How Do Reflection and Curvature Affect Images in Optics?

[jlmac2001]

1. Imagine that you are standing 5 ft from and looking directly toward, a brass ball 1 foot in diameterhanging in front of a pawn shop. Describe the image you would see in the ball.

-Would I have to find the an answer that tells me that it is inverted, real, virtual, etc?

2. What formula would I useto find the radius of curvature? if the magnification of a keratometer is found to be 0.037X when the object distance is se at 100 mm, what is the radius of curvature?

 

[dduardo]

Mirror problems are a little difficult to do in a forum. It really helps to have a picture. I highly suggest you do a google search on convex mirrors and look at the examples done in your physics book. For this types of problems DRAW A PICTURE and make sure you label everything

The equation you would use is
1/s' + 1/s = 1/f where 1/f = 2/R

R is the radius of curvature
f is the focus
s is the distance from your eye to the front of the mirror
s' is the distance from the front of the mirror to the image (may it be real or virtual)

m = -s'/s where m is the magnification

 

[M98Ranger]

1. If you are standing 5 ft away from a brass ball with a diameter of 1 foot, you would see an inverted, real image of the pawn shop in the ball. This is because the light from the pawn shop is reflected off the curved surface of the ball and forms an image on the opposite side of the ball. The size of the image would depend on the curvature of the ball and the distance between you and the ball.

2. To find the radius of curvature, you can use the formula 1/f = 1/u + 1/v, where f is the focal length, u is the object distance, and v is the image distance. In this case, the object distance is 100 mm and the magnification is 0.037X, so the image distance can be calculated by v = u/m = 100 mm/0.037 = 2703.7 mm. Then, using the formula, 1/f = 1/100 mm + 1/2703.7 mm = 0.0003709 mm^-1. Therefore, the radius of curvature would be r = 1/f = 2703.7 mm.