AP Physics - Problem Drill 21: Physical Optics 1. Which of these statements is incorrect? Question 01 (A) Visible light is a small part of the electromagnetic spectrum. (B) An electromagnetic wave is a combination of magnetic and electric fields. (C) In a vacuum all electromagnetic waves have the same velocity. (D) Red light has a shorter wavelength than blue light. (E) Radio waves have a lower frequency than x-rays. The electromagnetic spectrum covers a wide range of frequencies, only a small amount can be seen by humans. The name comes from the fact that these waves are made up of a magnetic and electric fields. The velocity of electromagnetic waves in a vacuum is the same, 3.14 x 10 8 m/s. D. Correct! Red light has a longer wavelength and lower frequency than blue light. Radio waves have long wavelengths and lower frequencies than x-rays. The electromagnetic spectrum covers a wide range of frequencies, only a small amount can be seen by humans. The name comes from the fact that these waves are made up of a magnetic and electric fields. They can be visualized as two perpendicular waves, one electric and one magnetic. The velocity of electromagnetic waves in a vacuum is the same, 3.14 x 10 8 m/s. The wavelength and frequency are different for different types of waves. In the visible spectrum, red light has a longer wavelength and lower frequency than blue light. Radio waves have long wavelengths and lower frequencies than x-rays. The correct answer is (D).
Question No. 2 of 10 2. Consider the interference colors seen in a film of gasoline on a wet street. What is the function of the water? Question 02 (A) A means of spreading the gasoline into a thin film. (B) A means of slowing the rapid evaporation of gasoline. (C) A chemical bond with the gasoline. (D) A second reflecting surface. (E) None of the above The gasoline would spread into a thin layer on many surfaces; water isn t needed to accomplish this. The rate of evaporation of gasoline is irrelevant here. Even if the rate was too great, you would simply see interference colors for a short period of time. There is not chemical bonding in this case, they are simply a mixture. D. Correct! Yes, there must be a second reflecting surface to provide two waves so that they can interfere. Some wavelengths of light are destructively interfered, leaving only certain colors that we observe. Consider wave superposition and interference. How could certain colors be produced from white light that is composed of all colors? Consider why all those colors aren t seen. The gasoline spreads into a thin film on the water. Some light reflects off the exterior surface. Other light penetrates the gasoline, and then reflects off the water surface. If the thickness of the oil film results in the light reflected from the water having travelled a whole number of wavelength of a particular color of light, there will be constructive interference and this color light will be reflected with a higher intensity than others. The specific color depends upon the thickness of the film of gasoline and on the angle at which the film is viewed. In most films, the thickness varies, so you see a variety of colors in the puddle. The correct answer is (D).
Question No. 3 of 10 3. How would the pattern of bright and dark fringes made in a Young s double slit experiment change if the light from each slit was shifted in phase by an amount equal to half a wavelength. Question 03 (A) No change would occur. (B) The bright fringes would become brighter. (C) The bright fringes would move closer. (D) The bright fringe would now be a dark fringe. (E) None of the above A. Correct! The appearance of bright and dark fringes depends upon interference. If each wave was changed by an equal ½ wavelength, they would be same relative to each other. Where there was a bright fringe before, there would still be one. Changing the phase or timing of the wave won t affect the brightness or energy content. This might be true if the wavelengths were changed, but in this case only the phase is changed. Therefore the spacing of the fringes wouldn t change. For there to be destructive interference one wave has to be out of phase with the other but in this case the phases have both shifted by the same amount so will still be in phase. Consider how the bright and dark fringes are produced. They occur because of constructive and destructive interference. Consider how changing each was by the same ½ wavelength phase would affect the fringes. The appearance of bright and dark fringes depends upon interference. For the bright fringes, two waves constructively interfere. They would have a relative phase difference of multiples of a wavelength. For the dark fringes, two waves destructively interfere. They would have a relative phase difference of ½ multiples of a wavelength. However, if each wave was changed by an equal ½ wavelength, they would be same relative to each other. Again, there would be no difference. Where there was a bright fringe before, there would still be one. Where there was a dark fringe before, there would be again. If only one wave were changed by some amount, then there would be a difference. The correct answer is (A).
Question No. 4 of 10 4. Which of these statements about Young s Double Slit experiment is incorrect? Question 04 (A) A pattern of dark and light regions is created on a screen due to constructive and destructive interference. (B) The angle between a horizontal at the center of the screen and the bright fringes on the screen is inversely proportional to the separation of the slits. (C) For the same experimental setup, if the wavelength of the incident light increases the distance between the bright regions will be closer together. (D) If two waves arrive at the same point on the screen and are 5/2 out of phase a dark region will be created. (E) The further away from the central bright area you move the lower the intensity of the bright regions. The dark regions occur when two waves interfere destructively, and the bright regions result from constructive interference. The formula to find the angle between the center horizontal and the bright fringes is sinθ = m. Where is the wavelength of incident light, m is an integer, and d is d the distance between the slits. So the angle is inversely proportional to the separation of the slits. C. Correct! The formula to find the angle between the center horizontal and the bright fringes is sinθ = m. Where is the wavelength of incident light, m is an integer, and d is d the distance between the slits. If d remains the same and wavelength increases the angle will increase and the bright regions are farther apart. The formula to find the angle between the center horizontal and the dark fringes 1 is. Where is the wavelength of incident light, m is an integer, and d sinθ = (m + ) 2 d is the distance between the slits. For destructive interference the waves must be some multiple of /2, in this case m is 5. The intensity of the bright regions will decrease the further you are from the central bright fringes. In Young s slit experiment a monochromatic light source is incident on a pair of slits a small distance, d, apart. A screen is placed a distance away from the slits, the distance to the screen is significantly larger than the slit separation. A series of light and dark regions will be seen on the screen. The dark regions result from destructive interference and the light regions from constructive interference of the waves from the two slits. The location of the bright and dark regions are given by the formulas Bright region: sinθ = m d Dark region: 1 sinθ = (m + ) 2 d, m is an integer or order of the region. Where θ is the angle between the horizontal and the region of interest (light or dark). The intensity of the light decreases for higher orders. The correct answer is (C).
Question No. 5 of 10 needed (3) Pick the answer (4) Go back to review the core concept tutorial as needed. 5. In a Young s double slit experiment, the seventh dark fringe is located.025m to the side of the central bright fringes which is 1.1 m away from the slits. The separation between the slits is 1.4x10-4 m. What is the wavelength of the light being used? Question 05 (A) 3.27x10-6 m (B) 2.4x10-5 m (C) 4.2x10-7 m (D) 4.5x10-7 m (E) Need the angle Don t forget to insert the correct m value. In this case, since the problem mentions the seventh dark fringe to the side of the central bright fringes, m Is 7. Don t forget to use the sine of theta, not just the angle theta, in the formula. C. Correct! Use the dark fringe formula. Sin θ = (m+1/2) /d. First you must find the angle by using the definition of tangent, opposite divided by adjacent side. Once you have the angle, substitute into the formula, and solve for wavelength. In this situation, since we are discussing the dark fringes, use the dark fringe formula that has the m+1/2 term in it. The other formula is for the light fringes. You can find the angle by using the equation for tan θ, the distance to the screen and the distance to the dark fringe. Known: Slit separation, d = 1.4x10-4 m Order of the dark fringes = 7 Distance from center to dark fringe =0.025m Distance to screen, L =1.1 m Unknown: Wavelength of Light, =? m Define: Formula for dark fringes: sin θ=(m+1/2) d θ 1.1 m 0.025 m Rearrange to find : sin θ d = (m+1/2) First you must find the angle. opp tan θ = adj Output: Substitute the distance values given..025m tan θ = = 0.023 1.1m This is just the tangent of the angle, take the inverse tangent to get the angle. -1 o tan (.023) = 1.3 Now use this angle in our dark fringe formula that we started with. Rearrange the formula for the wavelength since that is our unknown. Be sure to use m=7 and the other given information. -4 o 1.4 x 10 m sin(1.3 ) -7 = = 4.2 x 10 m (7 + 1/2) Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. The correct answer is (C).
Question No. 6 of 10 Instructions: (1) Read the problem and answer choices carefully (2) Work the problems on paper as 6. In a Young s double slit experiment, the angle that locates the second bright fringe is 2.0 The slit separation is 3.8x10-5 m. What is the wavelength of the light? Question 06 (A) 5.3x10-7 m (B) 6.6x10-7 m (C) 5.9x10-8 m (D) 6.4x10-8 m (E) 3.8x10-5 m Use an m value of 2 here, not 2.5. This problem involves a bright fringe so use whole integer values. B. Correct! Use the bright fringe formula wavelength, substitute, and solve. sin θ=m d Rearrange to solve for the unknown, Make sure your angle is in degrees, not radians. You may need to set this correctly on your calculator. Use the bright fringe formula, sin θ=m d. Remember to take the sin of the angle. Known: Angle of bright fringe, θ=2.0 Order of bright fringe, m=2 Separation of slits, d = 3.8x10-5 m Unknown: Wavelength of incident light, =? m Define: Use the bright fringe formula. sin θ=m d Rearrange to find wavelength d sinθ = m Output: There are no quantities that need to be converted. Simply substitute, and solve. 5 (3.8x10 m)(sin 2 ) 7 = = 6.6 x 10 m 2 Substantiate: Units are correct, sig figs are correct, magnitude is reasonable. Answer: B. 6.6 x 10-7 m The correct answer is (B).
Question No. 7 of 10 7. A diffraction grating with spacing of 1.6 x 10-6 m is used to create a diffraction pattern using a monochromatic light source of wavelength 600 nm. What is the angle of the first bright fringe and what is the distance from the central fringe to the first bright fringe if the screen is 12.5 cm away? Question 07 (A) 5.1 and 22 cm (B) 0.375 and 0.08 cm (C) 22 and 5.1 cm (D) 0.22 and 0.051 cm (E) 0.375 and 5.1 cm Check your units. Remember to use sin -1 to find the angle. C. Correct! Use the equation for bright fringes to find the angle and then use tan of the angle to find the distance to the fringe. Check that you are using exponents correctly. Check you are using sin -1 to find the angle. Known: Order of bright fringe, m=1 Separation of slits, d = 1.6 x 10-6 m Distance to screen, L = 12.5 cm Wavelength of light, = 600 nm Unknown: Angle of first bright fringe, θ =? Distance to first bright fringe, x = m θ x Define: First convert all units to SI units. Then use the formula for bright fringe sin θ=m d Next find x, use tan θ and rearrange 12.5 cm opp x Tan θ = = adj L Output: Convert cm to m, so x = L Tan θ 1 m 12.5 cm = 12.5 cm = 0.125 m 100 cm 1 m Convert nm to m, 7 600 nm = 600 nm = 6 x 10 m 9 1 x 10 nm -7 6.0 x 10 m sin θ= m = 1 = 0.375 d 6 1.6 x 10 m θ = 22 x = L Tan θ = 0.125 x Tan(22 ) = 0.051 m = 5.1 cm Substantiate: Units are correct, sig figs are correct, magnitudes are reasonable. The correct answer is (C).
Question No. 8 of 10 8. A diffraction grating with spacing of 1.6 x 10-6 m is used to create a diffraction pattern using a monochromatic light source of wavelength 400 nm on a screen that is 12.5 cm away. How far away will the 1 st fringe be for this wavelength, from the 1 st fringe for the 600 nm wavelength? Question 08 (A) 14.5 (B) 3.2 cm (C) 1.9 cm above (D) 1.9 cm below (E) 0 This the angle of the 1 st fringe, the question asks for a distance. The question asks for a difference in the position of the fringes. C Incorrect! Does the shorter wavelength make a larger or smaller angle and the longer wavelength? A sketch may help you. D. Correct! Use the same method as for the previous question and then calculate the difference in positions. Angle is proportional to wavelength so the shorter wavelength results in a smaller angle and the fringe will be below that for the 600nm light. Use the previous question as a guide and remember to take the difference in positions. Known: Order of bright fringe, m=1 Separation of slits, d = 1.6 x 10-6 m Distance to screen, L = 12.5 cm Wavelength of light, = 400 nm Unknown: Difference in position of 400nm and 600nm light, Δx=? cm θ x Define: First convert all units to SI units. Then use the formula for bright fringe: sin θ=m d 12.5 cm Next find x, use tan θ and rearrange: opp x Tan θ = = so x = L 2 Tan θ adj L2 Finally subtract this distance from the answer found for 600 nm light in the previous question. Δx= x-x 2 Output: Convert cm to m, 1 m 12.5 cm = 12.5 cm = 0.125 m 100 cm 1 m Convert nm to m, 7 400 nm = 400 nm = 4 x 10 m 9 1 x 10 nm -7 4.0 x 10 m sin θ= m = 1 = 0.250 d 6 1.6 x 10 m θ = 14.5 x = L Tan θ = 0.125 x Tan(14.5 ) = 0.032 m = 3.2 cm Δx= x-x 2 = 5.1 cm - 3.2 cm = 1.9 cm Substantiate: Units are correct, sig figs are correct, magnitudes are reasonable. The correct answer is (D).
Question No. 9 of 10 9. Which of these is not a correct statement? Question 09 (A) Sound and light waves travelling in air can be polarized. (B) The sun and incandescent bulb both emit unpolarized light. (C) A polarizer transmits light with only one orientation of electric field. (D) A pair of polarizers can be used to precisely adjust the intensity of light. (E) The average intensity of light transmitted by a pair of polarizer is dependent on the square of the cosine of the angle between the transmission axes of the polarizers. A. Correct! Polarization will only occur in transverse waves, where the vibrations or oscillations are perpendicular to the direction travel, electromagnetic waves can be polarized. Sound waves travelling in air are longitudinal waves so cannot be polarized. Light from the sun and from an incandescent bulb have electric fields that oscillates in more than one plane, these are unpolarized light sources. A polarizer will allow only one orientation of electric field to pass through. The transmission axis of the polarizer determines which orientation that is allowed through. By using two polarizers we can tune the intensity of the light that is transmitted. The polarizers reduce the amplitude of the transmitted light and intensity is proportional to the square of amplitude. Malus Law describes how the angle θ between a pair of polarizers affects the average intensity S of the transmitted light, S = S 0 Cos θ. Where S o is the intensity of the incident light. Polarization will only occur in transverse waves, where the vibrations or oscillations are perpendicular to the direction travel, electromagnetic waves can be polarized. Sound waves travelling in air are longitudinal waves so cannot be polarized. Light from the sun and from an incandescent bulb both have electric fields that oscillates in more than one plane, these are unpolarized light sources. A polarizer will allow only one orientation of electric field to pass through. The transmission axis of the polarizer determines which orientation that is allowed through. By using two polarizers we can tune the intensity of the light that is transmitted. The polarizers reduce the amplitude of the transmitted light and intensity is proportional to the square of the amplitude. Malus Law describes how the angle θ between a pair of polarizers affects the average intensity S of the transmitted light, S = S 0 Cos 2 θ. Where S o is the intensity of the incident light. The correct answer is (A).
Question No. 10 of 10 10. The intensity of light must be adjusted to accommodate the requirements of a photocell. In what orientation must two polarizing filters be placed so that the light intensity is 10.0% of the original value? Question 10 (A) 1.25 (B).32 (C) 45.0 (D) 71.6 (E) None of the above Be sure to do your calculations in degrees, not radians. You may need to set your calculator to do this. Use the cosine function in Malus law, not the sin function as in some of the other formulas from this tutorial. Use Malus law. The intensity of a pair of polarizers is equal to the original intensity times the cosine squared of the angle between the axes of the two polarizing filters. D. Correct! Use Malus law. The intensity of a pair of polarizers is equal to the original intensity times the cosine squared of the angle between the axes of the two polarizing filters. We are solving for the angle, so some algebraic rearranging is necessary. Substitute the known values and solve for the angled necessary. Use Malus law. The intensity of a pair of polarizers is equal to the original intensity times the cosine squared of the angle between the axes of the two polarizing filters. Known: Intensity of transmitted light S/S o = 10% =0.10 Unknown: Angle between polarizers, θ =? Describe: Use Malus law. 2 S = Socos θ We are solving for the angle, so some algebraic rearranging is necessary. S 2 cos θ S = o S cosθ S = o -1 S cos θ S = o Output: Substitute the known values and solve for the angled necessary. We are given the value of S/S o, 10% or 0.1-1 θ = cos 0.1 = 71.6 Note: This relatively large angle seems reasonable considering the small intensity of light that is left. 90 would give 0 intensity left. Substantiate: Units are correct, sig figs are correct, magnitudes are reasonable. The correct answer is (D).