Respuesta :
Answer:
The height of the image is 2.23 cm.
Explanation:
Given that,
Object distance = 10.0 cm
Focal length = -23.0 cm
Height of the object = 3.20 cm
We need to calculate the distance of image
Using image formula
[tex]\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}[/tex]
[tex]\dfrac{1}{v}=\dfrac{1}{-23.0}+\dfrac{1}{-10}[/tex]
[tex]\dfrac{1}{v}=-\dfrac{33}{230}[/tex]
[tex]v=-6.96\ cm[/tex]
We need to calculate the height of image
Using formula of magnification
[tex]m = \dfrac{-v}{u}=-\dfrac{h'}{h}[/tex]
[tex]-\dfrac{v}{u}=-\dfrac{h'}{h}[/tex]
Put the value into the formula
[tex]\dfrac{-6.96}{-10}=\dfrac{h'}{3.20}[/tex]
[tex]h'=\dfrac{-6.96\times3.20}{-10}[/tex]
[tex]h'=2.23\ cm[/tex]
Hence, The height of the image is 2.23 cm.
