Chapter 26: Reflection, Refraction and Geometrical Optics Selected Solutions

Selected Solutions

13. (a) To solve this problem, we divide the part of the building that can be seen into three parts, two of which are found by making use of similar triangles.

From the above diagram we can see that

Since we know the total height of the mirror, h, we add these two equations to yield

.

and use the facts that h = h₁ + h₂ = y₁ - y₂ and H = y₃ - y₄. Putting these results in the above equation gives

.

We can now determine the value of H to be

.

(b) Rearranging the above result for H gives

We can see that decreasing d increases the ratio D/d, and thus H. So, if the mirror is moved closer, the answer to part (a) will increase.

31. (a) Since the image is real, we know that the image distance is positive; the magnification must then be negative. Therefore,

d₁ = -md_o = -(-3)(22 cm) = 66 cm

(b) Using the mirror equation gives

45. Light travels at a constant speed v, therefore,

The total time then is given by

47. For the empty glass, with q_i being the angle between the vertical edge of the glass and the light ray, we have

and for the glass filled with water,
.

Making use of Snell's law, n_airsinq_i = n_wsinq_refr, we can say

So, we can now calculate H to be

67. (a) To project a real image onto the wall, a convex lens should be used. Since convex lenses have positive focal lengths, real images are formed behind the lens; therefore, the lens with focal length f₁ should be used.

(b) Here we need to find the image distance where d_o + d_i = 3.0 m. Therefore,

d_i = -md_o = -(-2)(3.0 m - d_i) = 6.0 m - 2d_i.

This gives,

.

The lens should be placed 2.0 m from the wall.

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