1.

Use the mirror equation to deduce that : a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f. [Note : The exercise helps you deduce algebraically properties of images that one obtains from explicit ray diagrams.]

Answer» 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)`
When `u gt f lt 2f`, we get
`m=(f)/(u-f)`
`=(f)/((gt f lt 2f)-f)`
`=(f)/((gt O lt f))= gt 1`
Hence `upsilon = f(m+1)=f(gt1+1)`
or `upsilon gt 2f.`
Since for concave mirror, f is negative, `upsilon` becomes negative.
It means image produced is real and beyound 2f.


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