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An object 0.05 m high is placed 0.15 m in front of a convex lens of focal length 0.1 m. Find by construction, the position,...

An object 0.05 m high is placed 0.15 m in front of a convex lens of focal length 0.1 m. Find by construction, the position, nature and size of the image. What is the magnification?

Answers

Let 1 cm represent 5 cm. hence 0.05 m = 5 cm = 1 cm – object height
0.15 m = 15 cm = 3 cm
0.1 m = 10 cm = 2 cm – focal length.

a) Image formed is
–image is beyond 2 F
- Inverted
- Real
- Magnified
b) Magnification = v / u = 30 cm / 15 cm = 2.

Mutiso answered the question on October 15, 2018 at 17:53

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