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Homework Help: Image Formation HW Questions - Help Needed

  1. May 3, 2005 #1
    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 text book 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.
    Last edited: May 4, 2005
  2. jcsd
  3. May 3, 2005 #2


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    You made an algebra mistake on the first one solving for q' from your last equation.

    For number 2, it is just telling you that p + q is constant and that the sum is d. The focal length is short enough so that the lens can either be closer to the object, or closer to the screen and produce an image. Look at the symmetry of the lens equation. What happens to the sum if the numbers for p and q are swapped?
    Last edited: May 3, 2005
  4. May 4, 2005 #3
    Could you possibly elaborate on where my error in my solution for Question 1 is, because I keep coming up with the same answer for q'. I re-solved the last equation 4 times and still result in the same answer.

    On Question 2: How is it that I go about starting to show that there exists two positions for the image. After looking at the lens equation, I see that by switching p and q results in the same solution.
  5. May 4, 2005 #4


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    I guess "algebra mistake" was excessively vague. Sorry. You need to be careful about the sign convention for distances when applying the formula. The focal length of a convex mirror is a negative number. That will make q' a _________ number.

    http://www.glenbrook.k12.il.us/gbssci/phys/Class/refln/u13l4d.html [Broken]

    For the second question, think about what the focal length of a lens would have to be if the object distance and the image distance were the same (i.e., the lens half way between the two). Then what would happen if you made the focal length longer? Could you still find a combination of distances that would solve the equation, remembering that you cannot change their sum? What if you make the focal length shorter? Could you adjust the distances to form an image still keeping the sum of the distances constant?

    Algebraically, you can substitute for the image distance in the lens equation with its equivalent expression in terms of the sum of the distances and the object distance. Then solve the equation for the object distance in terms of the focal length. What kind of equation do you get? What restrictions are there on the focal length for a solution to exist? If a solution(s) exists, what is it (are they)?
    Last edited by a moderator: May 2, 2017
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