Question

A thin convex lens of focal length $30.00cm$ forms an image $2.00cm$ high, of an object at infinity. A thin concave lens of focal length $20.00cm$ is placed $26.00cm$ from the convex lens on the side of the image. The height of the image now is

• A. $1.00cm$
• B. $1.25cm$
• C. $2.00cm$
• D. $2.50cm$

Answer 1

The correct option is D $2.50cm$

$\begin{array}{cc}\frac{1}{q}=\frac{1}{v}-\frac{1}{u}& \\ \frac{1}{-20}=\frac{1}{v}-\frac{1}{4}& \\ \frac{1}{4}-\frac{1}{20}=\frac{1}{v}& \\ v=& \frac{20}{4}\\ v=5cm& \end{array}$

$\frac{V}{u}=\frac{h_i}{h_0}$

$\frac{5}{4}=\frac{h_i}{2}$

$h_i=\frac{5}{4}\times 2$

$h_i=2.5cm$

so option $D$ is correct.