Physics 202Professor P. Q. Hung

311B, Physics Building

Physics 202 – p. 1/42

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The Human Eye

Physics 202 – p. 2/42

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The Human Eye

The Eye: A complex yet simple object withbasically: The cornea, Aqueous humor,Lens+ciliary muscles, vitrous humor, retina.

Basically, light travels say, in air, with n ≈ 1gets refracted at the cornea n = 1.38. Somefurther (small) refraction is done by theAqueous humor (n = 1.33), the lens (n = 1.40)and the vitrous humor (n = 1.34).

The ciliary muscles can change the radius ofcurvature of the lens so that the image can beproperly focused on the retina.

Physics 202 – p. 3/42

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The Human Eye

The Eye: A complex yet simple object withbasically: The cornea, Aqueous humor,Lens+ciliary muscles, vitrous humor, retina.

Basically, light travels say, in air, with n ≈ 1gets refracted at the cornea n = 1.38. Somefurther (small) refraction is done by theAqueous humor (n = 1.33), the lens (n = 1.40)and the vitrous humor (n = 1.34).

The ciliary muscles can change the radius ofcurvature of the lens so that the image can beproperly focused on the retina.

Physics 202 – p. 3/42

Optical Instruments

The Human Eye

The Eye: A complex yet simple object withbasically: The cornea, Aqueous humor,Lens+ciliary muscles, vitrous humor, retina.

Basically, light travels say, in air, with n ≈ 1gets refracted at the cornea n = 1.38. Somefurther (small) refraction is done by theAqueous humor (n = 1.33), the lens (n = 1.40)and the vitrous humor (n = 1.34).

The ciliary muscles can change the radius ofcurvature of the lens so that the image can beproperly focused on the retina. Physics 202 – p. 3/42

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The Human Eye

Physics 202 – p. 4/42

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Object at “infinity”: Ciliary muscles arerelaxed ⇒ the focal point is situated farthestfrom the lens.

Near object: Ciliary muscles are tense ⇒ thefocal point is situated closer to the lens.

Near Point: Nearest point to the eye for whicha near object is still focused, and beyond (i.e.as one gets closer and closer) which objectsappear fuzzy.

Far point: The farthest point from the eye forwhich an object is still focused. In general, itis practically infinity.

Physics 202 – p. 5/42

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Object at “infinity”: Ciliary muscles arerelaxed ⇒ the focal point is situated farthestfrom the lens.

Near object: Ciliary muscles are tense ⇒ thefocal point is situated closer to the lens.

Near Point: Nearest point to the eye for whicha near object is still focused, and beyond (i.e.as one gets closer and closer) which objectsappear fuzzy.

Far point: The farthest point from the eye forwhich an object is still focused. In general, itis practically infinity.

Physics 202 – p. 5/42

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Object at “infinity”: Ciliary muscles arerelaxed ⇒ the focal point is situated farthestfrom the lens.

Near object: Ciliary muscles are tense ⇒ thefocal point is situated closer to the lens.

Near Point: Nearest point to the eye for whicha near object is still focused, and beyond (i.e.as one gets closer and closer) which objectsappear fuzzy.

Far point: The farthest point from the eye forwhich an object is still focused. In general, itis practically infinity.

Physics 202 – p. 5/42

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Object at “infinity”: Ciliary muscles arerelaxed ⇒ the focal point is situated farthestfrom the lens.

Near object: Ciliary muscles are tense ⇒ thefocal point is situated closer to the lens.

Near Point: Nearest point to the eye for whicha near object is still focused, and beyond (i.e.as one gets closer and closer) which objectsappear fuzzy.

Far point: The farthest point from the eye forwhich an object is still focused. In general, itis practically infinity.

Physics 202 – p. 5/42

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The Human Eye

Physics 202 – p. 6/42

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The Human Eye: ExampleThe near-point distance of a given eye isN = 25 cm. Treating the eye as if it were a singlethin lens a distance 2.5 cm from the retina, findthe focal length of the lens when it is focused onan object (a) at the near point; (b) at infinity.

Physics 202 – p. 7/42

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The Human Eye: Example

(a) p = 25 cm, q = 2.5 cm.1f

= 1p

+ 1q

= 125 cm

+ 12.5 cm

= 0.44cm−1⇒

f = 2.3 cm.

(b) p = ∞, q = 2.5 cm.1f

= 1p

+ 1q

= 1∞

+ 12.5 cm

= 12.5 cm

⇒ f = 2.5 cm.The focal length for the “relaxed” state (b) islarger than that for a “tense” state.

Physics 202 – p. 8/42

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The Human Eye: Example

(a) p = 25 cm, q = 2.5 cm.1f

= 1p

+ 1q

= 125 cm

+ 12.5 cm

= 0.44cm−1⇒

f = 2.3 cm.

(b) p = ∞, q = 2.5 cm.1f

= 1p

+ 1q

= 1∞

+ 12.5 cm

= 12.5 cm

⇒ f = 2.5 cm.The focal length for the “relaxed” state (b) islarger than that for a “tense” state.

Physics 202 – p. 8/42

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The Camera: ExampleSame principle with one exception: The lens isfixed in shape and the focus is changed bymoving the lens.Example:A simple camera uses a thin lens with a focallength of 50.0 mm. What are the image distancesfor an object situated at 40.0 m and 3.00 m awayrespectively?

Physics 202 – p. 9/42

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The Camera: Example

Physics 202 – p. 10/42

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The Camera: ExampleFrom the thin lens equation, one gets

(a) For p = 40.0 m, one getsq = 5.006 cm.

(b) For p = 3.0 m, one getsq = 5.08 cm.

Physics 202 – p. 11/42

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The Camera: ExampleFrom the thin lens equation, one gets

(a) For p = 40.0 m, one getsq = 5.006 cm.

(b) For p = 3.0 m, one getsq = 5.08 cm.

Physics 202 – p. 11/42

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The Camera: Example

Difference in image distance:5.08 cm − 5.006 cm = 0.074 cm.

Suppose the distance between the lens andthe film is fixed and equal to one focal length,an object at 3 m away will have an imagebehind the film. The object at p = 40.0 m willbe, to a good approximation, an imagefocused on the film.

Physics 202 – p. 12/42

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The Camera: Example

Difference in image distance:5.08 cm − 5.006 cm = 0.074 cm.

Suppose the distance between the lens andthe film is fixed and equal to one focal length,an object at 3 m away will have an imagebehind the film. The object at p = 40.0 m willbe, to a good approximation, an imagefocused on the film.

Physics 202 – p. 12/42

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The Camera: f-number

f-number:f − number = f

D

D: Diameter of the lens.

For a fixed f , the intensity of light going intothe camera is found to be I ∝ D2. Largediameter ⇒ large intensity.

f-number: typically 1.4, 2, 2.8, 4, 5.6, .... Fromone f-number to the next higher one, the areaof the aperture is decreased by half.

Physics 202 – p. 13/42

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The Camera: f-number

f-number:f − number = f

D

D: Diameter of the lens.

For a fixed f , the intensity of light going intothe camera is found to be I ∝ D2. Largediameter ⇒ large intensity.

f-number: typically 1.4, 2, 2.8, 4, 5.6, .... Fromone f-number to the next higher one, the areaof the aperture is decreased by half.

Physics 202 – p. 13/42

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The Camera: f-number

f-number:f − number = f

D

D: Diameter of the lens.

For a fixed f , the intensity of light going intothe camera is found to be I ∝ D2. Largediameter ⇒ large intensity.

f-number: typically 1.4, 2, 2.8, 4, 5.6, .... Fromone f-number to the next higher one, the areaof the aperture is decreased by half.

Physics 202 – p. 13/42

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Lenses in combinationA series of two or more lenses can be veryuseful for a number of circumstances: eyesightcorrections, microscopes, telescopes, etc....

Physics 202 – p. 14/42

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Lenses in combination: ExampleOne simple example: Two converging lensesTwo converging lenses, separated by a distanceof 40.0 cm are used in combination. The focallengths are f1 = +10.0 cm and f2 = +12.0 cm. Anobject, 4.00 cm high, is placed 15.0 cm in front ofthe first lens. (see the picture in class.) Find theintermediate and final image distances, the totaltransverse magnification, and the height of thefinal image.

Physics 202 – p. 15/42

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Lenses in combination: Solution to example

Use the thin lens equation.

Let s = 40.0 cm be the distance between thetwo lenses then:1p1

+ 1q1

= 1f1

1p2

+ 1q2

= 1f2

p2 = s − q1.

p1 = 15.0 cm.⇒ q1 = +30.0 cm⇒ p2 = 40.0 cm − 30.0 cm = 10.0 cm ⇒

q2 = −60.0 cm

Physics 202 – p. 16/42

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Lenses in combination: Solution to example

Use the thin lens equation.

Let s = 40.0 cm be the distance between thetwo lenses then:1p1

+ 1q1

= 1f1

1p2

+ 1q2

= 1f2

p2 = s − q1.

p1 = 15.0 cm.⇒ q1 = +30.0 cm⇒ p2 = 40.0 cm − 30.0 cm = 10.0 cm ⇒

q2 = −60.0 cm

Physics 202 – p. 16/42

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Lenses in combination: Solution to example

Use the thin lens equation.

Let s = 40.0 cm be the distance between thetwo lenses then:1p1

+ 1q1

= 1f1

1p2

+ 1q2

= 1f2

p2 = s − q1.

p1 = 15.0 cm.⇒ q1 = +30.0 cm⇒ p2 = 40.0 cm − 30.0 cm = 10.0 cm ⇒

q2 = −60.0 cmPhysics 202 – p. 16/42

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Lenses in combination: Solution to example

Image 1: right of lens 1.Image 2: left of lens 2 and since the twolenses are separated by 40.0 cm, it is 20.0 cmto the left of lens 1.

Magnification for a single lens:M = −

q

p.

The total magnification for two lenses is:M = M1 × M2 = (− q1

p1

) × (− q2

p2

) =

(−30.0 cm15.0 cm

) × (−−60.0 cm10.0 cm

) = −12.0.The final image is inverted and twelve timeslarger!

Physics 202 – p. 17/42

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Lenses in combination: Solution to example

Image 1: right of lens 1.Image 2: left of lens 2 and since the twolenses are separated by 40.0 cm, it is 20.0 cmto the left of lens 1.

Magnification for a single lens:M = −

q

p.

The total magnification for two lenses is:M = M1 × M2 = (− q1

p1

) × (− q2

p2

) =

(−30.0 cm15.0 cm

) × (−−60.0 cm10.0 cm

) = −12.0.The final image is inverted and twelve timeslarger! Physics 202 – p. 17/42

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Lenses in combination: Another example

Physics 202 – p. 18/42

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Correcting eyesights

Myopia (Nearsightedness): Image of distantobject is formed in front of retina ⇒ blurredvision. The focal length for this case is tooshort.How do we correct it? Use a diverging lens.

Light rays diverge when incident on thediverging lens in front of the eye. Thediverging rays get refocused on the retina ifthe proper focal lengths are chosen. Thediverging lens forms a virtual image at the farpoint of a nearsighted eye.

Physics 202 – p. 19/42

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Correcting eyesights

Myopia (Nearsightedness): Image of distantobject is formed in front of retina ⇒ blurredvision. The focal length for this case is tooshort.How do we correct it? Use a diverging lens.

Light rays diverge when incident on thediverging lens in front of the eye. Thediverging rays get refocused on the retina ifthe proper focal lengths are chosen. Thediverging lens forms a virtual image at the farpoint of a nearsighted eye.

Physics 202 – p. 19/42

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Nearsightedness

Physics 202 – p. 20/42

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Correcting Nearsightedness

Physics 202 – p. 21/42

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Nearsightedness: ExampleA nearsighted person has a far point located only521 cm from the eye.Assuming that eye-glassesare to be worn 2 cm in front of the eye, find thefocal length needed for the diverging lenses ofthe glasses so the person can see distantobjects.

Physics 202 – p. 22/42

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Nearsightedness: Example

Virtual image at the far point to the left of thelens:q = −519 cm because the glasses are 2 cm infront of the eye.Here, p = ∞.

1f

= 1p+ 1

q= 1

∞+ 1

−519 cm⇒ f = −519 cm < 0. It

is negative as one would expect for adiverging lens.

Physics 202 – p. 23/42

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Nearsightedness: Example

Virtual image at the far point to the left of thelens:q = −519 cm because the glasses are 2 cm infront of the eye.Here, p = ∞.1f

= 1p+ 1

q= 1

∞+ 1

−519 cm⇒ f = −519 cm < 0. It

is negative as one would expect for adiverging lens.

Physics 202 – p. 23/42

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Farsightedness

Hyperopia (farsightedness): Image of nearbyobject is formed behind the retina.By nearby, one means an object which islocated between the eye and the near point.

Put a converging lens in front of the eye tobring the image back to the retina. Thisconverging lens acts in such a way as tocreate a virtual image at the near point. Seepicture.

Physics 202 – p. 24/42

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Farsightedness

Hyperopia (farsightedness): Image of nearbyobject is formed behind the retina.By nearby, one means an object which islocated between the eye and the near point.

Put a converging lens in front of the eye tobring the image back to the retina. Thisconverging lens acts in such a way as tocreate a virtual image at the near point. Seepicture.

Physics 202 – p. 24/42

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Farsightedness: ExampleA farsighted person has a near point located at210 cm from the eye. Obtain the focal length of aconverging contact lens that can be used to reada book held at 25.0 cm from the eye.

Contact lens ⇒ p = 25.0 cm

Virtual image at near point ⇒ q = −210 cm1f

= 1p+ 1

q= 1

25.0 cm+ 1

−210 cm⇒ f = 28.4 cm > 0

⇒ Converging lens

Physics 202 – p. 25/42

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Farsightedness: ExampleA farsighted person has a near point located at210 cm from the eye. Obtain the focal length of aconverging contact lens that can be used to reada book held at 25.0 cm from the eye.

Contact lens ⇒ p = 25.0 cm

Virtual image at near point ⇒ q = −210 cm1f

= 1p+ 1

q= 1

25.0 cm+ 1

−210 cm⇒ f = 28.4 cm > 0

⇒ Converging lens

Physics 202 – p. 25/42

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Farsightedness

Physics 202 – p. 26/42

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Correcting Farsightedness

Physics 202 – p. 27/42

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Refractive power

Refractive power of the lens:Refractive power = 1

f(meters) .Unit: Diopter

For the nearsightedness example, refractivepower = 1/ − 5.19 m = −0.193 diopter.

For the farsightedness example, refractivepower = 1/0.284 m = 3.52 diopter.

Physics 202 – p. 28/42

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Refractive power

Refractive power of the lens:Refractive power = 1

f(meters) .Unit: Diopter

For the nearsightedness example, refractivepower = 1/ − 5.19 m = −0.193 diopter.

For the farsightedness example, refractivepower = 1/0.284 m = 3.52 diopter.

Physics 202 – p. 28/42

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Refractive power

Refractive power of the lens:Refractive power = 1

f(meters) .Unit: Diopter

For the nearsightedness example, refractivepower = 1/ − 5.19 m = −0.193 diopter.

For the farsightedness example, refractivepower = 1/0.284 m = 3.52 diopter.

Physics 202 – p. 28/42

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Magnifying Glass

The eye sees blurry objects if they are locatedat distances less than the near point N. At N,the angle subtended is θ ≈ h0/N .

Put a convex lens of focal length f < N infront of the eye. As the object approaches thefocal point, the image which is upright andvirtual approaches infinity. This can befocused on. At F , the angle subtended isθ

′

≈ h0/f > θ.

Angular magnification: M = θ′

θ= N

f

Physics 202 – p. 29/42

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Magnifying Glass

The eye sees blurry objects if they are locatedat distances less than the near point N. At N,the angle subtended is θ ≈ h0/N .

Put a convex lens of focal length f < N infront of the eye. As the object approaches thefocal point, the image which is upright andvirtual approaches infinity. This can befocused on. At F , the angle subtended isθ

′

≈ h0/f > θ.

Angular magnification: M = θ′

θ= N

f

Physics 202 – p. 29/42

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Magnifying Glass

The eye sees blurry objects if they are locatedat distances less than the near point N. At N,the angle subtended is θ ≈ h0/N .

Put a convex lens of focal length f < N infront of the eye. As the object approaches thefocal point, the image which is upright andvirtual approaches infinity. This can befocused on. At F , the angle subtended isθ

′

≈ h0/f > θ.

Angular magnification: M = θ′

θ= N

f Physics 202 – p. 29/42

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Magnifying Glass

Physics 202 – p. 30/42

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Magnifying Glass

Physics 202 – p. 31/42

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Compound MicroscopeTwo lenses: Objective and eyepiece.Magnification:M = −

qN

fobjectivefeyepiece

Physics 202 – p. 32/42

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Compound Microscope

Physics 202 – p. 33/42

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Compound Microscope

Physics 202 – p. 34/42

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TelescopeTelescope: Objective and eyepieceMagnification:

M = θ′

θ=

fobjective

feyepiece

Physics 202 – p. 35/42

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Telescope

Physics 202 – p. 36/42

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Newtonian Telescope

Physics 202 – p. 37/42

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Telescope: ExampleA telescope has the following specifications:fo = 985 mm and fe = 5.00 mm. Find (a) theangular magnification, and (b) the approximatelength of the telescope.

M = fo

fe= 985 mm

5.00 mm= 197

L ≈ fo + fe = 990 mm

Physics 202 – p. 38/42

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Telescope: ExampleA telescope has the following specifications:fo = 985 mm and fe = 5.00 mm. Find (a) theangular magnification, and (b) the approximatelength of the telescope.

M = fo

fe= 985 mm

5.00 mm= 197

L ≈ fo + fe = 990 mm

Physics 202 – p. 38/42

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Lens Aberrations (Spherical)

Physics 202 – p. 39/42

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Lens Aberrations (Chromatic)

Physics 202 – p. 40/42

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Lens Aberrations (Chromatic)

Physics 202 – p. 41/42

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Lens Aberrations: Chromatic correction

Physics 202 – p. 42/42