Optical Instruments — Question 5

Question 5

Near-point correction maps a close object to a farther virtual image

An ideal eye can focus no closer than 1.00 m. A thin corrective lens at the eye plane is intended to let it view an object at 0.25 m while forming a virtual image at the eye near point. Ignore lens-eye separation.

1. Find the required focal length and optical power.

2. Find the corrective lens transverse magnification and the angular size received by the eye for an object of height h.

Question 5 - Solution

Near-point correction maps a close object to a farther virtual image

(a) Method and calculation

$$p=+0.25\,\mathrm m,\ q=-1.00\,\mathrm m\Rightarrow1/f=4-1=3\,\mathrm{m^{-1}}.$$$$\boxed{f=1/3\,\mathrm m,\quad\mathcal P=+3.00\,\mathrm{m^{-1}}}.$$

(b) Interpretation and further calculation

$$\boxed{m=-q/p=+4,\quad h_I=4h}.$$$$\boxed{\alpha\simeq h_I/(1.00\,\mathrm m)=h/(0.25\,\mathrm m)}.$$

Physical check

The correction changes the wavefront curvature so the eye can focus the same viewing direction; the simple transverse magnification is not itself an extra angular enlargement.