Question

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 1

Answer:

Given, two lenses having powers -4D and +6D form a combination.

The power of the combination of the lenses = -4D + 6D = 2D

P=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 cm

The 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 cm

Thus, the image formed is erect and 2.5 times magnified.