A needle placed at 45 cm from a lens forms an image on a screen placed 90 cm 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? The size of the needle is 5 cm?

Hint: here the object is placed at the left side of the lens, as per sign convention this distance will be taken in negative. The image is formed on the other side of the lens, as per sign convention this will be positive. So, we have object distance and image distance, we can make use of a thin lens formula to find out the focal length of the lens. If the focal length comes out to be positive then it is a convex lens and if negative then concave lens

Complete step by step solution:
Object distance, $u=-45cm$
Image distance,$v=90cm$
using lens formula,
$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u} \\
\Rightarrow \dfrac{1}{f}=\dfrac{1}{90}+\dfrac{1}{45} \\
\Rightarrow \dfrac{1}{f}=\dfrac{1+2}{90} \\
\Rightarrow \dfrac{1}{f}=\dfrac{3}{90} \\
\therefore f=30cm \\$
So, the focal length comes out to be positive and it is $30cm$. So, the lens used is convex lens.
Now given size of the object, \[{{h}_{0}}=5cm\]
$\Rightarrow \dfrac{{{h}_{i}}}{{{h}_{0}}}=\dfrac{v}{u} \\
\Rightarrow \dfrac{{{h}_{i}}}{5}=\dfrac{90}{-45} \\
\therefore {{h}_{i}}=-10cm \\$
So, the image is twice the object and image is inverted.

So, the image is twice the object and image is inverted.

Note: Magnification if greater than one then the image is enlarged and if it is less than one then image is diminished. If the magnification is negative then image is inverted and if magnification is positive then image is erect. If magnification is unity then size of the image is equal to the size of the object. While doing such problems always keep in mind cartesian sign conventions.