In: Physics

Anna holds a 14 D magnifier directly in front of her eye to get a close look at a 19-mm -diameter penny. Assume a typical 25 cm near point and a distance of 1.7 cm between the lens and the retina.

**What is the closest possible distance that she can hold the coin to have it appear in focus?**

Express your answer with the appropriate units.

The power of the magnifier is as follows:

Power \(=\frac{1}{f}=14 \mathrm{D}\)

\(\frac{1}{f}=14 \mathrm{D}\)

\(f=7.1 \mathrm{~cm}\)

The near point of the eye is \(25 \mathrm{~cm}\).

The distance between the lens and eye is \(1.7 \mathrm{~cm} .\)

Hence, the final image should be formed at a distance is as follows:

\(v=-(25 \mathrm{~cm}-1.7 \mathrm{~cm})\)

\(=-23.3 \mathrm{~cm}\)

Here, the negative sign indicates that the image is virtual.

Calculate the object distance is as follows:

The expression for the focal length of the magnifier is as follows:

\(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)

Substitute \(7.1 \mathrm{~cm}\) for \(f\) and \(-23.3 \mathrm{~cm}\) for \(v\) in the equation \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\) and solve for \(u\). \(\begin{aligned} \frac{1}{u} &=\frac{1}{f}-\frac{1}{v} \\ u &=\frac{(v)(f)}{v-f} \\ &=\frac{(-23.3 \mathrm{~cm})(7.1 \mathrm{~cm})}{-23.3 \mathrm{~cm}-7.1 \mathrm{~cm}} \\ &=5.44 \mathrm{~cm} \end{aligned}\)

Hence, the minimum distance of the coin from the magnifier is \(5.44 \mathrm{~cm}\).

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