An object of height 4 cm is kept to the left of and on the axis of the converging lens of focal length 10 cm as shown in the figure. A plane mirror is placed at ${45}^{\circ }$ to the lens axis 10 cm to the right of the lens (see figure). The position and size of the image formed by combination is:

First, let's consider the situation where the mirror is out of place.
We are given that the object is placed at a distance of 15cm from the lens and the focal length of the lens is 10 cm. So we can write:
$u=-15cm$ and $f=10cm$ convex lens.
$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}$

putting values:
$\frac{1}{v}=\frac{1}{10}+\frac{1}{(-15)}$

or

$v=\frac{(-15 X 10)}{(-15+10)}=+30cm$
Now, to determine the image size, a convex lens magnification formula is used:
$M=\frac{h_i}{h_o}=\frac{-v}{u}$

where $h_o$ is image length,
$h_i$ is the object length, v is the image range and u is the object range.
We may change this formula as:
$h_i=\frac{-v}{u}X h_o$

Where we know the values ​​are u = 15 cm, v = 30 cm, and h_o = 4cm.
To preserve these treasures,
$h_i=-\frac{30}{15} X 4 cm$

$h_i=2 X 4 cm$

$h_i=8cm$
So the image size is 8 cm. The built-in image will be converted.

Now, when we place the mirror on another lens focus light will appear on the screen and will create a realistic image when it is detected on the screen. The flight screen does not give any additional difference in the path it emits. It just sends light to the other side. If we produce its path behind a mirror, we will be able to see that the image behind the mirror is similar to the one created by the lens.