1. an image of a distant object formed by a single converging lens ____. a. can be focused on a...
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1. An image of a distant object formed by a single converging lens ____.
• A. can be focused on a screen.
• B. can be projected on a wall.
• C. is upside down.
• D. is real.
• E. all of the above.
1. An image of a distant object formed by a single converging lens ____.
• A. can be focused on a screen.
• B. can be projected on a wall.
• C. is upside down.
• D. is real.
• E. all of the above.
2. An image formed by a single diverging lens ____.
• A. is upside down.
• B. is larger than the object.
• C. can be projected on a wall.
• D. is virtual.
• E. all of the above.
2. An image formed by a single diverging lens ____.
• A. is upside down.
• B. is larger than the object.
• C. can be projected on a wall.
• D. is virtual.
• E. all of the above.
3. A magnifying glass is usually a ____.
• A. converging lens.
• B. combination of diverging and converging lenses.
• C. diverging lens.
3. A magnifying glass is usually a ____.
• A. converging lens.
• B. combination of diverging and converging lenses.
• C. diverging lens.
4. Which instrument is a human eye most similar to?
• A. camera
• B. microscope
• C. telescope
• D. slide projector
4. Which instrument is a human eye most similar to?
• A. camera
• B. microscope
• C. telescope
• D. slide projector
1. An object is placed 6.0 cm in front of a converging lens of focal length 4.0 cm. Draw a ray diagram to show the location and orientation of the image formed. What is its type and orientation?
• First draw a line parallel to the principle axis which refracts through the focal point.
• First draw a line parallel to the principle axis which refracts through the focal point.
• Then draw a line through the optical center of the lens. Where they cross is the image.
• This image is real and inverted (case 4). We use the equations to find the actual distance and size of the image.
2. The focal length of a double convex lens is 4.0 cm. An object 2.0 cm high is 10.0 cm from the lens. Draw a ray diagram of this situation. Calculate the location and size of the image. What is its type and orientation?
• First draw a line parallel to the principle axis which refracts through the focal point.
• First draw a line parallel to the principle axis which refracts through the focal point.
• Then draw a line through the optical center of the lens. Where they cross is the image.
• This image is real and inverted . We use the equations to find the actual distance and size of the image.
1 1 1 ------ = ------ + ------ f do di
1 1 1 ------ = ------ + ------ 4 10 di
di = 6.67 cm
hi di ------- = ------- ho do
hi 6.67 cm ------- = ----------- 2 cm 10 cm
hi = 1.3 cm, mag = 0.667
3. A double convex lens has a focal length of 5.0 cm. An object 0.75 cm high is 3.0 cm from the lens. Draw a ray diagram of this situation. Calculate the location and size of the image. What is its type and orientation?
• First draw a line parallel to the principle axis which refracts through the focal point.
• First draw a line parallel to the principle axis which refracts through the focal point.
• Then draw a line through the optical center of the lens. Where they cross is the image.
• Image is virtual, upright, and larger.
• Case 6, a magnifying glass.
1 1 1 ------ = ------ + ------ f do di
1 1 1 ------ = ------ + ------ 5 3 di
di = -7.5 cm
hi di
------- = ------- ho do
hi -7.5 cm --------- = ----------- 0.75 cm cm
hi = 1.88 cm, mag = -2.5
4. A concave lens has a focal length of 6.0 cm. An object is placed 7.0 cm from the lens. Draw a ray diagram of this situation. Calculate the location and size of the image. What is its type and orientation? What is the magnification of this lens?
• First draw a line parallel to the principle axis which refracts through the focal point.
• First draw a line parallel to the principle axis which refracts through the focal point.
• Then draw a line through the optical center of the lens. Where they cross is the image.
• This image is virtual, upright, and smaller. We use the equations to find the actual distance and size of the image.
1 1 1 ------ = ------ + ------ f do di
1 1 1 ------ = ------ + ------ -6 7 di
di = -3.2 cm
hi di
------- = ------- ho do
hi -3.2 cm --------- = ----------- ho 7 cm
mag = 0.46
5. What type of lenses are used to correct for farsightedness and nearsightedness?
• Farsightedness is corrected using a converging lens (convex).
• Nearsightedness is corrected using a diverging lens (concave).