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### Course: Physics library>Unit 15

Lesson 2: Mirrors

# Mirror equation example problems

In this video David solves a few exmaple problems involving concave and convex mirrors using the mirror equation and magnification equation.

## Want to join the conversation?

• Now I'm confused... Sal said the real image can be seen only if you put a screen there... In this video, they say the real image, in front of the concave, can be seen anyway, and it feels like you can reach out and touch it.... Which one is it?
• Sal didn't mean so.. In a real image, the light rays come to a point from a point of object and this happens to all other points of that object..So, if we put a screen at that distance, we can see a clear image and as the image is really created for every point it is called real image .

This image will still create if you don't put a screen there and if you go to mirror at that distance, you'll be able to see an image of that thing.. So you'll feel that there was something before you and you could touch it....

I hope you understand the matter..... :)
• when I did the problem where the object was 3cm from the mirror I got symmetrical results di=-12 but the object height was 12cm I think that is because the object height was quadrupled, Is this right and what would you see?
• At , David said "focal length" when I think he meant "focal point".
(1 vote)
• Both are correct. If the focal point is behind the mirror then the focal length will also be behind the mirror.
• what are sign conventions of every mirrors?
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• Concave Mirror +F -Do ±Di May be real and inverted or virtual and right side up
Convex Mirror -F +Do -Di Virtual and right side up
(1 vote)
• Can distances behind the mirror be negative?
(1 vote)
• Distance behind the mirror are considered to be negative.
(1 vote)
• David explains, that with a convex mirror and the eye and object on the left, the image is formed on the right (negative image length). I'm confused about why light forms an image through the mirror, is it that the mirror is only partially reflective? Thanks!
(1 vote)
• Nice question. The mirror is not partially reflective, which is generally the case unless otherwise mentioned.
If you construct the ray diagram for this case, you will see that the light rays diverge after reflection from the mirror, they do not move on a path that would make them intersect. And we know that image is formed when light rays intersect, so no image formation should take place, and yes, no real image is formed.
However, when you extend the rays to behind the mirror you will observe that the lines intersect at a single point, i.e. the rays appear to converge (/emanate from) a single point, so the rays are virtually intersecting, and yes, a virtual image is, in fact, formed! So in this case, a virtual image forms instead of a real image.
The image you see when you look into the mirror while brushing your teeth is also a virtual image, see how it appears to be behind the mirror, in a mirror world? This is why. Also, check out this video for more on virtual images:
(1 vote)
• I have a problem where I need to calculate the image distance if the image is between the focus and the object and the distance between the image ad the object is 15cm. Focus of the concave mirror is 20cm. If I try to do this problem even with this sign convention the results come out as complex numbers.
(1 vote)
• In our textbooks we have magnification formula like Г = H/h = f/d
My question is why we don't have minus before f/d which we have in video or just the concept is different and it is understood to be like that
(1 vote)
• Is radius of curvature of any curved mirror double its focal length?
And is there any equation for plane mirrors?