an object of height 5cm is placed perpendicular to the principle axis of a concave lens of focal length 10cm. if the distance of object from the optical centre of the lens is 20cm, determine the position, nature and size of the image formed. using lens formula?
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Height of object (h) = 5cm
Focal length = -10cm
Object distance (u) = -20cm
Image distance (v) = ?
Acc. To lens formula ,
1/v - 1/u = 1/f
Putting values of f and u and solving it u will get v = -6.66cm(position of image)
h'/h = v/u putting values of h, v and u and solving it, you get h' = 3.33cm(size of image)
Nature of image is virtual and erect
Hope it helps....
Focal length = -10cm
Object distance (u) = -20cm
Image distance (v) = ?
Acc. To lens formula ,
1/v - 1/u = 1/f
Putting values of f and u and solving it u will get v = -6.66cm(position of image)
h'/h = v/u putting values of h, v and u and solving it, you get h' = 3.33cm(size of image)
Nature of image is virtual and erect
Hope it helps....
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