Section   7.1Reflection

1. (Exercises 3 and 4) What is the minimum height for the reflecting portion of a mirror that can be used by a 7-ft 2-in. professional basketball star to see his complete (head-to-toe) image? This calculation is quite simple because the relationship between the minimum height of the mirror and the height of the person is just a factor of 2 (see Figure 7.4 in the textbook).

H_minimum = Height of person / 2 = 7 ft 2 in. / 2

H_minimum = 3.5 ft 1 in.

H_minimum = 3 ft 7 in.

Section   7.2Reflection and Dispersion

2. (Exercises 7 and 8) Light travels through glass at a speed that is 65% as fast as it travels through vacuum. What is the index of refraction of the glass?

First, we find the speed of light through the glass.

c_glass = c_vacuum (0.65) = 3.00 x 10⁸ m/s (0.65)

c_glass = 1.95 x 10⁸ m/s

Now we can find the index of refraction by using the definition given by Equation 7.1 in the textbook.

n_glass = c_vacuum / c_glass = 3.00 x 10⁸ m/s / 1.95 x 10⁸ m/s

n_glass = 1.54

Comparing this with the values for the index of refraction given in Table 7.1 in the textbook, we find that this is close to but not exactly the same index of refraction as that given for crown glass. The particular composition of glass determines its index of refraction, and this sample is evidently not the type commonly classified as crown glass.

Section   7.5Spherical Mirrors

3. (Exercises 11 and 12) A spherical concave mirror is used to form a real image of a 6.0 cm tall object that is located 15 cm in front of the mirror. The image is found where the light from the mirror is focused on a card located 30 cm in front of the mirror.

(a) What is the magnification, and how tall is the image?

(b) Calculate the focal length of the mirror.

(c) Determine the radius of curvature of the mirror.

(a) In this case the image distance (D_i) is 30 cm and the object distance (D_o) is 15 cm. Both are considered positive in this case. The magnification of the image is given by:

M = - D_i / D_o = - 30 cm / 15 cm = - 2

The "minus" sign means that the image is inverted, and the 2 means that it is twice the size of the object. In this case, the image will be 2 (6.0 cm), or 12 cm, tall.

(b) The focal length of the mirror can be found using the thin lens equation.

1 / f = 1 / D_o + 1 / D_i = 1 / 15 cm + 1 / 30 cm

= 2 / 30 cm + 1 / 30 cm = 3 / 30 cm = 1 / 10 cm

Now we have to be careful. At first glance, it appears that the answer is 1/10 cm, but it is not. This is equal to 1/f, not to the focal length (f). We must take the reciprocal of both sides of the above equation or, if you prefer, you can cross multiply to get the same mathematical result.

1/f = 1 / 10 cmand inverting both sides f = 10 cm

(c) The focal length of the mirror is 10 cm. The radius of curvature for any spherical mirror is just twice the focal length so we have:

R = 2f = 2 (10 cm) = 20 cm

The same type of analysis can be used on the image formed by a lens, as shown below. Section   7.6Lenses

4. (Exercises 21 and 22) An object is placed 10 cm in front of a convex lens that has a focal length of 15 cm. Determine the image distance and describe the image that is formed. Here D_o = 10 cm and f = +15 cm. The plus is important because it indicates the type of lens being used. In this case the lens is convex and will cause the incoming rays to converge, so the focal length is positive. A concave (or diverging) lens would have a negative focal length. The equation to use is the thin lens equation:

1 / f = 1 / D_o + 1 / D_i

but we must solve for 1 / D_i first.

1 / D_i = 1 / f - 1 / D_o = 1 / 15 cm - 1 / 10 cm

Using 30 as the lowest common denominator, these fractions become:

1 / D_i = 2 / 30 cm - 3 / 30 cm

1 / D_i = - 1 / 30 cm

Again we must invert both sides or cross multiply to get:

D_i = - 30 cm

The "minus" sign indicates that for a convex lens used in this configuration the image is virtual, because a real image would have a "plus" sign preceding the image distance. If we were to draw a ray tracing, we would find the image located on the same side of the lens as the object. This also shows that the image is virtual and that it cannot be focused on a screen.

To fully describe the image, we need to calculate its magnification as well.

M = - D_i / D_o = - (- 30 cm) / 10 cm = + 3

This tells us two things about the image. First, it is larger than the object by a factor of three, and secondly it is upright; that is, oriented upward with respect to the optic axis in the same direction as the object. The lens we have just described is being used as a simple magnifier, just like the one you might use to look at a butterfly wing or to read the fine print in a phone book.

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