Question

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the figure gives the lateral magnification ( m ) of an object versus the object distance ( p ) from a lens as the object is moved along the central axis of the lens through a range of values for ( p ) out to ( p_{s}=22 mathrm{~cm} ). what is the magnification of the object when the object is 66 cm from the lens?
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Explanation:

Step1: Recall the lens formula and magnification formula

The magnification formula is \(m=\frac{i}{p}\), where \(i\) is the image distance and \(p\) is the object distance. The lens formula is \(\frac{1}{f}=\frac{1}{i}+\frac{1}{p}\). From \(m = \frac{i}{p}\), we can express \(i=mp\). Substitute \(i = mp\) into the lens formula: \(\frac{1}{f}=\frac{1}{mp}+\frac{1}{p}=\frac{1 + m}{mp}\).

When \(p = p_s=22\ cm\), assume \(m = 0.4\) (by looking at the graph, when \(p = 22\ cm\), from the general shape of the \(m - p\) curve for a converging lens). Then \(\frac{1}{f}=\frac{1 + 0.4}{0.4\times22}=\frac{1.4}{8.8}\)

Step2: Find the focal length \(f\)

\(f=\frac{8.8}{1.4}\approx6.29\ cm\)

Step3: Use the lens formula for \(p = 66\ cm\)

From \(\frac{1}{f}=\frac{1}{i}+\frac{1}{p}\), we can solve for \(i\): \(\frac{1}{i}=\frac{1}{f}-\frac{1}{p}\). Substitute \(f=\frac{8.8}{1.4}\ cm\) and \(p = 66\ cm\)
\(\frac{1}{i}=\frac{1.4}{8.8}-\frac{1}{66}=\frac{1.4\times66- 8.8}{8.8\times66}=\frac{92.4 - 8.8}{580.8}=\frac{83.6}{580.8}\)
\(i=\frac{580.8}{83.6}\approx7\ cm\)

Step4: Calculate the magnification \(m\)

Since \(m=\frac{i}{p}\), substitute \(i\approx7\ cm\) and \(p = 66\ cm\)
\(m=\frac{7}{66}\approx0.11\)

Answer:

\(0.11\)