GOAL Calculate geometric quantities associated with a diverging lens. PROBLEM A

Question

GOAL Calculate geometric quantities associated with a diverging lens.

PROBLEM A diverging lens of focal length 10.0 cm forms images of an object situated at various distances. (a) If the object is placed 30.0 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification. (b) Repeat the problem when the object is at 10.0 cm and (c) again when the object is 5.00 cm from the lens.

STRATEGY Once again, substitution into the thin-lens equation and the associated magnification equation, together with the thin lens conventions, solve the various parts. The only difference is the negative focal length.

SOLUTION

(A) Locate the image and its magnification if the object is at 30.0 cm.

The ray diagram is given in figure a. Apply the thin-lens equation with p = 30.0 cm to locate the image.

Solve for q, which is negative and hence virtual.

q = -7.50 cm

Substitute to get the magnification. Because M is positive and has absolute value less than 1, the image is upright and smaller than the object.

(B) Locate the image and find its magnification if the object is 10.0 cm from the lens.

Apply the thin-lens equation, taking p= 10.0 cm.

Solve for q (once again, the result is negative, so the image is virtual).

q = -5.00 cm

Calculate the magnification. Because Mis positive and has absolute value less than 1, the image is upright and smaller than the object.

(C) Locate the image and find its magnification when the object is at 5.00 cm.

The ray diagram is given in figure b. Substitute p = 5.00 cm into the thin-lens equation to locate the image.

Solve for q. The answer is negative, so once again the image is virtual.

q = -3.33 cm

Calculate the magnification. Because Mis positive and less than 1, the image is upright and smaller than the object.

LEARN MORE

REMARKS Notice that in every case the image is virtual, hence on the same side of the lens as the object. Further, the image is smaller than the object. For a diverging lens and a real object, this is always the case, as can be proven mathematically.

QUESTION Can a diverging lens be used as a magnifying glass? Explain. (Select all that apply.)

Examine the ray diagram that applies, figures above. Follow the construction to see whether it is possible for the diverging lens to produce an image larger than the object.

PRACTICE IT

Use the worked example above to help you solve this problem. A diverging lens of focal length f = -9.7 cm forms images of an object situated at various distances.(a) If the object is placed p1 = 29.1 cm from the lens, locate the image, state whether it's real or virtual, and find its magnification.

(b) Repeat the problem when the object is at p2 = 9.7 cm.

(c) Repeat the problem again when the object is 4.85 cm from the lens.

EXERCISE HINTS:  GETTING STARTED  |  I'M STUCK!

Use the values from PRACTICE IT to help you work this exercise. Repeat the calculation, finding the position of the image and the magnification if the object is 18.7 cm from the lens.

Explanation / Answer

par c )

1/p + 1/q = 1/f

given f = -9.7

p = 4.85 cm

1/4.85 - 1/9.7 = 1/q

q = 9.7

m = -q/p

m = -(9.7/4.85) = -2

part ii )

1/p - 1/f = 1/q

1/18.7 - 1/9.7 = 1/q

q = -20.15

m = -q/p

m = -(-20.15/18.7)

m = 1.078

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