θ is the diffraction angle, λ is the wavelength, and m is the order number. The grating spacing d is the inverse of the number of lines per millimeter, so d = 1/450 mm = 1/450,000 m = 2.22 x 10^-6 m. For the grating to diffract light, sin(θ) must be less than or equal to 1. At the limit, for m=1 and λ=400 nm (4 x 10^-7 m), we have 2.22 x 10^-6 * 1 = 1 * 4 x 10^-7, which gives m = 2.22 x 10^-6 / 4 x 10^-7 = 5.55. Since m must be an integer, the highest possible order that can contain the entire visible spectrum is m=5. However, this calculation was based on a misunderstanding of the question's requirement for the highest order that contains the entire spectrum, not just the limit for the shortest wavelength. The correct approach is to use the condition that for the highest order, the diffraction angle for the longest wavelength (700 nm) should be less than 90 degrees, which corresponds to the first order where the spectrum starts to disappear. Using the grating equation, we find the order m that satisfies this condition for the longest wavelength, ensuring the entire spectrum is included up to the highest possible order without exceeding the physical constraints of diffraction.

Question 12

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

Accepted Answer

The formula for finding the angle of diffraction for a grating is given by:

nλ = d*sinθ

where n is the order of the maximum (in this case, n=2), λ is the wavelength of light, d is the spacing between the grating lines, and θ is the angle of diffraction.

We are given that the grating has 7000 lines per centimeter, which means that the spacing between the lines is:

d = 1/7000 cm = 10^-5 m

The second-order maximum occurs at an angle of θ = 62.4°. Plugging these values into the formula above, we get:

2λ = (10^-5)*sin(62.4°)

λ = (10^-5)*sin(62.4°)/2

λ ≈ 633 nm

Therefore, the wavelength of the light is 633 nm, which is answer choice C.

Question 13

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?

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 14

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

Accepted Answer

The slit spacing $d$ can be found using the diffraction grating equation for the first order maximum ($m = 1$), which is $d \sin(\theta) = m\lambda$. Here, $\lambda = 600 \, \text{nm} = 600 \times 10^{-9} \, \text{m}$, $\theta = 25.0^\circ$, and $m = 1$. Solving for $d$ gives $d = \frac{\lambda}{\sin(\theta)}$. Plugging in the values, $d = \frac{600 \times 10^{-9}}{\sin(25^\circ)}$, which calculates to approximately $1.42 \, \mu\text{m}$.

Question 15

When a grating having 5000 lines/cmis illuminated by monochromatic light,the angle between the central maximum and the fourth order maximum is 47.2°.What is the wavelength of the light?

Accepted Answer

The grating equation is d*sin(θ) = m*λ, where d is the distance between grating lines (1/5000 cm), θ is the angle, m is the order, and λ is the wavelength. Solving for λ with m=4 and θ=47.2° gives approximately 367 nm.

Question 16

A metallic sheet has a large number of parallel slits,5.0 mm wide and 20 cm apart.This sheet is used as a 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?

Accepted Answer

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

Question 17

A soap bubble with air on both sides has an index of refraction of 1.33.What minimum thickness of the wall of this bubble will ensure maximum reflectance of normally incident light having a wavelength of 530 nm?

Accepted Answer

The answer of A soap bubble with air on both...

Question 18

Light in air of wavelength 500 nm illuminates a soap film that has an index of refraction of 1.3 and air on both sides.What is the minimum thickness of this film that will produce cancellation in the reflected light when the light is incident normally on the film?

Accepted Answer

The answer of Light in air of wavelength 500 nm...

Question 19

The distance between the ruled lines on a grating is 1900 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?

Accepted Answer

The angular width of the gap between the first and second order spectra is determined by the difference in angles for the longest wavelength of the first order and the shortest wavelength of the second order. Using the grating equation d*sin(θ) = m*λ, for m=1 and m=2, the difference in angles is calculated to be approximately 3.3°.

Question 20

Light is incident perpendicularly from air onto a liquid film that is on a glass plate.The liquid film is $199 \mathrm {~nm}$ thick,and the liquid has index of refraction 1.60.The glass has index of refraction 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 nm and 700 nm.

Accepted Answer

The answer of Light is incident perpendicularly from air onto...

Quiz 24: Wave Optics