MCAT Physics and Math Review
Chapter 8: Light and Optics
1. If a light ray has a frequency of 5.0 × 1014 Hz, in which region of the electromagnetic spectrum is it located?
2. A child stands between two mirrors with his arms out, perpendicular to the mirrors. One plane mirror is 5 m away from his left hand and another plane mirror is 7 m away from his right hand. How far apart are the two images produced by the mirrors if the child has an arm span of 0.5 m?
1. 2 m
2. 12 m
3. 12.5 m
4. 24.5 m
3. An object is placed at the center of curvature of a concave mirror. Which of the following is true about the image?
1. It is real and inverted.
2. It is virtual and inverted.
3. It is virtual and upright.
4. It is real and upright.
4. When monochromatic light is refracted as it passes from air to glass, which of the following does NOT remain the same? (Note: Assume that the wave is fully transmitted.)
5. A ray of light (f = 5 × 1014 Hz) travels from air into crystal into chromium. If the indices of refraction of air, crystal, and chromium are 1, 2, and 3, respectively, and the incident angle is 30°, then which of the following describes the frequency and the angle of refraction in the chromium?
1. 5 × 1014 Hz; 9.6°
2. 5 × 1014 Hz; 57°
3. 1.0 × 1010 Hz; 9.6°
4. 1.0 × 1010 Hz; 57°
6. A source of light (f = 6.0 × 1014 Hz) passes through three plane polarizers. The first two polarizers are in the same direction, while the third is rotated 90° with respect to the second polarizer. What is the frequency of the light that comes out of the third polarizer?
1. Light will not pass through the third polarizer
2. 3.0 × 1014 Hz
3. 6.0 × 1014 Hz
4. 9.0 × 1014 Hz
7. Which phenomenon would cause monochromatic light entering the prism along path AB to leave along path CD?
8. Which of the following describes the image formed by an object placed in front of a convex lens at a distance smaller than the focal length?
1. Virtual and inverted
2. Virtual and upright
3. Real and upright
4. Real and inverted
9. A submarine is inspecting the surface of the water with a laser that points from the submarine to the surface of the water and through the air. At what angle will the laser not penetrate the surface of the water but rather reflect entirely back into the water? (Assume nwater = 1.33 andnair = 1.)
10.A student is analyzing the behavior of a light ray that is passed through a small opening and a lens and allowed to project on a screen a distance away. What happens to the central maximum (the brightest spot on the screen) when the slit becomes narrower?
1. The central maximum remains the same.
2. The central maximum becomes narrower.
3. The central maximum becomes wider.
4. The central maximum divides into smaller light fringes.
11.Which of the following are able to produce a virtual image?
1. Convex lens
2. Concave lens
3. Plane mirror
1. I only
2. III only
3. II and III only
4. I, II, and III
12.Monochromatic red light is allowed to pass between two different media. If the incident angle in medium 1 is 30° and the incident angle in medium 2 is 45°, what is the relationship between the speed of the light in medium 2 compared to that in medium 1?
13.A scientist looks through a microscope with two thin lenses with m1 = 10 and m2 = 40. What is the overall magnification of this microscope?
14.Imagine that a beam of monochromatic light originates in air and is allowed to shine upon the flat surface of a piece of glass at an angle of 60° with the normal. The reflected and refracted beams are perpendicular to each other. What is the index of refraction of the glass?
15.Which of the following will not result in the splitting of white light into its component colors?
1. Dispersion through a prism
2. Diffraction through a grating
3. Refraction within a thin film
4. Reflection from an ideal convex mirror
Answers and Explanations
It is unnecessary to memorize the entire electromagnetic spectrum for Test Day; however, it is important to know that the visible spectrum runs from 400–700 nm. We can calculate the wavelength of this light ray:
This wavelength falls within the visible spectrum has a red-orange color.
In plane mirrors, the image is as far away from the mirror as the object is. In other words, the image produced by the left mirror is 5 m away from the mirror because the child is standing 5 m away from the mirror. Similarly, the right mirror produces an image that is 7 m away from the center of the mirror. To calculate how far away the two images are, take into consideration not only the image distance but also the distance of the object (the child) from the mirrors and the child’s arm span of 0.5 m. Therefore, the images are 5 + 5 + 0.5 + 7 + 7 = 24.5 m apart.
One could solve this question with a ray diagram, but be wary about using ray diagrams on Test Day. It is easy to make small mistakes that cause the light rays not to intersect. Therefore, solve the question using the sign convention. If the object is at the center of curvature, its distance is 2f. We can plug into the optics equation:
Because i is positive, the image is real. For single mirrors or lenses, all real images are inverted.
As light rays travel from one medium to another, their wavelengths change. Even if we did not know this immediately, we can determine the answer through process of elimination. Frequency and period are inverses of each other, so if either of these quantities changes, the other would have to change as well, eliminating choices (B) and (D). Further, because the wave is fully transmitted, there is no absorption or reflection, and the amplitude (which is related to intensity) should not change, eliminating choice (C). When light is refracted, its speed changes; although the frequency does not change, the wavelength does.
This question contains two parts—we have to determine the frequency and the angle of refraction of the light ray. The first part, however, is straightforward because the frequency of a light ray traveling from one medium to another does not change. Because the frequency must be 5 × 1014 Hz, we can eliminate choices (C) and (D). For the angle of refraction, we can either calculate it or determine it using logic. First, the light ray goes from air into crystal; that is, from a low index of refraction to a higher one. According to Snell’s law, the angle of refraction will be smaller than the incident angle (closer to the normal). When the light ray moves from crystal to chromium, it again goes from a lower index of refraction into a higher one, thus making the angle of refraction even smaller, eliminating choice (B). This question could also be answered by calculation using Snell’s law, but the calculations are time consuming and unnecessary.
Plane-polarized light is light in which the electric fields of all the waves are oriented in the same direction. Light passing through the first two polarizers will only contain rays with their electric field vectors in the same direction. When it reaches the third polarizer, however, the light will not be able to pass through because all the light rays will be oriented in the direction dictated by the first and second polarizers.
Even though the light is traveling through a prism, the change in the light’s direction is caused by refraction, not dispersion. Dispersion involves the breaking up of polychromatic light into its component wavelengths because the degree of refraction depends on the wavelength. We are told that the incident light is monochromatic or, in other words, of only one wavelength; therefore, light will not be dispersed, eliminating choice (A). Diffraction, choice (C), describes the spreading of light waves as they pass through a small opening. Polarization, choice (D), is the alignment of the electric field component of light waves.
The image produced by a convex lens can be either real or virtual. It is real if the object is placed at a distance greater than the focal point, and virtual if the object is placed at a distance less than the focal point (between the focal point and the lens). Remember that for a single mirror or lens, an image that is real must be inverted and one that is virtual must be upright. In this question, the object is placed in front of the focal point, so the image must be virtual and, therefore, upright. We could also determine this from the optics equation. If f > o, then is negative, and i is therefore negative (virtual).
This question is testing our understanding of total internal reflection. As the laser beam travels from water to air—that is, from a higher to a lower index of refraction—the angle of refraction increases. At the critical angle (θc), the angle of refraction becomes 90°; at this point, the refracted ray is parallel to the surface of the water. When the angle of incidence is greater than the critical angle, all the light is reflected back into the water. The question is asking for the critical angle:
The inverse sine of 0.75 must be slightly higher than 48.59° is the exact answer.
This question is testing our understanding of diffraction. When light passes through a narrow opening, the light waves spread out; as the slit narrows, the light waves spread out even more. When a lens is placed between the narrow slit and the screen, a pattern consisting of alternating bright and dark fringes can be observed on the screen. As the slit becomes narrower, the central maximum (the brightest and most central fringe) becomes wider. This can also be seen in the equation for the position of dark fringes in a slit–lens setup (a sin θ = nλ). As a, the width of the slit, decreases, sin θ must increase because nλ is constant for a given fringe. If sin θ increases, θ necessarily increases, implying that the fringes are spreading further apart.
All images produced by plane mirrors will be virtual, so statement III is true. The same goes for diverging species (convex mirrors and concave lenses), so statement II is true. Converging species (concave mirrors and convex lenses) can produce real or virtual images, depending on how far the object is from the species, so statement I is also true.
First, the color of the light is irrelevant here; the ratio would be the same even if the specific color were not mentioned. Second, recall Snell’s Law: n1 sin θ1 = n2 sin θ2. Although we don’t know the value of n for either medium, you do know the simple relationship . Replacing n in Snell’s law, and canceling out c from both sides, we get:
The overall magnification of a system of multiple lenses is simply the product of each lens’s magnification. In this case, that is 10 × 40 = 400.
Drawing a diagram is best here. Because the angle given is with respect to the normal, you know that the incident angle must equal 60°. You know that the reflected beam will have an angle of 60° relative to the normal. Therefore, the reflected beam will make an angle of 30° with the plane of the glass. If the reflected and refracted beams are perpendicular to each other, the refracted beam will make a 60° angle with the plane of the glass. θrefracted is therefore 30° relative to the normal.
Using n1 sin θ1 = n2 sin θ2, we have
Light can be split into its component colors by dispersion, such as that through a prism, eliminating choice (A). Diffraction by a diffraction grating will also separate colors by their wavelengths, eliminating choice (B). The refraction of light within a thin film also leads to light dispersion as the different colors are refracted at slightly different angles in the film, eliminating choice (C). A mirror with significant aberration could lead to a separation of light into its component colors, but we are told that this is an ideal mirror. Thus, choice (D) is the correct answer.