an object of height 4 cm is placed at a distance of 18 cm from concave mirror having focal length 12 cm. find the position, nature and height of the image.

Dear Student

f=-12 cm

u=-18cm

h_o=4cm

By mirror formula
$$\begin{gathered} \frac{1}{f} =\frac{1}{u}+\frac{1}{v} \\ \frac{1}{v} =\frac{1}{f}-\frac{1}{u} =-\frac{1}{12}-\frac{1}{(-18)} =\frac{1}{18}-\frac{1}{12} \\ \frac{1}{v} =\frac{12-18}{18\times 12} =-\frac{6}{18\times 12} =-\frac{1}{36} \\ v = -36 cm \end{gathered}$$

When the image distance is negative, the image is behind the mirror so image form behind the mirror. Image is real.
m=- v/u

m=- (-36)/(-18) = -2

Magnification is neagtive hence image is inverted.

Also, m=h_i/h_o

2=h_i/4cm

h_i=8 cm
 

hence image is magnified i.e. 8cm high

​​​​​​​Regards