Question

A thin converging lens forms the real image of certain real object magnified m times. The magnification of real image become n, when lens is moved nearer to object by distance x.

Answer 1

Answer:

I don't know the answer

Answer 2

The focal length of the converging lens is:

mnx/(m-n)

Explanation:

If the object distance is u and image distance is v then

We know that the magnification of mirror is given by

$m=-\frac{v}{u}$

By mirror formula, if focal length is f

$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$

$\implies \frac{1}{v}=\frac{1}{f}-\frac{1}{u}$

$\implies \frac{1}{v}=\frac{u-f}{uf}$

$\implies v=\frac{uf}{u-f}$

Therefore,

$m=-\frac{f}{u-f}$

$\implies m(f-u)=f$

$\implies mu=mf-f$

$\implies u=f-\frac{f}{m}$

When the lens is moved nearer to the object by distance x

The new object distance will be u-x

Thus, the new image distance

$v'=\frac{(u-x)f}{u-x-f}$

Also,

$n=-\frac{v'}{u-x}$

Thus,

$n=-\frac{f}{u-x-f}$

$\implies n(u-x-f)=-f$

$\implies n(f-\frac{f}{m}-x-f)=-f$

$\implies n(\frac{f}{m}+x)=f$

$\implies \frac{f}{m}+x=\frac{f}{n}$

$\implies \frac{f}{n}-\frac{f}{m}=x$

$\implies f(\frac{m-n}{mn})=x$

$\implies f=\frac{mnx}{m-n}$

Hope this answer is helpful.

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