image formation by lenses - department of physicscwells/phys1070/optics2.pdf · image formation by...

Image formation by Lenses

n1 n2 NOTE: r1 is radius of r1 lens surface which

r2 faces medium 1nL (analogously for r2)

light Find object image t relations and

here: r1 > 0, r2 < 0 magnification

For object in medium 1 (to left of lens)

Image formation involves two refractions

1. Light rays from object first refract atsurface between medium 1 and lens to forman image (as described in fishbowl example).

2. Image from 1. (whether real or virtual)serves as object for second refraction atsurface between lens and medium 2.

Use the thin-lens approximation: t is much smallerthan object and image distances.

RESULTS: (for details: derivation in text pp 3-10)

Object - Image distance relation

n1/p + n2/q = P SAME as for singlesurface except:

Power

P = P1 + P2 = (nL-n1)/r1 + (n2–nL)/r2

the power is the SUM of the powers of the tworefracting surfaces of the lens.

Magnification

M = -(n1q)/(n2p) PRODUCT ofmagnifications producedby each surface.

Result is exactly the same as for the singlerefracting surface because the nL andintermediate image (which becomes the object forsecond refraction) distances cancel.

SIGN CONVENTIONS GIVENPREVIOUSLY STILL APPLY

Focal points of lenses:

1st (or primary) focal point: Object positionwhich produces image at infinity, q = ∞∞.

Call this position f1: n1/p + n2/q = P⇒ n1/f1 = P

⇒ n1/P = f1 in object space

2nd (secondary) focal point: Image positionproduced by an object at infinity, p = ∞∞.

Call this position f2: n1/p + n2/q = P⇒ n2/f2= P

⇒ n2/P = f2 in image space

n1 n2

Y light r1

O r2 Y’ I

p q

f1 f2

Simple Common Case

Often the medium on both sides of the lens is thesame: n1 = n2 = n. Then:

LENSMAKER’S EQUATION

n/p + n/q = P where P = (nL-n)(1/r1 - 1/r2)

Now, the focal lengths are equal but on oppositesides of the lens.

f1 = f2 = f = n/P

and the LENSMAKER’S EQUATION can be re-written:

1/p + 1/q = 1/f

The “n” cancels out ⇒ Equation holds for ANY “n”.

A similar thing happens to the magnification

Magnification

M = -q/p

Two types of lenses

Consider a lens with the same medium on bothsides:

P = (nL-n)(1/r1 - 1/r2) and with nL > n.

For example a glass lens in air (nL = 1.5, n = 1)

1. Converging or Positive Lenses P > 0

light r1

r1 > 0, r2 < 0 r2 BICONVEX

light r1

r1 > 0, r2 > 0r2 but r2 > r1

so 1/r2 < 1/r1

light r1

r1 < 0, r2 < 0 r2 r1 > r2

Positive lenses: thicker in the middle than edges

2. Diverging or Negative Lenses P< 0

NB : r1 is radius for side FACING light

light r2

r1 < 0, r2 > 0 r1 BICONCAVE

light r1

r1 < 0, r2 < 0r2 r1 < r2

so 1/r2 < 1/r1

light r1

r1 > 0, r2 > 0 r2 but r2 < r1

Negative lenses are thinner in the middle thanat edges

Power of lens is independent of which side is facinglight (orientation).

Results from LENSMAKER’S EQUATION

1.) Positive lens, f > 0

case p > f (object further than focal point)

1/p < 1/f1/q = 1/f - 1/p > 0 ⇒ q > 0

i) image downstream from lens.ii) Magnification = -q/p = inverted

case p < f (object closer than focal point)

1/p > 1/f1/q = 1/f - 1/p < 0 ⇒ q < 0

iii) image upstream from lens.iv) Magnification = -q/p = upright

2.) Negative lens, f < 0

1/q = 1/p – 1/f

both terms negative, hence q < 0.

Solved Problem A 30 cm overhead slide is to beprojected onto 2.0 m high screen 3m away. Whatfocal length lens should be used and at whatdistance should the object be placed to fill thescreen?

2 mneglect distance image

heightq = 3.0 m

p0.30 m = object height

Real image projected on screen would be invertedif not for presence of mirror above lens.

Magnification = -2/0.30 = -6.67

Since M = -6.67 = -q/p = -3/p⇒ p = 0.45 m

Hence lens must be positioned 0.45 m above lens

focal length: 1/f = 1/p + 1/q = 1/0.45 + 1/3 = 2.56 df = 1/2.56 = 0.39 m

Combination of lenses

e.g. microscope, telescope, eye plus spectacles

Method: Image from first lens becomes“object” for second lens.

Solved Example (txt p.3-24) Two converging lensesare separated by 0.45 m. Their focal lengths are0.13 m for the first lens and 0.08 m for thesecond. Find the magnification and image positionof an object placed 0.15 m in front of first lens.

0.15 m 0.45 m

lens 1 lens 2

For 1st lens: object distance p1 = 0.15 mfocal distance f1 = 0.13 m

1/f1 = 1/p1 + 1/q1 ⇒ 1/q1 = 1/0.13 – 1/0.15 = 1.0 m-1

⇒ q1 = 1/1.0 = 1.0 m[ Real image since p1 > f1]

Light

0.15 m 0.45 m 0.55 m

lens 1 lens 21.0 m = q1

NB: Distance from surface of lens 2 to object forlens 2 (Image of lens 1) is with the flow of light.THEREFORE p2 IS NEGATIVE.

For 2nd lens: obj. dist. = p2 = -(1 - 0.45) = - 0.55 m focal distance f2 = 0.08 m

1/f2 = 1/p2 + 1/q2

⇒ 1/q2 = 1/0.08 – 1/(-0.55) = 14.3 m-1

⇒ q2 = 1/14.3 = 0.070 m

Since q2 > 0, distance from lens 2 to image is withthe flow of light.

Light Final image

0.15 m 0.45 m 0.55 m

lens 1 lens 2 q2

1.0 m = q1

Magnification:M = M1M2 = (-q1/p1)(-q2/p2)

= (-1/0.15)(-0.070/(-0.55))

M = -0.85 Final image is inverted

Other examples: text problem 3-9

Accomodation of the eye

AH VH Index of refraction (n)Air 1

Cornea cornea 1.38aqueous humour 1.34

Lens Lens 1.41vitreous humuor 1.34

1. Largest change in n occurs at air-corneainterface most refraction occurs there.

2. For sharp vision light must focus on fovea

3. Must be able to focus light from objects atdifferent distances.

Accomodation: ability of eye to focus on objectsat varying distances

Mainly caused by changes in power of lens due tochanges in its curvature caused by ciliary muscles.

Simplified model: “Reduced eye” uses a singlerefracting surface with variable power

1.67 mm

d n= 1.34 O

n = 1 25 mm

To give correct results distances must bemeasured from a plane 1.67 mm behind cornea:e.g.: if object-cornea distance = d, then use:

p = d – 1.67 mmand q = 25 – 1.67 = 23.3 mm

Power of reduced eye is:P = 1/p + n/q = 1/(d+0.00167) + 1.34/(0.0233)

OVER WHAT RANGE MUST POWER VARY?

For large distances, d →∞, the eye relaxes.

P = P0 = 0 + 1.34/(0.0233) = 57.5 diopter.

For shorter distances, d, the lens mustACCOMMODATE to focus. Its power mustincrease above P0.

P = Pd = 1/(d+0.00167) + 1.34/(0.0233)

= 1/(d+0.00167) + 57.5 diopter

The nearest point at which a typical human eye canfocus (its near point) is 25 cm: d = 0.25 m

Pd = 1/(0.25 + 0.00167) + 57.5 = 61.5 diopter

Therefore, the ciliary muscles need to be able tocause a 61.5 – 57.5 = 4 diopter change to focusover the typical range.

Eye defects and their Correction

Hyperopia: “farsightedness” cannot see nearbyobjects clearly.

Curvature of cornea/lens OR cornea-retinadistance is too small: Image forms behind retina.

Uncorrected corrected

converginglens required

Myopia: “nearsightedness” cannot see distantobjects clearly.

Curvature of cornea/lens OR cornea-retinadistance is too large: Image is in front of retina

Uncorrected corrected (with -ve lens)

Hyperopia or Hypermetropia

Near point: minimum distance clearly focussed.

The near point of a hyperopic is larger than 25 cm.

Solved exampleA person’s near point is 1 m (they can only seeclearly beyond 1 m). What power of lens isrequired so the person can see clearly at 25 cm.

Need a lens to form an image at 1 m (where theperson can see it) when object is 25 cm from eye.That is, if p = 0.25 m, q = -1 m

Image must bein front of lens

q surface to imagedistance against

light flow of light. p (q is –ve)

P = 1/p + 1/q = 1/0.25 – 1/1 = 3 diopter

positive or converging lens

Myopia

Far point: maximum distance clearly focussed.

The far point of a myopic is less than infinity.

Solved exampleA person’s near point is 0.5 m (they can only seeclearly within 1 m). What power of lens is requiredto correct the myopia. (far point of eye plusspectacle = infinity).

Need a lens to form an image at 0.5 m (where theperson can see it) when object is very far fromeye.That is, if p = ∞, q = -0.5 m

Image must bein front of lens

p = ∞ surface to imagedistance against

light flow of light. q (q is –ve)

P = 1/p + 1/q = 1/∞ – 1/0.5 = - 2 diopter [-ve lens]

AstigmatismDue to a cornea which is not spherical but is moresharply curved in one plane than others.

vertical plane these planes may havedifferent curvatures

front horizontal plane

Cannot focus simultaneously on vertical andhorizontal lines. Corrected with lenses ofcompensating curvatures.

ExampleCornea is more sharply curved (has a smaller radiusof curvature) in vertical plane.

Can compensate with appropriate cylindrical lens

increase horizontal or decrease verticalOPTICALPOWERwith

+ve lens -ve lenscurved horizontally curved vertically