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

Click For Summary

Homework Help Overview

The discussion revolves around the formation of an image by a concave mirror when sunlight, which consists of parallel rays, strikes it. The original poster presents a scenario involving an object placed at a specific distance from the mirror and seeks to determine the image's location and height.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the mirror equation and ray diagram techniques to find the image distance and height. They express confusion about using a ray diagram and whether it can yield accurate results without being to scale.

Discussion Status

Participants have provided guidance on the use of ray diagrams, suggesting that keeping the diagram to scale could help verify calculations. There is acknowledgment of the original poster's findings, but no explicit consensus on the correctness of the results has been reached.

Contextual Notes

The original poster mentions specific measurements and the need to understand the relationship between object distance, image distance, and focal length in the context of concave mirrors. There is an emphasis on the importance of visual representation through ray diagrams.

imatreyu
Messages
79
Reaction score
0

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!
 
Last edited:
Physics news on Phys.org
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
 

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
19
Views
1K
  • · Replies 5 ·
Replies
5
Views
937
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
7
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
13
Views
4K
  • · Replies 5 ·
Replies
5
Views
2K