Question

A farsighted woman breaks her current eyeglasses and is using an old pair whose refractive power is +5/3 diopters. Since these eyeglasses do not completely correct her vision, she must hold a newspaper 42.00 cm from her eyes in order to read it. She wears the eyeglasses 2.00 cm from her eyes.

a.What is the focal length of the lenses?

b.Are the lenses converging or diverging? Why?

c.How far is her near point from her eyes?

d.In order to be able to read the newspaper at 25cm from her eyes what must her new prescription (diopters) be?

Answer 1

According to the thin-lens equation:

the reciprocal of the image distance  $d_{i}$ is

$\frac{1}{d_{i}}=\frac{1}{f}-\frac{1}{d_{0}}$

(a) So for focal length(f), the refractive power in diopters is the reciprocal of the focal length in meters,

so we have

$Refractive power=\frac{1}{f}$

$=>f=\frac{1}{Refractive power}=\frac{1}{1.660diopters}=0.6024m$

$or f=60.24cm$

therefore Focal length is 60.24cm

Remembering that the eyeglasses are worn 2.00 cm from the eyes,

so now we can apply the thinlens equation as below

$\therefore \frac{1}{d_{i}}=\frac{1}{f}-\frac{1}{d_{0}}=\frac{1}{60.24cm}-\frac{1}{42cm-2cm}$

$=>\frac{1}{d_{i}}=-0.00840/cm$

$=>d_{i}=-119cm$

(c) Therefore the magnitude of this value for di is 119 cm and gives the location of the woman’s near point with respect to the eyeglasses.

Since the eyeglasses are worn 2.00 cm from the eyes, the near point is located at a distance of 119+2=121cm from the eyes

(d) The object distance (where she wants to read the paper at, 25cm) to be 2cm closer to the glasses' lens than her eyes, and her near point (the distance she can read the paper at without glasses) to similarly be 2cm closer to the glasses' lens than her eyes.

in this case

$\frac{1}{f}= \frac{1}{d_{i}}+\frac{1}{d_{0}}$

$\frac{1}{f}= \frac{1}{-(42-2)}+\frac{1}{(25-2)}=54.11cm$

Here the distance of the object is negative because it's to the left of the lens of the eyeglass.

Now in order to get the answer in diopters, we can use the equation $R(refractive power) = \frac{1}{f (in meters)}$

which is 1/0.5411m, or 1.848 diopters (which should be rounded to an even 2)

hope this will help you and hope to get positive rating.

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