*d*

_{o1},

*d*

_{i1}, and

*f*

_{1}.

This image 1 will then become the object 2 for lens 2, which will produce the (final) image 2, and the thin lens equation is again applied to relate

*d*

_{o2},

*d*

_{i2}, and

*f*

_{2}.

Keep in mind that the intermediate image 1 from the first lens is subsequently "fed" to the second lens as its object 2.

*If*the eye suffers from a visual defect, then by itself it will only be able to see an object 2 located at a distance

*d*

_{o2}. If used correctly, a contact lens will take an object 1 (located at a nominal distance

*d*

_{o1}), and produce an image 1 that will be located at a distance

*d*

_{o2}=

*d*

_{i1}that the eye is able to focus on.

So knowing the nominal object distance

*d*

_{o1}, and the actual distance

*d*

_{o2}an uncorrected eye can focus on, the focal length

*f*

_{1}of the contact lens can be solved for using only one thin lens equation.

*P*"diopters," which are merely the inverse of the focal length

*f*(measured in meters).

The sign convention for refractive power is the same as for focal lengths: positive values for converging contact lenses, and negative values for diverging contact lenses.

*P*("diopters") listed there.

*d*

_{o2}= +5.0 m (positive object distances correspond to being in

*front*of the eye) that this student's unaided eye can focus on, instead of the nominal farthest distance of +∞.

What this means is that the contact lens will take an object located (in front of it) at

*d*

_{o1}= +∞, and produce a virtual image (in front of it!) at

*d*

_{i1}= –5.0 m. This virtual image 1 becomes the object 2 (

*d*

_{o2}= +5.0 m) for the eye, which is able to see 5.0 m in front of it.

Solving for the focal length

*f*

_{1}of the contact lens,

(1/

*d*

_{o1}) + (1/

*d*

_{i1}) = (1/

*f*

_{1}),

(1/(+∞)) + (1/(–5.0 m)) = (1/

*f*

_{1}),

*f*

_{1}= –5.0 m,

as (1/(+∞)) = 0. Remember that contact lenses are prescribed in diopters rather than focal lengths, such that the refractive power is:

*P*= (1/

*f*

_{1}) = (1/(–5.0 m)) = –0.20 m

^{–1}or –0.20 D.

*d*

_{o2}= +0.60 m (positive object distances correspond to being in

*front*of the eye) that this student's unaided eye can focus on, instead of the nominal reading distance of 0.25 m.

What this means is that the contact lens will take an object located (in front of it) at

*d*

_{o1}= +0.25 m, and produce a virtual image (in front of it!) at

*d*

_{i1}= –0.60 m. This virtual image 1 becomes the object 2 (

*d*

_{o2}= +0.60 m) for the eye, which is able to see 0.60 m in front of it.

Solving for the focal length

*f*

_{1}of the contact lens,

(1/

*d*

_{o1}) + (1/

*d*

_{i1}) = (1/

*f*

_{1}),

(1/(+0.25 m)) + (1/(–0.60 m)) = (1/

*f*

_{1}),

*f*

_{1}= +0.43 m,

Remember that contact lenses are prescribed in diopters rather than focal lengths, such that the refractive power is:

*P*= (1/

*f*

_{1}) = (1/(+0.43 m)) = +2.3 m

^{–1}or +2.3 D.

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