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A convex mirror in an amusement park has a radius of curvature of 3.00m. A man stands in front of the mirror so that his image is half as tall as his actual height. At what distance must the man focus his eyes in order to see his image?

I calculated the focal length using f=(-.5)(3.00m)

then I tried to find the image distance(di) but I get stuck.

What am I supposed to be doing here?

  • physics -

    You have correctly determined that f = -1.5 m for this mirror. From the magnification of 1/2, and the fact that it is a virtual image, you also know that
    di/do = -1/2.

    Therefore 1/di + 1/do = 1/di - 1/2di =
    1/2di = 1/f = -2/3

    Therefore di = -3/2 m and do = 3 m

    Since the observer is the "object" and his image is on the other side of the mirror, he must focus his eyes 4.5 meters away.

    Check my thinking. I could have made a mistake somewhere

  • physics -

    you kind of just droped 1/di there. where did it go?

    I just don't get how you got to:

    1/di - 1/2di

    and then you went from

    1/di - 1/2di = 1/ just droped 1/di. That doesnt make any since.

  • physics -

    1/di could also be written 2/2di then if you take away 1/2di you are left with 1/2di. It's like saying 1 minus 1/2 equals 1/2

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