A man uses a thin convex lens of focal length | vedclass

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Question

(A) For a magnifying glass,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Here,$f = +5 \, cm$. The image distance $v$ must be

Vedclass · Accepted Answer

$4.17 \, cm, 5 \, cm$

Vedclass · Answer

(A) For a magnifying glass,the lens formula is given by $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$.Here,$f = +5 \, cm$. The image distance $v$ must be between the near point and the far point of the eye.For the closest distance of the object $(u_{min})$,the image must be formed at the near point of the eye,so $v = -25 \, cm$.Using the lens formula: $\frac{1}{-25} - \frac{1}{u} = \frac{1}{5} \implies -\frac{1}{u} = \frac{1}{5} + \frac{1}{25} = \frac{5+1}{25} = \frac{6}{25}$.Thus,$u = -\frac{25}{6} \approx -4.17 \, cm$. The magnitude is $4.17 \, cm$.For the farthest distance of the object $(u_{max})$,the image must be formed at infinity for relaxed vision,so $v = -\infty$.Using the lens formula: $\frac{1}{-\infty} - \frac{1}{u} = \frac{1}{5} \implies 0 - \frac{1}{u} = \frac{1}{5}$.Thus,$u = -5 \, cm$. The magnitude is $5 \, cm$.Therefore,the closest and farthest distances are $4.17 \, cm$ and $5 \, cm$ respectively.

$A$ man uses a thin convex lens of focal length $5 \, cm$ as a magnifying glass. He reads a book by holding it close to his near point of $25 \, cm$. What are the closest and farthest distances at which he can keep the book from the lens to read clearly?

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