Use the mirror equation to deduce that : c) the virtual image produced by a convex mirror is always diminished in size and is located between the focu
Use the mirror equation to deduce that :
c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
[Note : The exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]
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The mirror equation is `(1)/(u)+(1)/(v)=(1)/(f)`
But `m=(upsilon)/(u)=(f)/(u-f)=(upsilon-f)/(f)`
Hence `m=(f)/(u-f)`
`upsilon =f(m+1)`
c) For relation, `m=(f)/(u-f)` and f being positive for convex mirror, m will always be negative and less than one. Hence virtual image formed will always be diminished.
For relation, `m=(upsilon -f)/(f)` and m being negative, `upsilon` will always be less than f.
Hence image will be formed between pole and focus.
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