Assume that light travels from left to right as it passes through the lens. An object
(a) Calculate the image distance with THREE SIG FIGS. cm.
(b) The image is
(c) The image is located -
(d) Calculate the magnification, m with THREE SIG FIGS.
(e) Calculate the image height, where the height can be positive or negative depending on the image's orientation (THREE SIG FIGS). cm
(f) The image is
a
The image distance is 
b
The image is real and inverted because the image distance in positive
c
The image is located to the left side
d
The magnification is 
e
The image height is 
f
The image is real and inverted because the image distance in positive
Explanation:
In the Solution We assume that the lens is converging
The lens Equation is mathematically represented as
Where f is the focal length = 4.20 cm
v is the image distance = ?
u is the object distance = + 1.5 (since is to the left i.e the source of light )
Making v the subject in the equation above
Mathematically magnification is represented as
Mathematically the image height is represented as
Where 
is the image height
is the object height = 1.5
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