Answered

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A 30cm tall pug is standing 120cm in front of a convex lens that creates an image 30 cm on the opposite side of the lens. What is the focal length of the lens, and the magnification of the pug?

Sagot :

Answer:

The focal length of the lens = 24 cm

The magnification of the pug = 0.25

Explanation:

The height of the pug = 30 cm

The pug's distance from the convex lens, u = 120 cm

The image distance from the convex lens, v = 30 cm

Calculate the focal length using the equation below

[tex]\frac{1}{f}=\frac{1}{u}+\frac{1}{v}[/tex]

Substitute u = 120 cm, v = 30 cm, and solve for f

[tex]\begin{gathered} \frac{1}{f}=\frac{1}{120}+\frac{1}{30} \\ \frac{1}{f}=\frac{1+4}{120} \\ \frac{1}{f}=\frac{5}{120} \\ f=\frac{120}{5} \\ f=24\text{ cm} \end{gathered}[/tex]

The focal length of the lens = 24 cm

The magnification of the pug = v/u

Magnification = 30/120

Magnification = 0.25

The magnification of the pug = 0.25

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