HURRY....IMPORTANT QUESTION. ...
A 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 20cm the distance of the object from the lens is 15 cm find the nature position and size of the image
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HURRY....IMPORTANT QUESTION. ...
A 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 20cm the distance of the object from the lens is 15 cm find the nature position and size of the image
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Focal length of the convex lens is always positive, so, f = 20 cm
Distance of Object from lens (u) =-15 cm
We need to find nature , position (v) and size of image (h')
So, v/u=h'/h
Magnified as h'>h
Virtual and erect (as v is negative and h' is positive)
Verified answer
Height of object ho = 4 cmFocal lenght of convex lens = 20 cm Object
distance u = -15 cm
Image distance v = ?
Image height hi = ? Lens
formula: 1/v - 1/u = 1/f 1/v =
1/f + 1/u = 1/20 + 1/(-15) =
-1/60
⇒ v = - 60 cm
The image distance is 60 cm on the same side of the lens as the object.
Magnification m = v/u
= (-60)/(-15)
= 4
= hi/ho
= hi/4
⇒ hi = 4 x 4 = 16 cm The
image is enlarged and erect. So a virtual, erect, enlarged image is formed at 60 cm from the optic centre on the same side of the lens as the object is positioned.