Question 33.12: Farsighted eye. Sue is farsighted with a near point of 100 c......
Farsighted eye. Sue is farsighted with a near point of 100 \mathrm{~cm}. Reading glasses must have what lens power so that she can read a newspaper at a distance of 25 \mathrm{~cm} ? Assume the lens is very close to the eye.
APPROACH When the object is placed 25 \mathrm{~cm} from the lens, we want the image to be 100 \mathrm{~cm} away on the same side of the lens (so the eye can focus it), and so the image is virtual, as shown in Fig.33-29, and d_{\mathrm{i}}=-100 \mathrm{~cm} will be negative. We use the thin lens equation (Eq. 33-2) to determine the needed focal length. Optometrists’ prescriptions specify the power (P=1 / f, Eq. 33-1) given in diopters \left(1 \mathrm{D}=1 \mathrm{~m}^{-1}\right).
P=\frac{1}{f} (33-1)
\frac{1}{d_{\mathrm{o}}}+\frac{1}{d_{\mathrm{i}}}=\frac{1}{f} (33-2)
Given that d_{\mathrm{o}}=25 \mathrm{~cm} and d_{\mathrm{i}}=-100 \mathrm{~cm}, the thin lens equation gives
\frac{1}{f}=\frac{1}{d_{\mathrm{o}}}+\frac{1}{d_{\mathrm{i}}}=\frac{1}{25 \mathrm{~cm}}+\frac{1}{-100 \mathrm{~cm}}=\frac{4-1}{100 \mathrm{~cm}}=\frac{1}{33 \mathrm{~cm}} .
So f=33 \mathrm{~cm}=0.33 \mathrm{~m}. The power P of the lens is P=1 / f=+3.0 \mathrm{D}. The plus sign indicates that it is a converging lens.
NOTE We chose the image position to be where the eye can actually focus. The lens needs to put the image there, given the desired placement of the object (newspaper).