Q.A myopic adult has a far point at 0.1 m. His power of accomodation is 4 diopters.

Given: far point $= 0.1\ \text{m}$, accommodation $= 4\ \text{D}$, eye-lens-to-retina distance $v = 0.02\ \text{m}$.

(i) The correcting lens must image infinity at the far point: $u = \infty$, $v = -0.1\ \text{m}$.

(ii) The relaxed eye focuses the far point on the retina ($u = -0.1\ \text{m}$, $v = +0.02\ \text{m}$):

Full accommodation adds $4\ \text{D}$: $P_{\max} = 64\ \text{D}$. Near point $u_n$:

$64 = \frac{1}{0.02} - \frac{1}{u_n} = 50 - \frac{1}{u_n} \Rightarrow u_n = -\frac{1}{14}\ \text{m} \approx -0.071\ \text{m}.$ …

Data: far point $= 0.1\ \text{m}$; power of accommodation $= 4\ \text{D}$; image distance eye-lens to retina $v = 2\ \text{cm} = 0.02\ \text{m}$.

(i) Lens to see distant objects

A myopic eye focuses parallel light in front of the retina; the correcting lens must make an object at infinity appear at the far point. With $u = \infty$ and $v = -0.1\ \text{m}$ (virtual, on the object side):

So a diverging lens of power $-10\ \text{D}$ (focal length $-10\ \text{cm}$) is required.

(ii) Near point without glasses

Treat the eye as a single lens forming the image on the retina, $v = +0.02\ \text{m}$. When relaxed it is focused on the far point ($u = -0.1\ \text{m}$):

At maximum accommodation the eye adds $4\ \text{D}$:

The near point is the closest object still imaged on the retina:

So the unaided near point is about $7.1\ \text{cm}$ from the eye.

(iii) Near point with glasses …

Method: Solving Eye-Defect (Myopia/Hypermetropia) Problems With the Lens Formula

Use this method for any question involving a defective eye's far point, near point, or power of accommodation, whether or not corrective spectacles are involved.

Steps

Step 1: Model the eye as a single lens imaging onto a fixed retina distance

Treat the eye's own lens as forming a real image at a fixed image distance $v$ (the given eye-lens-to-retina distance), for whatever object distance $u$ it is looking at. Every calculation below uses the thin-lens power formula

with all distances in metres so $P$ comes out in dioptres.

Step 2: For the corrective-lens power, use the far point as the target image location

A corrective lens must take an object at infinity ($u=\infty$) and form a virtual image exactly at the eye's own far point (myopia) — or bring a nearby object's image out to the near point (hypermetropia). Apply $P = \frac{1}{v} - \frac{1}{u}$ with $v$ = the far/near point distance (negative if virtual, by sign convention) and $u=\infty$ (so $1/u=0$) to isolate $P$.

Step 3: For the eye's own near point (no glasses), add the accommodation power to the relaxed-eye power …