An object is placed at a distance of 60 cm from a concave lens of focal length 30cm. find the position and nature of the image?

using the lens formula: 1/f = 1/u + 1/v

focal length f = -30 (negative because it is concave lens)

object distance u = 60

image distance v= unknown

1/-30 = 1/60 + 1/v

v = -20

So, the image is 20cm from lens (on the same side along with the object), and it is virtual (because of negative sign) and erect (concave lens must produce erect images).

answer: i'm not sure which is which for i get confused on it all the time but i can tell you that the answer lies with the convex and concave devices. i hope this you. have a good day and best of luck to you!

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