Question

A 4.48 cm tall toy soldier is placed 33.8 cm in front of a diverging lens with a focal length of -13 cm. What is the height of the image created by the lens?

Answer 1

We have-

Height of the object(toy soldier)=h_o= + 4.48 cm

The distance of the object from the lens=u= - 33.8 cm

The distance of the image from the lens=v

The focal length of the lens= f = - 13 cm

Height of the image(toy soldier)=h_i​​​​=?

Calculating v:

We know that the lens formula is written as:

$\frac1f=\frac1u+\frac1v$

Putting the known values and solving for v:

$\frac1{-13}=\frac1{-33.8}+\frac1v\\ \\ \Rightarrow \frac1v =-\frac{1}{13}+\frac1{33.8} \\ \\ \Rightarrow \frac1v =\frac{-33.8+13}{33.8*13}=-0.0473373\\ \\ \Rightarrow v=\frac1{-0.0473373}=-21.125 \;cm$

Calculating h_i:

Now we know that magnification,m of the lens is given by:

$m=\frac{h_i}{h_o}=-\frac vu$

Now putting the known values and calculation h_i:

$h_i=-\frac {v*h_o}u\\ \\ \Rightarrow h_i=-\frac{-21.125*4.48}{-33.8}=2.8\;cm$

The height of the image(toy soldier) created by the lens is 2.8 cm.

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