PHYS 1112
Chapter 23 Problems

Selected Solutions

  1. Since this is a concave mirror, and the image is real, we are given that M = -4. Since p = 30 cm, this means

    M = -4 = -q/p
    ==> q = 4 p = 4(30 cm) = 120 cm

    Now that we know the image distance, we can plug this information into the mirror equation to get

    1/q + 1/p = 2/R
    1/(120 cm) + 1/(30 cm) = 1/(120 cm) + 4/(120 cm) = 5/(120 cm) = 1/(24 cm) = 2/R

    Thus, R = 2(24 cm) = 48 cm.

  2. Since the object (the goldfish) is in the tank, let us take n1 = 1.33 and n2 = 1.00. If that is the case, then R is negative and p is positive (the radius of curvature and the object are in front of the surface). Using p = 10.0 cm and R = -15.0 cm, we get

    1.33/(10.0 cm) + 1.00/q = (1.00 - 1.33)/(-15.0 cm) = .33/(15.0 cm)
    1.00/q = .33/(15.0 cm) - 1.33/(10.0 cm) = .66/(30.0 cm) - 3.99/(30.0 cm) = -3.33/(30.0 cm)
    q = -(30.0 cm)/3.33 = -9.0 cm

    Thus, the goldfish appears to be 9.0 cm inside of the bowl.

  3. a) We know that f = 20.0 cm for this particular thin lens. If the object is placed at a distance of p = 40.0 cm, then

    1/(40.0 cm) + 1/q = 1/(20.0 cm)
    1/q = 1/(20.0 cm) - 1/(40.0 cm) = 1/(40.0 cm)

    Thus, q = 40.0 cm. If we use this in the magnification equation, we get M = -q/p = 1(40.0 cm)/(40.0 cm) = -1. Hence, the image is real, inverted, the same size, and a distance of 40.0 cm behind the lens.

    b) For p = 20.0 cm,

    1/(20.0 cm) + 1/q = 1/(20.0 cm)
    1/q = 0

    Thus , q is equal to infinity. This means that no image is formed, and the rays are parallel.

    c) For p = 10.0 cm,

    1/(10.0 cm) + 1/q = 1/(20.0 cm)
    1/q = 1/(20.0 cm) - 1/(10.0 cm) = -1/(20.0 cm)

    Thus, q = -20.0 cm. For the magnification, we get M = - q/p = -(-20.0 cm)/(10.0 cm) = 2. Hence, the image is upright, magnified by 2, and 20.0 cm in front of the lens.

If you find an error in these, or just have a comment, send an e-mail message to jpratte@kennesaw.edu