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Sunlight falls on a concave mirror, where is the image formed?

  1. Apr 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Sunlight falls on a concave mirror and forms an image 3.0 cm from the mirror. If an object 24 mm high is placed 12.0 cm from the mirror, where will its image be formed?

    a. Use a ray diagram
    b. Use the lens/ mirror equation
    c. How high is the image?

    2. Relevant equations

    (1/f)= (1/di) + (1/do)

    3. The attempt at a solution

    Knowing that in the case of the sun, all rays of sunlight are essentially parallel to the principle axis because of the huge distance between the sun and the earth, I assume that all light of the sum converge at the focal point. Thus, the image's location as formed by the sun marks the focal point of the mirror. Therefore, f= 3cm, di= ?, do=12cm.

    I use the lens/mirror equation:
    (1/f)= (1/di) + (1/do)
    1/3cm = (1/di) + 1/12cm
    1/4cm = 1/di
    di= 4 cm

    Consulting the ray diagram that I drew, I see that the image would be inverted and is between the focal point and the object's location .. .

    ^ Is that correct?
    If so, is it possible to find this value simply by drawing a not-to-scale diagram?
    I dont understand how I can use a ray diagram. . .

    I'm so confused; thank you in advance!!!
    Last edited: Apr 12, 2010
  2. jcsd
  3. Apr 12, 2010 #2


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    Homework Helper
    Gold Member

    If you want to find the actual values using a ray diagram, keep things to scale. This would be a good time for graph paper and a straight edge. And since you have to draw the ray diagram anyway, you can use it double check your answer you found using the equation. :tongue:

    Your coursework might have some examples of ray diagrams to follow. An Internet search on Ray Diagrams for Mirrors might prove fruitful too (for example purposes).
  4. Apr 12, 2010 #3
    Oh, okay, thank you! I'll do that now. . .
  5. Apr 12, 2010 #4
    Well, I did it. . .and it seems to verify my answer as 4 cm. . .

    I hope this is right!
  6. Apr 12, 2010 #5


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    Homework Helper
    Gold Member

    Seems okay to me! :approve:
  7. Apr 12, 2010 #6
    Yess! Thank you! :D
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