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Double A
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The Following two questions are giving me the worst time. Maybe someone else can lead me in the right direction.
Question 1: You unconsciously estimate the distance to an object from the angle it subtends in your field of view. This angle [tex]\theta[/tex] in radians is related to the linear height of the object h and to the distance d by [tex]\theta[/tex] = h/d. Assume that you are driving a car and another car, 1.50 m high, is 24.0 m behind you. (a) Suppose your car has a flat passenger-side rearview mirror, 1.55 m from your eyes. How far from your eyes is the image of the car following you? (b) What angle does the image subtend in your field of view? (c) What if? Suppose instead that your car has a convex rearview mirror with a radius of curvature of magnitude 2.00 m. How far from your eyes is the image of the car behind you? (d) What angle does the image subtend at your eyes? (e) Based on its angular size, how far away does the following car appear to be?
My Solution so far:
Given: [tex]\theta = \frac{h}{d}[/tex] h=1.50 m, p=24.0 m, q=1.55 m
(a) d=p+q = 24.0 m + 1.55 m = 25.6 m
(b) [tex]\theta = \frac{h}{d} = \frac{1.50 m}{25.6 m} = 0.0586 rad[/tex]
(c) R = 2.00 m
[tex]f = \frac{R}{2} = \frac{2.00 m}{2} = 1.00 m[/tex]
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q'}[/tex]
[tex]\frac{1}{1.00 m} = \frac{1}{24.0 m} + \frac{1}{q'}[/tex]
q' = 1.04 m
From this I find that it should be about 2.59 m from the eyes by adding q' with q. However, the real answer should be 2.51 m. This is where I get stuck and wrong answers through the rest of the problem.
Question 2: An object is at a distance d to the left of a flat screen. A converging lens with focal length f < d/4 is placed between object and screen. (a) Show that two lens positions exist that form an image on the screen, and determine how far these positions are from the object. (b) How do the two images differ from each other?
My Solution: I'm having a hard time starting this problem. This maybe from that I find it confusion when it says that their are two positions where an image can form on the screen. I looked through my textbook of the corresponding section but found nothing of help.
Hope someone can lead me in the correct direction. Any input and help is much appreciated.
Question 1: You unconsciously estimate the distance to an object from the angle it subtends in your field of view. This angle [tex]\theta[/tex] in radians is related to the linear height of the object h and to the distance d by [tex]\theta[/tex] = h/d. Assume that you are driving a car and another car, 1.50 m high, is 24.0 m behind you. (a) Suppose your car has a flat passenger-side rearview mirror, 1.55 m from your eyes. How far from your eyes is the image of the car following you? (b) What angle does the image subtend in your field of view? (c) What if? Suppose instead that your car has a convex rearview mirror with a radius of curvature of magnitude 2.00 m. How far from your eyes is the image of the car behind you? (d) What angle does the image subtend at your eyes? (e) Based on its angular size, how far away does the following car appear to be?
My Solution so far:
Given: [tex]\theta = \frac{h}{d}[/tex] h=1.50 m, p=24.0 m, q=1.55 m
(a) d=p+q = 24.0 m + 1.55 m = 25.6 m
(b) [tex]\theta = \frac{h}{d} = \frac{1.50 m}{25.6 m} = 0.0586 rad[/tex]
(c) R = 2.00 m
[tex]f = \frac{R}{2} = \frac{2.00 m}{2} = 1.00 m[/tex]
[tex]\frac{1}{f} = \frac{1}{p} + \frac{1}{q'}[/tex]
[tex]\frac{1}{1.00 m} = \frac{1}{24.0 m} + \frac{1}{q'}[/tex]
q' = 1.04 m
From this I find that it should be about 2.59 m from the eyes by adding q' with q. However, the real answer should be 2.51 m. This is where I get stuck and wrong answers through the rest of the problem.
Question 2: An object is at a distance d to the left of a flat screen. A converging lens with focal length f < d/4 is placed between object and screen. (a) Show that two lens positions exist that form an image on the screen, and determine how far these positions are from the object. (b) How do the two images differ from each other?
My Solution: I'm having a hard time starting this problem. This maybe from that I find it confusion when it says that their are two positions where an image can form on the screen. I looked through my textbook of the corresponding section but found nothing of help.
Hope someone can lead me in the correct direction. Any input and help is much appreciated.
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