Question

A 4.00-cm tall plant has placed a distance of 45.7 cm from a concave mirror having a focal length of 15.2 cm. Determine the image distance and image height.

Answer 1

$\LARGE{\bf{\underline{\underline{GIVEN:-}}}}$

• Height of object (h) = 4 cm.
• Object distance (u) = –45.7 cm. [u is always negative].
• Focal length (f) = –15.2 [f is always negative for concave mirror].

$\LARGE{\bf{\underline{\underline{TO \ FIND:-}}}}$

• Image distance (v) = ?
• Image height (h') = ?

$\LARGE{\bf{\underline{\underline{SOLUTION:-}}}}$

We shall use the mirror formula:

$\mapsto \sf{\red{\dfrac{1}{f}=\dfrac{1}{v}+\dfrac{1}{u} }}$

Where:

• f is focal length.
• v is image distance.
• u is object distance.

Substitute the values.

$\displaystyle \sf \to -\frac{1}{15.2} =-\frac{1}{45.7} +\frac{1}{v}$

$\to \sf v \approx -22.8 cm.$

Image height can be given by magnification:

$\displaystyle \sf \frac{h'}{h} =\frac{v}{u}$

Where:

• h' is image height.
• h is object height.
• v is image distance.
• u is object distance.

$\to \displaystyle \sf \frac{h'}{4} =\frac{-22.8}{-45.7}$

$\sf \to h' \approx 2 cm.$

• IMAGE DISTANCE IS -22.8 cm.
• IMAGE HEIGHT IS 2 cm.