GOAL Calculate geometric quantities for a sequential pair of lenses. PROBLEM Two

Question

GOAL Calculate geometric quantities for a sequential pair of lenses.

PROBLEM Two converging lenses are placed 20.0 cm apart, as shown in figure a, with an object 30.0 cm in front of lens 1 on the left. (a) If lens 1 has a focal length of 10.0 cm, locate the image formed by this lens and determine its magnification. (b) If lens 2 on the right has a focal length of 20.0 cm, locate the final image formed and find the total magnification of the system.

STRATEGY We apply the thin-lens equation to each lens. The image formed by lens 1 is treated as the object for lens 2. Also, we use the fact that the total magnification of the system is the product of the magnifications produced by the separate lenses.

SOLUTION

(A) Locate the image and determine the magnification of lens 1.

See the ray diagram, figure b. Apply the thin lens equation to lens 1.

Solve for q, which is positive and hence to the right of the first lens.

q = +15.0 cm

Compute the magnification of lens 1.

(B) Locate the final image and find the total magnification.

The image formed by lens 1 becomes the object for lens 2. Compute the object distance for lens 2.

p = 20.0 cm 15.0 cm = 5.00 cm

Once again apply the thin lens equation to lens 2 to locate the final image.

q = 6.67 cm

Calculate the magnification of lens 2.

Multiply the two magnifications to get the overall magnification of the system.

M = M1M2 = (0.500)(1.33) = 0.665

LEARN MORE

REMARKS The negative sign for M indicates that the final image is inverted and smaller than the object because the absolute value of M is less than 1. Because q is negative, the final image is virtual.

QUESTION If lens 2 is moved so it is 40 cm away from lens 1, the final image would be: (Select all that apply.)

PRACTICE IT

Use the worked example above to help you solve this problem. Two converging lenses are placed d2 =23.8 cm apart, as shown in figure a, with an object d1 = 35.7 cm in front of lens 1 on the left.(a) If lens 1 has a focal length of f1 = 11.9 cm, locate the image formed by this lens and determine its magnification.

(b) If lens 2 on the right has a focal length of f2 = 23.8 cm, locate the image formed and find the total magnification of the system.

EXERCISEHINTS:  GETTING STARTED  |  I'M STUCK!

If the two lenses in the figure are separated by 11.9 cm, locate the final image and find the magnification of the system. [Hint: The object for the second lens is virtual!] (Use the necessary information from the Practice It portion of the problem.)

Explanation / Answer

(a)

p1 = 35.7 cm

f1 = 11.9 cm

1/p1 + 1/q1 = 1/f1

1/25.7 + 1/q1 = 1/11.9

q1 = 22.16 cm <<---answer

M1 = -q1/p1 = -22.16/35.7 = -0.621 <<-answer

(b)

for lens 2

q1 < d2

object distance = p2 = d2-q1 = (23.8-22.2) = 1.6 cm

1/p2 + 1/q2 = 1/f2

1/1.6 + 1/q2 = 1/23.8

q2 = -1.72 cm <<----answer

M2 = -q2/p2 = 1.72/1.64 = 1.05

total magnification = M = M1*M2 = -0.621*1.05 = -0.652 <<----answer

if d2 = 11.9 cm

q1 > d2

q2 = -(22.2-11.9) = -10.3 cm

1/p2 + 1/q2 = 1/f2

-1/10.3 + 1/q2 = 1/23.8

q2 = 7.19 cm

M2 = -q2/q1 = 7.19/10.3 = 0.698

total amgnification M = M1*M2 = -0.621*0.698 = -0.433   <<-----answer

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