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

AI Thread Summary
Sunlight converges at the focal point of a concave mirror, which is 3 cm from the mirror. An object 12 cm away from the mirror produces an image that can be calculated using the mirror equation, yielding a distance of 4 cm for the image. A ray diagram confirms that the image is inverted and located between the focal point and the object. The discussion emphasizes the importance of using a scale for accurate ray diagrams. Overall, the calculations and ray diagram support the conclusion that the image is formed 4 cm from the mirror.
imatreyu
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Homework Statement



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?

Homework Equations



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

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 don't understand how I can use a ray diagram. . .

I'm so confused; thank you in advance!
 
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imatreyu said:
If so, is it possible to find this value simply by drawing a not-to-scale diagram?
I don't understand how I can use a ray diagram. . .

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. :-p

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).
 
Oh, okay, thank you! I'll do that now. . .
 
Well, I did it. . .and it seems to verify my answer as 4 cm. . .

I hope this is right!
 
imatreyu said:
Well, I did it. . .and it seems to verify my answer as 4 cm. . .

I hope this is right!

Seems okay to me! :approve:
 
Yess! Thank you! :D
 
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