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

Click For Summary
SUMMARY

The discussion centers on the formation of an image by a concave mirror when sunlight, treated as parallel rays, strikes it. The focal length (f) is determined to be 3 cm, and using the mirror equation (1/f) = (1/di) + (1/do), the image distance (di) is calculated as 4 cm. The object height is 24 mm, and the image is confirmed to be inverted and located between the focal point and the object. The ray diagram drawn by the user corroborates the calculated image distance.

PREREQUISITES
  • Understanding of concave mirrors and their properties
  • Familiarity with the mirror equation (1/f) = (1/di) + (1/do)
  • Ability to construct and interpret ray diagrams
  • Basic knowledge of image characteristics (inverted, size, location)
NEXT STEPS
  • Study the principles of ray diagrams for concave mirrors
  • Practice using the mirror equation with different object distances
  • Explore the effects of varying the focal length on image formation
  • Learn about real vs. virtual images in concave mirrors
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding image formation using concave mirrors.

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
975
  • · Replies 5 ·
Replies
5
Views
591
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 11 ·
Replies
11
Views
1K
Replies
7
Views
2K
Replies
22
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
Replies
13
Views
4K