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Chapter 31: Problem 21

By what factor is the image magnified for an object 1.5 focal lengths from a converging lens? Is the image upright or inverted?

Short Answer

Expert verified
The magnification of the image can be computed by finding the ratio of the image to the object distance. Its sign will give the relative orientation of the image.

Step by step solution

01

Identify given inputs

The object is placed at a distance of 1.5f (where f is the focal length) from the lens. So, \(u = -1.5f\) (negative sign because the object is placed on the left of the lens).
02

Calculate the image distance

Rewrite the lens formula and solve for \(v\), \(v = \frac{1}{{1/f+1/u}}\). Substitute \(f\) and \(u\) into the formula to compute \(v\), which represents the image distance.
03

Compute the magnification

Magnification (m) is calculated using the formula \(m=-v/u\). Substitute \(v\) and \(u\) into the magnification equation.
04

Determine the nature of the image

If the magnification is positive, the image is upright, and if the magnification is negative, the image is inverted. Apply this rule to the calculated magnification of the image to determine whether the image is upright or inverted.

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