Question

What magnification will be produced by a lens of power –4.00 D (such as might be used to correct myopia) if an object is held 29 cm away?

Answer 1

Given data in this problem :

Power of the lens $=-4.00 \mathrm{D}$ Distance $d=29 \mathrm{~cm}$

to find out magnification firstwecalculatethe focallengththatis $=\frac{100}{\text { Power }}$ Focal $-$ length $=\frac{100}{-4.00}=-25 \mathrm{~cm}$

now calculate the image distance $u$ that is $1 / f=1 / d+1 / u$

$u=f \cdot d / d-f$

put all these value :

$u=-25(29) / 29+25$

$u=725 / 54=-13.43 \mathrm{~cm}$

image distance $=-13.43 \mathrm{~cm}$ so magnification $=-$ image distance $/$ distance magnification $=13.43 / 29=0.46$

so magnification is 0.46

Magnification of the lens is +0.46.