Question 33.13: Nearsighted eye. A nearsighted eye has near and far points o......

(a) For an object at infinity \left(d_{\mathrm{o}}=\infty\right), the image must be in front of the lens 17 \mathrm{~cm} from the eye or (17 \mathrm{~cm}-2 \mathrm{~cm})=15 \mathrm{~cm} from the lens; hence d_{\mathrm{i}}=-15 \mathrm{~cm}. We use the thin lens equation to solve for the focal length of the needed lens:

\frac{1}{f}=\frac{1}{d_{\mathrm{o}}}+\frac{1}{d_{\mathrm{i}}}=\frac{1}{\infty}+\frac{1}{-15 \mathrm{~cm}}=-\frac{1}{15 \mathrm{~cm}} .

So f=-15 \mathrm{~cm}=-0.15 \mathrm{~m} or P=1 / f=-6.7 \mathrm{D}. The minus sign indicates that it must be a diverging lens for the myopic eye.

(b) The near point when glasses are worn is where an object is placed \left(d_{0}\right) so that the lens forms an image at the “near point of the naked eye,” namely 12 \mathrm{~cm} from the eye. That image point is (12 \mathrm{~cm}-2 \mathrm{~cm})=10 \mathrm{~cm} in front of the lens, so d_{\mathrm{i}}=-0.10 \mathrm{~m} and the thin lens equation gives

\frac{1}{d_{\mathrm{o}}}=\frac{1}{f}-\frac{1}{d_{\mathrm{i}}}=-\frac{1}{0.15 \mathrm{~m}}+\frac{1}{0.10 \mathrm{~m}}=\frac{-2+3}{0.30 \mathrm{~m}}=\frac{1}{0.30 \mathrm{~m}}

So d_{\mathrm{o}}=30 \mathrm{~cm}, which means the near point when the person is wearing glasses is 30 \mathrm{~cm} in front of the lens, or 32 \mathrm{~cm} from the eye.