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.