An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the le

Name: Problem 91, Chapter 34: University Physics with Modern Physics
Uploaded: 2019-08-07T17:36:24-08:00
Duration: 5 min 28 s
Channel: Bruce Edelman
Description: An object to the left of a lens is imaged by the lens on a screen 30.0 cm to the right of the lens. When the lens is moved 4.00 cm to the right, the screen must be moved 4.00 cm to the left to refocus the image. Determine the focal length of the lens.

Question

An object to the left of a lens is imaged by the lens on a screen 30.0 $\mathrm{cm}$ to the right of the lens. When the lens is moved 4.00 $\mathrm{cm}$ to the right, the screen must be moved 4.00 $\mathrm{cm}$ to the left to refocus the image. Determine the focal length of the lens.

   An object to the left of a lens is imaged by the lens on a screen 30.0 $\mathrm{cm}$ to the right of the lens. When the lens is moved 4.00 $\mathrm{cm}$ to the right, the screen must be moved 4.00 $\mathrm{cm}$ to the left to refocus the image. Determine the focal length of the lens.

University Physics with Modern Physics

Hugh D. Young, Roger… 12th Edition

Chapter 34, Problem 91

Instant Answer

Solved by Expert Bruce Edelman

08/07/2019

Step 1

Let's assume that the object is located at a distance of $d_o$ to the left of the lens. Show more…

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An object to the left of a lens is imaged by the lens on a screen 30.0 $\mathrm{cm}$ to the right of the lens. When the lens is moved 4.00 $\mathrm{cm}$ to the right, the screen must be moved 4.00 $\mathrm{cm}$ to the left to refocus the image. Determine the focal length of the lens.

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Key Concepts

Two-Position Method for Fixed Object–Image Separation

When an object and a screen are kept fixed, there can be two distinct lens positions that produce a sharp image on the screen. In this scenario, the sum of the object distance and the image distance remains constant. The two different placements of the lens lead to a pair of equations via the lens formula whose difference depends on the displacement of the lens. This symmetry in positions is exploited to eliminate the object distance and solve directly for the focal length, as seen in problems where adjustments in the lens and screen positions yield known displacements.

Thin Lens Equation

This equation, 1/f = 1/s + 1/s?, is fundamental in geometrical optics. It relates the focal length (f) of a thin lens to the object distance (s) and the image distance (s?), allowing us to predict where an image will form for a given object placement or to determine the lens’s focal length when s and s? are known.

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