Question

In: Physics

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

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.

Solutions

Expert Solution

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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