Test: Light and Electromagnetic Radiation - MCAT MCQ

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.

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.

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.

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.

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 (θ_c), 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.

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.

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.

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.

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.

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.