posted by Jon on .
1)An ____ lens is used to produce a real image.
The link you gave me confused me b/c the closest thing I could find was convex. and you said concave mirrors produce virtual images. but the question doesnt ask for virtual they ask for real.
2)What is the nature of an image formed by a concave mirror of focal length 10 cm for a real object placed 3 cm from the mirror?
Virtual and inverted.
1. I didn't say anything about concave mirrors. THe question was on lenses. There are two types: concave (diverging), and convex (converging). If concave lenses produce only virtual images, what does that imply about convex?
Here is my previous brief response to the same question:
<<concave lenses form virtual images. They are often called diverging lenses, because they spread light apart. >>
What I recommend is that you take a few cases of convex and concave lenses, put an object somewhere, and examine the image. Is it real, or virtual? These ought to come to you readily.
After you get that, do the same for mirrors.
That ray diagram would answer question two. By nature of the image, they are looking for whether it is real or virtual, inverted or erect, and enlarged or not. A ray diagram answers that readily, in which you can discern your answer is not right on one count. You ought to state if it is enlarged or not.
If you want to check your diagram (dont cheat), http://www.glenbrook.k12.il.us/GBSSCI/PHYS/mmedia/optics/rdcmd.html
1) "what does that imply about convex?"
it implies that they produce real images
2)they are only looking for 2 things.
A)real and inverted
B)real and upright
C)virtual and inverted
D)virtual and upright
my 1st answer was A but someone said no
I was right on 2 answers that I had that I was told were wrong.
1)Specular reflection obeys the laws of reflection while diffuse reflection does not. True or false
I said false and I asked dwrls I believe, to explain why he said that was true
and 2)A rod of length 5.0 cm lies along the axis of a concave mirror of radius of curvature 20 cm. The end of the rod closer to the mirror is 15 cm from the mirror. Find the length of the image of the rod.
all I had to do was find the focal length which was 10 cm