Answer :
Answer:
The position of the image = 7.2 cm.
Explanation:
Given:
A Converging lens:
Object distance [tex](d_o)[/tex] = 12 cm = -12 cm (sign convention)
Focal length [tex](f)[/tex] = 18 cm
We have to find the position of the image.
Let the position of the image be [tex](d_i)[/tex] .
Sign convention:
The focal length of converging (convex) lens is always positive,while object distance is negative.
Using lens formula:
⇒ [tex]\frac{1}{object\ distance} + \frac{1}{image\ distance} =\frac{1}{focal\ length}[/tex]
⇒ [tex]\frac{1}{d_o} + \frac{1}{d_i} = \frac{1}{f}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} - \frac{1}{-d_o}[/tex]
⇒ [tex]\frac{1}{d_i} =\frac{1}{f} + \frac{1}{d_o}[/tex]
⇒ [tex]d_i=\frac{d_o\times f}{d_o+f}[/tex]
⇒ [tex]d_i=\frac{12\times 18}{12+18}[/tex]
⇒ [tex]d_i=\frac{216}{30}[/tex]
⇒ [tex]d_i=7.2[/tex] cm
So the position of the image = 7.2 cm.