Physics 6C. Cameras and the Human Eye. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

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1 Physics 6C Cameras and the Human Eye

CAMERAS A typical camera uses a converging lens to ocus a real (inverted) image onto photographic ilm (or in a digital camera the image is on a CCD chip).

2 CAMERAS A typical camera uses a converging lens to ocus a real (inverted) image onto photographic ilm (or in a digital camera the image is on a CCD chip). Light goes through the aperture, and the EXPOSURE is jointly proportional to the Intensity o the light, the Area o the aperture, and the Time it is open. Thus or a given object, i the diameter o the opening is doubled, the time should be ¼ as long to give the same exposure. You need this idea or the homework.

3 Example: Problem 25.2 Camera A has a lens with an aperture diameter o 8.0mm. It photographs an object using the correct exposure time o /30 seconds. What exposure time should be used with camera B, which has a lens with an aperture diameter o 23.mm?

4 Example: Problem 25.2 Camera A has a lens with an aperture diameter o 8.0mm. It photographs an object using the correct exposure time o /30 seconds. What exposure time should be used with camera B, which has a lens with an aperture diameter o 23.mm? We want the total exposure (amount o light energy reaching the ilm) to be the same in both cases. Since exposure is proportional to time and aperture AREA, we need a smaller amount o time or the larger lens.

5 Example: Problem 25.2 Camera A has a lens with an aperture diameter o 8.0mm. It photographs an object using the correct exposure time o /30 seconds. What exposure time should be used with camera B, which has a lens with an aperture diameter o 23.mm? We want the total exposure (amount o light energy reaching the ilm) to be the same in both cases. Since exposure is proportional to time and aperture AREA, we need a smaller amount o time or the larger lens. We can set up a ormula: ( area) A (time) A = (area) B (time ) B

6 Example: Problem 25.2 Camera A has a lens with an aperture diameter o 8.0mm. It photographs an object using the correct exposure time o /30 seconds. What exposure time should be used with camera B, which has a lens with an aperture diameter o 23.mm? We want the total exposure (amount o light energy reaching the ilm) to be the same in both cases. Since exposure is proportional to time and aperture AREA, we need a smaller amount o time or the larger lens. We can set up a ormula: (area) A (time) A = (area) B (time) B (area) (time) A B = (time) A (area) B 2 (diam ) (time) A B = (time) 2 A (diamb) 2 (8.0mm) (time) ( B = sec) 2 30 (23.mm) (time) B = sec Here I have spelled out all the details o the proportionality. You can save several steps i you understand that area is proportional to the square o the diameter.

7 The Human Eye Ideally, the lens o the eye ocuses the light rays rom an object on the retina at the back o the eyeball. The lens is somewhat lexible, and its ocal length can be adjusted by contracting or relaxing the muscles around the lens. This is called accomodation. The two most common vision problems involve the curvature o the eye s lens. When the lens cannot adjust enough, the light will ocus in ront o or behind the retina, creating a blurry image.

The Human Eye Myopia (nearsightedness) Even when ully relaxed, the lens is too curved (i.e. the light is bent too much). I the object is ar away, the image ormed by the lens is in ront o the retina.

8 The Human Eye Myopia (nearsightedness) Even when ully relaxed, the lens is too curved (i.e. the light is bent too much). I the object is ar away, the image ormed by the lens is in ront o the retina. The FAR POINT is the arthest distance that can be seen clearly without correction. To correct, the light is intercepted by a diverging lens beore it gets to the eye. The diagram shows a distant object and a myopic eye. The diverging lens orms a virtual image at the ar point. It is this virtual image that the eye sees, and an image is ormed on the retina, where it can be properly interpreted by the brain.

9 The Human Eye Hyperopia (arsightedness) The lens will not curve enough to ocus on very close objects. I the object is too close, the image ormed by the lens is behind the retina. The NEAR POINT is the closest distance that can be seen clearly without correction. To correct, the light is intercepted by a converging lens beore it gets to the eye. Everybody has a near point try to ind yours: Focus on your inger, and move it closer and closer until you just barely can t see your ingerprint clearly. This is your near point. The diagram shows a very close object and a hyperopic eye. The converging lens orms a virtual image at the near point. It is this virtual image that the eye sees, and an image is ormed on the retina, where it can be properly interpreted by the brain.

10 We need to deine one term lens POWER. The power o a lens (units are called DIOPTERS) is just the reciprocal o the ocal length, in meters. Thus a lens with a power o +0.5 diopters will be a converging lens with ocal length 2 meters. I you need a ormula, here it is: Lens = Power Make sure the ocal length is in meters.

11 Here s an example: Doctor Bob needs glasses. He is both myopic and hyperopic, so he will require biocals. His range o clear vision is rom 35cm to 85cm. Anything outside that range seems a bit blurry to him. What prescription(s) will the optometrist give him? Assume that Bob would like to be able to ocus on objects as close as 20 cm, and as ar away as the weather will allow (objects at ininity).

12 Here s an example: Doctor Bob needs glasses. He is both myopic and hyperopic, so he will require biocals. His range o clear vision is rom 35cm to 85cm. Anything outside that range seems a bit blurry to him. What prescription(s) will the optometrist give him? Assume that Bob would like to be able to ocus on objects as close as 20 cm, and as ar away as the weather will allow (objects at ininity). First let s deal with the myopia. The problem is that anything arther than 85cm looks blurry. So ideally Bob would like to look at something that is very ar away (object distance = ) and his corrective lenses will orm an image at 85cm so that his eye can ocus on it.

13 Here s an example: Doctor Bob needs glasses. He is both myopic and hyperopic, so he will require biocals. His range o clear vision is rom 35cm to 85cm. Anything outside that range seems a bit blurry to him. What prescription(s) will the optometrist give him? Assume that Bob would like to be able to ocus on objects as close as 20 cm, and as ar away as the weather will allow (objects at ininity). First let s deal with the myopia. The problem is that anything arther than 85cm looks blurry. So ideally Bob would like to look at something that is very ar away (object distance = ) and his corrective lenses will orm an image at 85cm so that his eye can ocus on it. We can use our standard ormula or this: = S + S'

14 Here s an example: Doctor Bob needs glasses. He is both myopic and hyperopic, so he will require biocals. His range o clear vision is rom 35cm to 85cm. Anything outside that range seems a bit blurry to him. What prescription(s) will the optometrist give him? Assume that Bob would like to be able to ocus on objects as close as 20 cm, and as ar away as the weather will allow (objects at ininity). First let s deal with the myopia. The problem is that anything arther than 85cm looks blurry. So ideally Bob would like to look at something that is very ar away (object distance = ) and his corrective lenses will orm an image at 85cm so that his eye can ocus on it. We can use our standard ormula or this: = S Power + S' Notice how the image at ininity gives a ocal length equal to the near point distance. = m Also note the image distance it is negative. = 0.85m =.8 Diopters It is a virtual image it will always be negative.

15 Here s an example: Doctor Bob needs glasses. He is both myopic and hyperopic, so he will require biocals. His range o clear vision is rom 35cm to 85cm. Anything outside that range seems a bit blurry to him. What prescription(s) will the optometrist give him? Assume that Bob would like to be able to ocus on objects as close as 20 cm, and as ar away as the weather will allow (objects at ininity). First let s deal with the myopia. The problem is that anything arther than 85cm looks blurry. So ideally Bob would like to look at something that is very ar away (object distance = ) and his corrective lenses will orm an image at 85cm so that his eye can ocus on it. We can use our standard ormula or this: = S Power + S' Notice how the image at ininity gives a ocal length equal to the near point distance. = m Also note the image distance it is negative. = 0.85m =.8 Diopters It is a virtual image it will always be negative. Now or the hyperopia. Anything closer than 35cm looks blurry. He should be able to have an object at 20cm, and his glasses will orm an image at his near point. Here is the calculation:

16 Here s an example: Doctor Bob needs glasses. He is both myopic and hyperopic, so he will require biocals. His range o clear vision is rom 35cm to 85cm. Anything outside that range seems a bit blurry to him. What prescription(s) will the optometrist give him? Assume that Bob would like to be able to ocus on objects as close as 20 cm, and as ar away as the weather will allow (objects at ininity). First let s deal with the myopia. The problem is that anything arther than 85cm looks blurry. So ideally Bob would like to look at something that is very ar away (object distance = ) and his corrective lenses will orm an image at 85cm so that his eye can ocus on it. We can use our standard ormula or this: = S Power + S' Notice how the image at ininity gives a ocal length equal to the near point distance. = m Also note the image distance it is negative. = 0.85m =.8 Diopters It is a virtual image it will always be negative. Now or the hyperopia. Anything closer than 35cm looks blurry. He should be able to have an object at 20cm, and his glasses will orm an image at his near point. Here is the calculation: = + = m 0.20m 0.35m Power = Diopters Again note the image distance it is negative. It is a virtual image it will always be negative.

Physics 6C. Cameras and the Human Eye. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB

Physics 6C. Cameras and the Human Eye. Prepared by Vince Zaccone For Campus Learning Assistance Services at UCSB Physics 6C Cameras and the Human Eye CAMERAS A typical camera uses a converging lens to focus a real (inverted) image onto photographic film (or in a digital camera the image is on a CCD chip). Light goes