1. Aberrations in Optical Components (lenses, mirrors)
Chromatic Aberrations: Because the index of refraction of the lens material depends slightly on wavelength, the focal length also depends on wavelength, so different wavelengths will form images at slightly different places.
(Doesn’t occur for mirrors, since  = ’ for all wavelengths)
2) Spherical Aberration (both lenses and mirrors):
We have made the paraxial approximation, that all rays
are near the principal axis. As rays get further from the
principal axis, their focus point changes, so if the lens or
mirror collects light through a large diameter aperture,
the image will be blurry.
To sharpen the image, use a smaller diameter aperture
in front of or behind the lens (or in front of the mirror)
to restrict the light to being close to the axis;
in a camera this requires a longer exposure
time to compensate for the reduced amount
of light entering the camera/second.
aperture

2. Camera
A converging lens forms a real image on the film or CCD detector.
In more expensive cameras, the distance between the lens and CCD/film (q) is adjustable so that 1/q = 1/f - 1/p to keep the image sharp. An expensive camera will also have an adjustable aperture so one can limit spherical aberrations, compensating by increasing the exposure time.
or film

3. Less expensive cameras have a fixed lens position
i.e. (lens-aperture distance = f)
and aperture, with diameter D <<f.
The image will be in reasonable focus on the CCD/film as long as the image distance p >> f.
(i.e. 1/q = 1/f – 1/p  1/f if p >> f)
(In the limit, D 0, you don’t even need a lens: the camera is a pin-hole camera: only “one ray” from each point on the object reaches the CCD/film. These are sometimes used for surveillance, because difficult to detect.) f
Cameras in cell phones typically replace the mechanical shutter with an “electronic shutter” which turns the pixels in the CCD detector on/off; this can sacrifice resolution (because of timing issues). Also, better cell phone cameras have an autofocus, that moves the lens slightly to get better focus.

4. Eye Pupil is aperture which controls how much light enters eye.
Real image formed on retina where light sensors (rods and cones) send signals through optic nerve to brain.
Refraction occurs in outer surface (tears, cornea), aqueous humor, and lens.
Ciliary muscles can cause the lens to contract,
decreasing its focal length, to keep image in focus
on retina.

5. 1/f = 1/p + 1/q, to keep q = size of eyeball.
For a “normal” eye, parallel rays from  are focused on the retina when the lens is “relaxed” (i.e. not compressed). Therefore, the normal eye can focus on . f
For rays from a closer object to be focused on the retina, the lens must be compressed to decrease f:
1/f = 1/p + 1/q, to keep q = size of eyeball.
The shortest distance that the eye can focus on = near point. For a normal eye, near point  25 cm.
If the object is closer than the near point, it will be focused outside the eye; i.e. the image on the retina will be blurry.

6. an
A farsighted person’s near point is larger than 25 cm. This can happen because the eyeball is “too short” or because the lens cannot compress (and reduce its focal length) enough. (The latter often happens with old age as the lens becomes less flexible or the ciliary muscle weakens.)

7. A farsighted person’s image can be corrected by placing a converging lens in front of the eye, which will bring the focus onto the retina.

8. One needs f(lens) > object distance, so that the lens forms a virtual image at the near point of the eye when the object is 25 cm from the eye. -
Consider the case where the lens is very close to the eye (e.g. contact lens) so that image and object distances can be measured from the eye.
One wants the position of the virtual image of the correctional lens (q)
q = - patient’s near point when p = 25 cm (normal near point):
 1/f = 1/25cm - 1/near point.
Example, if near point = 50 cm, one needs 1/f = 1/25cm - 1/50cm = +1/50cm
f = 50 cm

9. For a nearsighted eye, the focal length of the relaxed eye-lens (i. e
For a nearsighted eye, the focal length of the relaxed eye-lens (i.e. its largest value) is less than the length of the eyeball, so the eye cannot focus on objects at infinity (i.e. the eyeball is “too long”.

10. The most distant object that can be brought to focus on the retina (i
The most distant object that can be brought to focus on the retina (i.e. by the relaxed eye-lens) is at the “far point”.
far point
[For the normal eye, the far point = .]

11. an

13. The diverging lens should form a virtual image at the eye’s far point for an object at “infinity”.
Image at far point -q p
For example, if the diverging lens is very close to the eye (e.g. contact lens) so all distances can be measured from the eye, 1/p + 1/q = 1/f
1/ - 1/far pt. = 1/f
f = - far pt.
diverging lens

14. Magnifying Glass
Recall that a converging lens will form a magnified virtual image if the object is inside the focal point (p < f). Since 1/q = 1/f - 1/p  q = -pf/(f-p). Therefore, the lateral magnification M = -q/p = f/(f-p) > 1.

15. It is more useful to consider the “angular magnification”, which compares how large a viewer perceives the image to be with how the large the object could appear without use of the lens; i.e. compares the sizes of the images (with and without the magnifying glass) on the retina.
For an object a distance p from the eye:
Since the minimum value of p for which the eye can form a clear image is the near point, the image on the retina will have its maximum size when the object is at the eye’s near point:
maximum size (without magnifier) =
0  h/LNP
LNP
The size of the image of an object on the retina    h/p (tan  ).

16. (LNP/LFP) (LFP/f +1) < m < (LNP/f + 1)
L = -q
The angle for the image formed by the magnifying glass is
 = h’/L = h/p
To see this image clearly, one needs L to be between the eye’s near point (LNP) and its far point (LFP).
Since L = -q = pf/(f-p)  p = Lf /(L+f)
so LNPf/(LNP+f) < p < LFPf/(LFP+f)
Therefore, the angular magnification
m = /0 = (h/p) / (h/LNP) = LNP/p
(LNP/LFP) (LFP/f +1) < m < (LNP/f + 1)
image at near point:
maximum m
image at far point: relaxed eye

17. (LNP/LFP) (LFP/f +1) < m < (LNP/f + 1)
For normal eye, LNP = 25 cm and LFP =  
25 cm/f < m < 25cm/f + 1
relaxed eye
maximum m (image at NP = 25 cm)
Example: If f = 6 cm, < m < 5.167
If f = 2.5 cm < m < 11
The smaller f [i.e. the “fatter” the lens], the larger m.

18. Problem (36. 66): A lens with focal length = 5
Problem (36.66): A lens with focal length = 5.00 cm is to be used as a magnifier. Where should the object be placed to obtain maximum magnification with a clear image for a normal eye, and what is the magnification?
25 cm/f < m < 25cm/f + 1

19. Problem (36. 66): A lens with focal length = 5
Problem (36.66): A lens with focal length = 5.00 cm is to be used as a magnifier. Where should the object be placed to obtain maximum magnification with a clear image for a normal eye, and what is the magnification?
25 cm/f < m < 25cm/f + 1
Maximum size clear image: q = - near point = -25 cm
1/p = 1/f – 1/q = 1/5cm – (-1/25cm) = +6/25cm
p = cm
m = 1 +25/5 = 6
[For image at far point (q = -), p = 5 = 5.00 cm, and m = 25/5 = 5]