Question

one student uses a lens of focal length +50 cm and another of -50cm
state the nature and find the power of each lens
which of the two lenses will give always give a virtual erect and diminished image irrespective of the position of the object

Answer 1

focal length is always in metre so therefore
50cm =0.5m
when focal length is -0.5m means the lens is concave
P=1/f
P=1/0.5
P=1×10/5
P= 2.0 ( the focal length of the lens is in negative which means the lens is concave and the power of concave lens is always negative)

P=-2.0D
When focal length is +50cm
50cm =0.5m
P=1/f
P=1/0.5
P=1×10/5
P=2.0+ ( the focal length of the lens is positive which means the lens is convex and its power is always positive)
P=+2.0 D

Answer 2

Answer:

focal length is always in metre so therefore

50cm =0.5m

when focal length is -0.5m means the lens is concave

P=1/f

P=1/0.5

P=1×10/5

P= 2.0 ( the focal length of the lens is in negative which means the lens is concave and the power of concave lens is always negative)

P=-2.0D

When focal length is +50cm

50cm =0.5m

P=1/f

P=1/0.5

P=1×10/5

P=2.0+ ( the focal length of the lens is positive which means the lens is convex and its power is always positive)

P=+2.0 D

Explanation: