Question

An object sits on the axis of a converging spherical mirror. It is over twice the distance from the mirror as the focal length. (i.e. the distance to the object is more than 2 times f.) Describe the image by telling how big it is and how far it is from the mirror relative to the object. Tell whether it is upright or upside down and whether it is real or virtual. Explain how you arrived at your answers...

Answer 1

As we know when object is twice the distance of focal length then image also formed at the same location of object.

But if object is more then twice then image is between f and 2f.

Image is real, upside down and diminished.

All this can be proved by the mirror formula and magnification.

By mirror formula

1/v + 1/u = 1/f

1/v + 1/(-2f) = 1/(-f)

v = - 2f

But if u>2f then by the equation v<2f

As negative so real image.

By magnification = hi/ho = - v/u

As in right side both v and u are negative so overall is positive but there is also a negative sign so hi is negative. So image is upside down.

The ratio v/u is less then one so height of image is less then object. So smaller then object.