Transcript Ex. 53 PowerPoint
High School
by SSL Technologies
Part 2 /2
PART-2 /2
Physics Ex-53 THE EYE The eye is an optical “instrument”. It contains a converging lens used to focus images on the “retina” (a kind of screen at the back of the eye). Images on the retina are inverted.
Physics Ex-53
PART-2
DEFECTS IN VISION NEARSIGHTEDNESS The ability to see objects clearly which are near but objects far away appear blurred.
Nearsightedness, known as of light focus in front myopia , is caused by the fact that rays of the retina of the eye. Myopia is corrected by using a diverging lens as illustrated in the diagram below.
FARSIGHTEDNESS The ability to see objects clearly which are far but objects close to the eye appear blurred.
Farsightedness, known as of light focus behind the retina of the eye. Hyperopia is corrected by using a converging lens.
hyperopia , is caused by the fact that rays
CORRECTING DEFECTS IN VISION PART-2
Physics Ex-53
PART-2
Physics Ex-53 Reminder In applying the lens equation, be sure to use the signs carefully : d o is the object distance and is always positive f is the focal length and is positive for convex lenses but negative for concave lenses d i is the image distance and is positive for real images but negative for virtual images
Question-1
Physics Ex-53 A lens forms an image 10 cm high. If the object is 4 cm in height and situated 8 cm away, what is the focal length of the lens? Remember: A negative d i indicates a virtual image.
8 Note that while the standard unit for length in a physics formula is the meter, since in this case we have lengths on both sides of the equation, we need not convert to meters as the units automatically cancel out.
Question-2
Physics Ex-53 The focal length of a concave lens is 10 cm. An object, whose height is 2 cm, is placed 15 cm in front of the lens. Determine the characteristics of the image. Type (real or virtual) Virtual Location: When the magnification is less than one, the image is smaller than the object. Height: Attitude (upright/inverted): Negative image distance means a virtual image.
Question-3
Physics Ex-53 An object whose height is 4 cm is placed 6 cm in front of a converging lens. If the focal length of the lens is 8 cm, determine the characteristics of the image. Type (real or virtual) Virtual Location: 24 cm Magnification: _______________ Height: 4 16 cm When the magnification is greater than one, the image is greater than the object. Attitude (upright/inverted):
Question-4
Physics Ex-53 The focal length of a camera is 10 cm. The lens forms an image that is 4 cm high when the negative (film) is 12 cm from the lens. a) What is the object distance?
Question-4
Physics Ex-53 The focal length of a camera is 10 cm. The lens forms an image that is 4 cm high when the negative (film) is 12 cm from the lens. b) What is the object height? d o = 60 cm (previously calculated)
The negative sign indicates inversion.
Question-4
Physics Ex-53 The focal length of a camera is 10 cm. The lens forms an image that is 4 cm high when the negative (film) is 12 cm from the lens. h o = 20 cm (previously calculated) c) What is the magnification factor?
The negative sign indicates inversion.
Question-5
Physics Ex-53 A 35 mm slide (the object) is placed 8.2 cm from a projection lens whose focal length is 8 cm. Determine: a) The image distance.
Question-5
Physics Ex-53 A 35-mm slide (the object) is placed 8.2 cm from a projection lens whose focal length is 8 cm. Determine: b) The image height.
The negative sign indicates inversion.
Question-5
Physics Ex-53 A 35-mm slide (the object) is placed 8.2 cm from a projection lens whose focal length is 8 cm. Determine: c) The magnification factor.
The negative sign indicates inversion.
Question-6
Physics Ex-53 The focal length of a magnifying glass is 10 cm. The lens is used to view a stamp that is 2.0 cm in height. If the stamp is placed 6.0 cm away from the magnifying glass, calculate the height of the image.
Question-7
Reminder : concave lenses have a negative focal length.
Physics Ex-53 An object is 4 cm from a concave lens whose focal length is 12 cm. Where will the image be located?
The negative sign indicates a virtual image.
Question-8
Physics Ex-53 An object 3 cm high is located 30 cm from a concave lens whose focal length is 15 cm. Determine: a) The image distance.
The negative sign indicates a virtual image.
Question-8
Physics Ex-53 An object 3 cm high is located 30 cm from a concave lens whose focal length is 15 cm. Determine: b) The magnification.
Question-8
Physics Ex-53 An object 3 cm high is located 30 cm from a concave lens whose focal length is 15 cm. Determine: c) The size of the image.
Question-9
Define each of the following terms : a) Myopia: The problem of not seeing far objects clearly.
Physics Ex-53 b) Hyperopia: The problem of not seeing near objects clearly.
Question-10
Physics Ex-53 Illustrated below are two common eye problems or State the name of each defect and draw the appropriate lens in order to correct the problem.
defects .
Myopia Hyperopia