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?

## 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:

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

$$=-23.3 \mathrm{~cm}$$

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 $$5.44 \mathrm{~cm}$$.