f= = s = Hz m Thus (B) is the correct answer.

Size: px

Start display at page:

Download "f= = s = Hz m Thus (B) is the correct answer."

Byron Griffith
5 years ago
Views:

1 MCAT Physics Problem Solving Drill 17: Electromagnetic Radiation Question No. 1 of 10 Question 1. Violet light has a wavelength of 700 nm. What is the frequency of this radiation? Question #01 (A) Hz (B) Hz (C) Hz (D) None of the above Frequency is the speed of light divided by the wavelength, not the wavelength divided by the frequency. B. Correct! Frequency is the speed of light divided by the wavelength. The question asked for the frequency, not the angular frequency. One of the above choices is correct. Use the formula that relates the speed of light, c, the frequency, f, and the wavelength, λ. c=fλ Solve the preceding equation for f. c f= λ Finally, substitute numerical values m/s f= = s = Hz m Thus (B) is the correct answer.

2 Question No. 2 of 10 Question 2. One radio wave has a frequency twice as large as another radio wave. What is the ratio of the wavelength of the first radio wave to the wavelength of the second radio wave? Question #02 (A) 1:4 (B) 1:2 (C) 1:1 (D) 2:1 Although the first radio wave will have a smaller frequency, this is not the correct proportion. B. Correct! The ratio of wavelengths will be the inverse of the ratio of frequencies. See the solution below. The ratio of wavelengths depends upon the ratio of frequencies, which is not 1:1. Although this is the ratio of frequencies, the ratio of wavelengths is the inverse of this proportion. The speed of light is constant for both radio waves: c=f λ =f λ We are interested in the ratio λ 1 /λ 2, which can be expressed in terms of the frequencies: λ λ f = f Since we know the following: f 1 =2, f 2 the ratio λ 1 /λ 2 can be found: λ1 1 = λ 2. 2 Thus the correct answer is (B).

3 Question No. 3 of 10 Question 3. Though 700 nm light is in the visible region of the electromagnetic spectrum, it is very near. Question #03 (A) the radio wave region (B) the microwave region (C) the infrared region (D) the ultraviolet region Use the formula to find the B field at a distance from a current carrying wire. The magnetic field equal the permittivity of free space times the current divided by 2 pie times the distance away from the wire. B. Incorrect. Although microwaves have a longer wavelength than visible light spectrum, adjacent to red color region is the infrared region. C. Correct! A wavelength of 700 nm corresponds to the red region of the visible spectrum. This is closest to the infrared region. The ultraviolet region is closest to the low-wavelength end of the visible spectrum, ~ 400 nm. A wavelength of 700 nm corresponds to the red region of the visible spectrum. This is closest to the infrared region. On the other hand, wavelengths close to 400 nm correspond to violet light, which is closest to UV region. Microwaves have wavelengths even larger then infrared.

4 Question No. 4 of 10 Question #04 Question 4. Redshift is a phenomenon in which light from an object is shifted towards the red end of the spectrum: when an object is receding from the observer, the wavelength observed from the source is observed to increase. Which of the following phenomena is responsible for the redshift of electromagnetic radiation? (A) Doppler effect (B) ionization (C) reflection (D) refraction A. Correct! Redshift is the Doppler effect applied to electromagnetic radiation. B. Incorrect. Ionization is a process in which an electron is taken away from an atom due to electromagnetic radiation being imparted upon the atom. This has no relation to the wavelength of a source changing. Although reflection is a wave phenomenon, generally the wavelength of light is not affected upon reflection. Although wavelength may change upon refraction, which is a wave phenomenon, this depends upon the index of refraction of the material in between the source and the observer. Electromagnetic radiation is a wave phenomenon. Thus the Doppler effect, which is a wave phenomenon encountered when studying sound, can be applied to light. According to the Doppler effect, when an object receding from the observer, the frequency will be lower. Equivalently, when an object is receding from the observer, the wavelength will be higher, or shifted towards the red end of the spectrum. Although the equations describing the Doppler effect for sound and light are different (in the case of light, beyond the scope of MCAT physics!), they are conceptually similar.

5 Question No. 5 of 10 Question 5. Which of the following types of electromagnetic radiation has the largest speed? (A) radio waves (B) ultraviolet waves (C) X rays (D) they are all equal Question #05 Although radio waves have a relatively large wavelength, and a small frequency they are not necessarily the fastest. B. Incorrect. Ultraviolet waves have a higher frequency than visible light. However, they are not necessarily the fastest wave listed. C. Incorrect! Alhtough X rays have the largest frequency listed, they are not necessarily the fastest wave. D. Correct. They all travel at the same speed in any particular medium. They may have different wavelengths and frequencies, but they have the same speed. They all travel at the same speed in any particular medium. They may have different wavelengths and frequencies, but they have the same speed. This is an important characteristic of all electromagnetic radiation.

6 Question No. 6 of 10 Question 6. The wavenumber, k,, is given in terms of the wavelength, λ, as k,= 2π/λ. Also, the angular frequency ω, can be equal to 2π f. What is the speed of a light wave in terms of k and its angular frequency ω? Question #06 (A) ωk (B) ω/k (C) k/ω (D) 2πω/k This quantity would be equal to 4π 2 f/λ, which is not correct and not even in the correct units for speed. B. Correct! This combination is the only answer choice provided that equals fλ for a light wave. This is the inverse of the relation desired. Although this is in the units of speed, this has an unwanted additional factor of 2π. The wave number, k, is defined in terms of the wavelength, λ, as: k=2π/λ. Recall that the angular frequency, ω, is given in terms of f as: ω=2πf. Since we know: λf=c, write the wavelength in terms of the wave number, and the frequency in terms of the angular frequency to obtain: 2π ω =c k 2π. The above expression simplifies (cancel out factors of 2π), so finally obtain: ω =c k. Thus answer (B) is the correct answer choice.

7 Question No. 7 of 10 Question 7. Glass is opaque to what type of electromagnetic radiation? Question #07 (A) Ultraviolet (B) Infrared (C) Visible light (D) Both ultraviolet and infrared A. Incorrect! Although glass is opaque to ultraviolet, that is true of another type too. B. Incorrect. Although glass is opaque to infrared, that is true for another type of electromagnetic radiation too. Glass is transparent for visible light. D. Correct. Both of those types don t get transmitted through glass. This is one of the causes for the greenhouse effect. Visible light is transmitted through glass. This heats the interior. The heated materials emit infrared light which is trapped inside since glass is opaque to infrared light. Ultraviolet too is mostly stopped by glass.

8 Question No. 8 of 10 Question #08 Question 8. Based on the validity of the following statements: I. Electric and magnetic fields are perpendicular in an electromagnetic wave II. Electromagnetic waves are transverse waves III. Electromagnetic waves require a medium to propagate which of the following answer choices is correct? (A) I only (B) II only (C) I and II (D) II and III Although statement I is true, statement II is also true. B. Incorrect. Although statement II is true, statement I is also true. C. Correct! This is the correct answer choice, since statements I and II are correct, and statement III is false. See the solution below. Statement III is false, because electromagnetic waves can travel in vacuum (no medium). Statement I is correct, since electric and magnetic fields travel perpendicular to each other while oscillating. Statement II is also true, since the oscillations occur perpendicular to direction of propagation.

9 Question No. 9 of 10 Question 9. Consider the following types of electromagnetic radiation: Radio, infrared, X rays Which of those has the highest frequency? Question #09 (A) Radio waves (B) Infrared waves (C) X rays (D) They all have the same frequency Radio waves have one of the lowest frequencies. They also have a very large wavelength. Some radio waves can be the size of a house! B. Incorrect! Infrared waves have a relatively low frequency. In terms of frequency, they are just below the visible light we observe with our eyes. C. Correct! X rays are very energetic. They also have a very high frequency and carry a lot of energy. The electromagnetic spectrum is made up of a wide variety of different frequency radiations. Of the types listed, X rays have the highest frequency. This also means they have the smallest wavelength. All of the types listed have the same speed. Radio waves Infrared Ultraviolet Gamma rays Micro waves Visible light X rays

10 Question No. 10 of 10 Question 10. If a laser beam is sent to a reflecting surface on the moon and it returns in 2.5 seconds, how far away is the moon at that point in time? Question #10 (A) 3x10^8 m (B) 1.25 m (C) 7.510^8 m (D) ^8 m This may be the value for the speed of light in a vacuum, but it isn t the distance to the moon. B. Incorrect! This would be a ridiculously small value for the distance to the moon. Although this distance is in the correct ballpark, it isn t the actual distance to the moon in this problem. This distance is the distance to the moon and back. D. Correct. With the information given in the problem, this represents the distance to the moon. The 2.5 seconds given represents the time for the laser beam to go to the moon, reflect, and return to earth. Half of this time is the one way trip, 1.25 seconds. Use the known speed of light, 3.0x10^8 m/s and determine the unknown distance. d v= t 8 8 d=t x v=(1.25s)(3.0x10 m/s)=3.75x10 m

Wave - Particle Duality of Light

Wave - Particle Duality of Light Properties of Light Objectives Explain wave-particle duality State the speed of light Describe electromagnetic waves and the electromagnetic spectrum Explain how light interacts with transparent and opaque