Question

A 4cm tall bulb is placed a distance of 45.7cm from a concave mirror having a focal lenght of 15.2cm determine image distance and image size.

Answer 1

Given,

• $\displaystyle\sf{Object\:Height\:(h_o)=4\:cm}$

• $\displaystyle\sf{Object\:Distance\:(u)=45.7\:cm}$

• $\displaystyle\sf{Focal\:Length\:(f)=15.2\:cm}$

Let,

• Image distance be denoted as v cm

• Height of object be denoted as h₀ .

• Height of image be denoted as $\bold{\displaystyle\sf{h_i}}$ .

Using the mirror equation ,

$\boxed{\displaystyle\sf{\frac{1}{f}=\frac{1}{v}+\frac{1}{u}}}$

$\boxed{\displaystyle\sf{\frac{1}{v}=\frac{1}{f}-\frac{1}{u}}}$

Applying the values to the equation ,

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

$\implies\displaystyle\sf{\frac{1}{v}=0.0657-0.0218}$

$\implies\displaystyle\sf{\frac{1}{v}=0.0439}$

$\implies\displaystyle\sf{v=\frac{1}{0.0439}}$

$\implies\displaystyle\sf{v=22.8\:cm}$

$\boxed{\displaystyle\sf{Image\:Distance=22.8\:cm}}$