Question

An object of height 4cm is placed at a distance of 20cm from concave lens of f =10 CM. use lens formula to determine it's position of image

Answer 1

Hey mate..
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Given,

Size ( height ) of the object , h = 4 cm

Object Distance , u = - 20cm

Image Distance , v = ?

Focal length of a concave lens , f = -10cm

From Lens Formula we have,
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$\frac{1}{f } = \frac{1}{v } - \frac{1}{u} \\ \\ = > \frac{1}{f} + \frac{1}{u } = \frac{1}{v} \\ \\ = > \frac{1}{ - 10} + \frac{1}{( - 20)} = \frac{1}{v} \\ \\ = > \frac{ - 2 - 1}{20} = \frac{1}{v} \\ \\ = > \frac{ - 3}{20} = \frac{1}{v} \\ \\ = > \frac{20}{ - 3 } = v \\ \\ = > v = - 6.67cm (approx.)$

So,

The image is virtual and erect and it is formed at the distance of 6.67 m on the same side of the object.

Hope it helps !!

Answer 2

Answer: 6.66cm

hope it helped