high school part 2 /2 by ssl technologies physics ex-53 click the eye the eye is an optical...
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High School
Part 2 /2
by SSL Technologies
Physics Ex-53
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THE EYE
The eye is an optical “instrument”. It contains a converging lensused to focus images on the “retina” (a kind of screen at theback of the eye). Images on the retina are inverted.
PART-2 /2
Physics Ex-53
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FARSIGHTEDNESS The ability to see objects clearly which arefar but objects close to the eye appear blurred.
Farsightedness, known as hyperopia, is caused by the fact that raysof light focus behind the retina of the eye. Hyperopia is correctedby using a converging lens.
DEFECTS IN VISION
NEARSIGHTEDNESS The ability to see objects clearly which arenear but objects far away appear blurred.
Nearsightedness, known as myopia, is caused by the fact that raysof light focus in front of the retina of the eye. Myopia is correctedby using a diverging lens as illustrated in the diagram below.
PART-2
Physics Ex-53
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PART-2
CORRECTING DEFECTS IN VISION
Physics Ex-53
Reminder
In applying the lens equation, be sure to use the signs carefully :
do is the object distance and is always positive
f is the focal length and is positive for convex lenses but negative for concave lenses
di is the image distance and is positive for real images but negative for virtual images
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Physics Ex-53Question-1
A lens forms an image 10 cm high. If the object is 4 cm in heightand situated 8 cm away, what is the focal length of the lens?
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Note that while the standard unit for length in a physics formulais the meter, since in this case we have lengths on both sidesof the equation, we need not convert to meters as the units
automatically cancel out.
Remember: A negative di indicates a virtual image.
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Physics Ex-53Question-2
The focal length of a concave lens is 10 cm. An object, whoseheight is 2 cm, is placed 15 cm in front of the lens. Determinethe characteristics of the image.
Type (real or virtual): _______________
Location: _______________
Magnification: _______________
Height: _______________
Attitude (upright/inverted): _______________
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Virtual
6 cm
0.4
0.8 cm
Upright
Negative image distance means a virtual image.
When the magnification isless than one, the image
is smaller than the object.
Physics Ex-53Question-3
An object whose height is 4 cm is placed 6 cm in front of aconverging lens. If the focal length of the lens is 8 cm,determine the characteristics of the image.
Type (real or virtual): _______________
Location: _______________
Magnification: _______________
Height: _______________
Attitude (upright/inverted): _______________
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Virtual
24 cm
4
16 cm
Upright
When the magnification isgreater than one, the imageis greater than the object.
Physics Ex-53
The focal length of a camera is 10 cm. The lens forms an imagethat is 4 cm high when the negative (film) is 12 cm from the lens.
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a) What is the object distance?
Physics Ex-53Question-4
The focal length of a camera is 10 cm. The lens forms an imagethat is 4 cm high when the negative (film) is 12 cm from the lens.
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b) What is the object height?
The negative signindicates inversion.
do = 60 cm (previously calculated)
Physics Ex-53
The focal length of a camera is 10 cm. The lens forms an imagethat is 4 cm high when the negative (film) is 12 cm from the lens.
Question-4
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c) What is the magnification factor?
The negative signindicates inversion.
ho = 20 cm (previously calculated)
Physics Ex-53Question-5
A 35 mm slide (the object) is placed 8.2 cm from a projectionlens whose focal length is 8 cm. Determine:
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a) The image distance.
Physics Ex-53Question-5
A 35-mm slide (the object) is placed 8.2 cm from a projectionlens whose focal length is 8 cm. Determine:
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b) The image height.
The negative signindicates inversion.
Physics Ex-53Question-5
A 35-mm slide (the object) is placed 8.2 cm from a projectionlens whose focal length is 8 cm. Determine:
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c) The magnification factor.
The negative signindicates inversion.
Physics Ex-53Question-6
The focal length of a magnifying glass is 10 cm. The lens is usedto view a stamp that is 2.0 cm in height. If the stamp is placed6.0 cm away from the magnifying glass, calculate the height ofthe image.
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Physics Ex-53Question-7
An object is 4 cm from a concave lens whose focal lengthis 12 cm. Where will the image be located?
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Reminder : concave lenses have a negative focal length.
The negative sign indicates a virtual image.
Physics Ex-53
An object 3 cm high is located 30 cm from a concave lens whosefocal length is 15 cm. Determine:
Question-8
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a) The image distance.
The negative sign indicates a virtual image.
Physics Ex-53Question-8
An object 3 cm high is located 30 cm from a concave lens whosefocal length is 15 cm. Determine:
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b) The magnification.
Physics Ex-53Question-8
An object 3 cm high is located 30 cm from a concave lens whosefocal length is 15 cm. Determine:
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c) The size of the image.
d) The type of image: ___________________
e) The attitude of the image: ___________________
Virtual
Upright
Physics Ex-53Question-9
Define each of the following terms :
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a) Myopia:
The problem of not seeing far objects clearly.
b) Hyperopia:
The problem of not seeing near objects clearly.
Physics Ex-53Question-10
Illustrated below are two common eye problems or defects.State the name of each defect and draw the appropriate lensin order to correct the problem.
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Myopia
Hyperopia
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