An object 2cm height is placed at a distance of 30cm in front of a concave lens of focal length 15cm . Find the position, nature and size of the image.

An object 2cm height is placed at a distance of 30cm in front of a con

1.	An object 2cm height is placed at a distance of 30cm in front of a concave lens of focal length 15cm . Find the position, nature and size of the image.
Answer» Given, TWO lenses having powers -4D and +6D form a combination.The power of the combination of the lenses = -4D + 6D = 2DP=1fThus, the equvalent focal length of the combination, f=12=0.5 m=50 cmObject distance, U = −30 cmLet the IMAGE distance be vFrom lens formula1v−1u=1f1v=1f+1u=150−130=3−5150=−2150v=−75 cmThe image is formed on the same side of the lens as the OBJECT is.Thus, the image is virtual.Magnification, M = vu=−75−30=2.5Height of the object = 2 cmM=Height of the imageHeight of the object=2.5Height of the image=2.5×2=5 cmThus, the image formed is ERECT and 2.5 times magnified.

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An object 2cm height is placed at a distance of 30cm in front of a concave lens of focal length 15cm . Find the position, nature and size of the image.​

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An object 2cm height is placed at a distance of 30cm in front of a concave lens of focal length 15cm . Find the position, nature and size of the image.