Question

A thin converging lens is formed with one surface convex and the other plane. Does the position of image depend on whether the convex surface or the plane surface faces the object?

Answer 1

for a lens different radius of curvature R1 position of object

$\frac{1}{V}-\frac{1}{u}=(u-1)(\frac{1}{R1}-\frac{1}{R2})$

$R1=+R,R2=\infty$

$\frac{1}{V}-\frac{1}{u}=(u-1)(\frac{1}{R}-\frac{1}{\infty })$

$\frac{1}{V}-\frac{1}{u}=\frac{(u-1)}{R}$

$R1=\infty ,R2=-R$

$\frac{1}{V}-\frac{1}{u}=(u-1)(\frac{1}{\infty }-\left(\frac{-1}{R}\right))$

$\frac{1}{V}-\frac{1}{u}=\left(\frac{u-1}{R}\right)$