Question

A needle placed $45cm$ from a lens forms an image on a screen placed $90cm$ on the other side of the lens. Identify the type of the lens and determine its focal length. What is the size of the image if the size of the needle is $5.0cm$?

Answer 1

u=-45cm

v=90cm

Since the image is virtual therefore, it must be a convex lens

$\begin{array}{c}\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\\ \frac{1}{f}=\left(\frac{1}{90}\right)-\left(-\frac{1}{45}\right)\\ f=30cm\\ \\ m=\frac{v}{u}=\left(\frac{90}{-45}\right)=-2\\ m=\frac{h_i}{h_o}\\ -2=\frac{h_i}{5}\\ h_i=-10cm\end{array}$