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Sagot :
Given:
the height of the object is
[tex]h_0=5\text{ cm}[/tex]The distance of the object is
[tex]d_0=-20\text{ cm}[/tex]The focal length of the lens is
[tex]f=15\text{ cm}[/tex]Required: the distance of the image and height of the image.
Explanation:
the lens formula is given by
[tex]\frac{1}{f}=\frac{1}{d_i}-\frac{1}{d_0}[/tex]Plugging all the values in the above relation, we get:
[tex]\begin{gathered} \frac{1}{15\text{ cm}}=\frac{1}{d_i}-\frac{1}{-20\text{ cm}} \\ \frac{1}{d_i}=\frac{1}{15\text{ cm}}-\frac{1}{20\text{ cm}} \\ \frac{1}{d_i}=\frac{4-3}{60\text{ cm}} \\ d_i=60\text{ cm} \end{gathered}[/tex]Thus, the distance of the image is 60 cm.
now calculate the height of the image
we know that
[tex]\frac{h_i}{h_0}=\frac{d_i}{d_0}[/tex]substitute all the values in the above relation, we get:
[tex]\begin{gathered} h_i=5\text{ cm}\times\frac{60}{-20} \\ h_i=-15\text{ cm} \end{gathered}[/tex]Thus, the height of the image is 15 cm.
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