A 0.38 m tall object is placed 0.24 m from a converging lens with a 0.19 m focal length. How tall is the image? (Remember that negative height just means the image is inverted)

Respuesta :

In order to find the image's height, first let's find the image's position, using the formula below:

[tex]\frac{1}{f}=\frac{1}{d_o}+\frac{1}{d_i}[/tex]

Where f is the focal length, do is the object's distance and di is the image's distance.

So, using f = 0.19 and do = 0.24, we have:

[tex]\begin{gathered} \frac{1}{0.19}=\frac{1}{0.24}+\frac{1}{d_i}\\ \\ \frac{1}{d_i}=\frac{1}{0.19}-\frac{1}{0.24}\\ \\ \frac{1}{d_i}=5.263-4.167\\ \\ \frac{1}{d_i}=1.096\\ \\ d_i=\frac{1}{1.096}\\ \\ d_i=0.9124 \end{gathered}[/tex]

Now, to calculate the image's height, we can use the formula below:

[tex]\begin{gathered} \frac{h_i}{h_o}=\frac{-d_i}{d_o}\\ \\ \frac{h_i}{0.38}=\frac{-0.9124}{0.24}\\ \\ h_i=\frac{-0.9124\cdot0.38}{0.24}\\ \\ h_i=-1.445\text{ m} \end{gathered}[/tex]

Therefore the image's height is 1.445 m.

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