If a concave lens is placed in the path of co | vedclass

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(A) For a concave lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Here,the incident rays are converging,so the object is v

Vedclass · Accepted Answer

$0$ and $f$

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(A) For a concave lens,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Here,the incident rays are converging,so the object is virtual. The distance of the point of convergence from the pole is $u$,which is positive $(u > 0)$.The focal length of a concave lens is negative,so we take $f_{lens} = -f$ (where $f > 0$).The lens formula becomes $\frac{1}{v} - \frac{1}{u} = -\frac{1}{f}$,which simplifies to $\frac{1}{v} = \frac{1}{u} - \frac{1}{f} = \frac{f - u}{uf}$.For a real image,the image distance $v$ must be positive $(v > 0)$.Since $u > 0$ and $f > 0$,for $v$ to be positive,the numerator $(f - u)$ must be positive.This implies $f - u > 0$,or $u Therefore,the distance $u$ must lie between $0$ and $f$.

If a concave lens is placed in the path of converging rays,a real image will be produced if the distance of the pole from the point of convergence of incident rays lies between ($f =$ magnitude of focal length of lens).

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