## GR0877 Problem 74

**PROBLEM STATEMENT**: *This problem is still being typed*.

**SOLUTION**: (**E**) Using mirrow equation one has (note the minus sign). From this equation . The only choice that corresponds to this distance is (E), but if you don’t want to invoke mirrow equation just drop two rays onto the mirror, say, one from the tip of the object parallel to the lens’ axis and the second from the tip of the object to the point where the lens’ axis intersects the mirror, and you will see where the image is formed (and why it is upright).

*Found a typo? Comment!*

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Since it’s a virtual image, shouldn’t the image distance l be negative? I know your solution leads to the correct answer. I am wondering about the sign convention. Thank you!

Actually, it’s in reverse. When we write down the mirror equation we pretend we don’t know what type of the image we’ll get (real or virtual). What we definitely know is that the mirror is convex (thus the minus sign in the right-hand side of the equation) and that the object is in front of the mirror (thus the plus sign before ). Plugging into the equation, we get (minus). And only now by the sign convention we claim that the image is virtual (in the back of the mirror).

Something good to remember here (that I completely forgot on the test) is that for spherical mirrors, the focal length is half the radius of curvature. With that in mind, one can use the equation:

1/(image distance) +1/(object distance) = 1/f

lol That’s what I actually used here.

Actually it’s good to know that Convex mirrors always create virtual images, while Concave mirror can create both, dependent on object distance.

So, virtual images = image to the right. We only have two choices to guess from without any calculation. (C) and (E)