Physics 1C Week 4 Questions Fall 1998
1. A man 6 ft. tall standing 10 ft. from a 3 ft. high vertical mirror can see his entire image in it. If his eyes are 4 in. below the very top of his head, how high above the floor is the bottom edge of the mirror?
2. A fish looking straight upwards towards the surface receives a cone of rays and sees a circle of light filled with the images of sky and ships and whatever else is up there (see figure below). This bright circular field is surrounded by darkness. Explain what is happening and compute the full angle of the cone.
3. If a spherical mirror is immersed in water, how does its focal length change?
4. If a photographic film is placed at the location of a real image, the film will record the image. Can this be done with a virtual image? How might one record a virtual image?
5. Fancy bathroom mirrors used for shaving/makeup are: a) convex, b) concave, or c) flat
The image is: a) larger, b) smaller than the object
The image is: a) erect, b) inverted
The image is: a) real, b) virtual
6. The law requires that mirrors must be mounted on the sides of cars. You use them to view traffic. These mirrors are: a) convex, b) concave, or c) flat
The image is: a) larger, b) smaller than the object
The image is: a) erect, b) inverted
The image is: a) real, b) virtual
Draw to scale a ray diagram showing center of curvature, focal point(s), principal rays, etc.
Physics 1C Week 4 Solutions Fall 1998, Page 1
1.
From the figure below, the top of the mirror must be 70" high and the bottom 34" high. The size of the mirror is 36" = 3 ft. which is half the height of the man. This is true whatever distance you stand away from the mirror, and is also independent of where your eyes are.
2. Light entering the water at glancing incidence is transmitted at the critical angle, and those rays limit the cone of light from above the water reaching the fish; sinc = 1/1.333; c = 49o and the cone-angle is twice this, or = 98o. What does the fish see outside this cone angle?
3. The focal length of the mirror does not change in water.
4. Film placed at a virtual image gets no light and will not record an image. Make image by refocusing light from virtual image, e.g. with eye, camera, or any positive lens, to make a real image. Record the real image.
5. The idea is to have an erect image which is larger than life. A flat mirror won't magnify. A convex mirror always forms erect, virtual, smaller images of real objects, i.e. not quite what we want. For real objects, a concave mirror can make either a real inverted image (for object distances greater than f), or a virtual erect magnified image (for object distances less than f). So we put our face closer to the mirror than the focal length of the mirror, and view a large, erect, virtual image.
More mathematically: for a curved mirror, we have 1/s + 1/s' = 1/f, so that s' = sf/(s-f), and m = s'/s = f/(f-s).
For f < 0 (convex mirror) we have m < 1. Only a concave mirror (f > 0) can yield m > 1. We want the image to be larger than the object. A concave mirror will produce an erect or inverted image, depending on whether f > s or f < s. We prefer an erect image, and so we choose f > s, and therefore s' < 0 (virtual).
Physics 1C Week 4 Solutions Fall 1998, Page 2
6. Here we want the driver to be able to view as much as possible and the image must be the right way up. Such "wide-angle" viewing is accomplished through a low-magnification mirror. So we need 0 < m < 1.
Since m = f/(f-s), this can only be achieved if f < 0, i.e. if the mirror is convex. So the image is smaller, erect, and, since s' = -ms = 0, the image is virtual.