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The object is 10 cm in front of a mirror and its image is found at 5.0cm behind the mirror.1.What is the focal length of the mirror?2. What is the magnification of the image formed?3. What type of image is formed?4. What kind of mirror is used?

Respuesta :

[tex]\begin{gathered} \frac{1}{f}=\frac{1}{v}+\frac{1}{u} \\ \\ \frac{1}{f}=\frac{1}{5}+\frac{1}{-10} \\ \\ \frac{1}{f}=\frac{1}{10} \end{gathered}[/tex]

1. result focal length f=10cm

[tex]\begin{gathered} m=\frac{-v}{u} \\ \\ m=\frac{-5}{-10}=\frac{1}{2} \end{gathered}[/tex]

2. result magnification m= 0.5 cm

3. virtual erect and diminished image is formed

4. convex mirror

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