Question

A diverging lens of focal length -18 cm is placed to the left of a converging lens of focal length 12 cm. The image of an object placed 24 cm to the left of the diverging lens forms on a screen located 34 cm behind the converging lens. a) What is the separation between the lenses? b) If the object is 2 cm tall, how tall will the final image be? c) Will the image be upright or inverted? d) Draw a ray diagram of the problem, using at least two beams for each lens and showing the location and orientation of both images.

          A diverging lens of focal length -18 cm is placed to the left of a converging lens of focal length 12 cm. The image of an object placed 24 cm to the left of the diverging lens forms on a screen located 34 cm behind the converging lens.
a) What is the separation between the lenses?
b) If the object is 2 cm tall, how tall will the final image be?
c) Will the image be upright or inverted?
d) Draw a ray diagram of the problem, using at least two beams for each lens and showing the location and orientation of both images.

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Added by Timothy A.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Monisha Shukla

06/04/2022

Step 1

Solving for \( x \), we get: \[ \frac{1}{-18} = \frac{1}{x} - \frac{1}{-24} \] \[ -\frac{1}{18} = \frac{1}{x} + \frac{1}{24} \] \[ -\frac{1}{18} - \frac{1}{24} = \frac{1}{x} \] \[ -\frac{7}{72} = \frac{1}{x} \] \[ x = -\frac{72}{7} \approx -10.29 \, \text{cm} \] Show more…

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A diverging lens of focal length -18 cm is placed to the left of a converging lens of focal length 12 cm. The image of an object placed 24 cm to the left of the diverging lens forms on a screen located 34 cm behind the converging lens. a) What is the separation between the lenses? b) If the object is 2 cm tall, how tall will the final image be? c) Will the image be upright or inverted? d) Draw a ray diagram of the problem, using at least two beams for each lens and showing the location and orientation of both images.

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