A pharmacy student has eyes who have a relaxed refractive power of 48 diopters a

Question

A pharmacy student has eyes who have a relaxed refractive power of 48 diopters and her retina is 2.35 cm from her lens.

a). If she is nearsighted, what is her far point? If she is farsighted, what is her near point?

b). Let's say she wants to be able to focus clearly on objects that are from 200 m to 6 cm away. What is the power of the corrective lens she should wear assuming her glasses are 2 cm from her eye?

Corrective Power =

**Please work out each step, I've had so many problems with this and I've tried several ways so I don't know what I'm doing wrong.

Explanation / Answer

refractive power P=48 D

power P=48 D

focal length, f=100/p (in cm)

distance between lense and retine v=2.35 cm

a)

she is nearsighted,

because P is +ve

use,

1/u+1/2.35=1/f

1/u+1/2.35=48/100

u=18.36 cm

near pont distance is u=18.36 cm

b)

200 m to 6cm

2cm fro her eye

distance between glass and eye is,d=2cm

power of the eye lense P1=48 D

power of the corrective lense is,P2

effective power,

P=P1+P2+d*P1*P2

1/v=48+P2+0.02*48*P2

(1/0.0235)=(48)+(P2)+(0.02*48*P2)

===> P2=-2.78 D

corrective power P2=-2.78 D

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