concave mirror

[cowgiljl]

if a double concave mirror with a focal length f = -20 inches is used , where does the image seem to be located , if the image is 30 feet away from the lens

first i converted the 30ft to inches and got 360inches
used formula 1/p + 1/q = 1/f
.0028 + 1/q = -.05
1/q = -.053
q = -18.9 inches

is this correct

[Doc Al]

I assume you mean a double concave lens, not mirror. (A diverging lens.) And I also assume you are asked to find the image distance when the object is 30 feet away. If so, you are correct. But what does the negative sign mean? (The negative sign is correct; I'm just checking that you know where the image is.)

[cowgiljl]

not sure really where the image is i didn't think you could have a negitive distance

joe

[Soilwork]

the sign in front tells you where the image is situated.
either in front if it's virtual or at the back if it's real.
so in your case I would say that the image is virtual because the sign is negative.
I think anyway haha :)

[Doc Al]

not sure really where the image is i didn't think you could have a negative distance

Just as I suspected! :eek:

To use the lens equation, you need to know the sign convention:

For converging lenses f is +, for diverging f is -
Object distance is always + (except for virtual objects)
Image distance is + for real images, - for virtual images

If the object is on the left of the lens, a real image (positive image distance) is on the right of the lens, while a virtual image (negative image distance) would be on the left.

In this case the image is virtual: it's on the same side of the lens that the object is on.

[cowgiljl]

thanks for that information because I was going to ask the question tuesday