physics
posted by hunar .
An object is placed 3cm in front of a diverging lens of focal length of 6cm. the heihgt of the object is 1.5 cm. calculate the image distance and height

1/61/3=1/2
image distance=2
hi/ho=di/do
hi/1.5=(2)/3
hi=(2/3)x(3/2)
hi=1 
i want to know the formula according to conventions as i saw two different systems as i studied.

For convex and concave lens and for concave and convex mirror
the formula is
1/ di + 1/do = 1/f
m = hi/ho =  di /do
Note that different conventions are in practice .
Accordingly the formula will change .
For the above formula the following sign conventions are used.
• f is + if the lens is a double convex lens (converging lens)
• f is  if the lens is a double concave lens (diverging lens)
• di is + if the image is a real image and located on the opposite side of the lens.
• di is  if the image is a virtual image and located on the object's side of the lens.
• hi is + if the image is an upright image (and therefore, also virtual)
• hi is  if the image an inverted image (and therefore, also real)
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For concave and convex mirror
• f is + if the mirror is a concave mirror
• f is  if the mirror is a convex mirror
• di is + if the image is a real image and located on the object's side of the mirror.
• di is  if the image is a virtual image and located behind the mirror.
• hi is + if the image is an upright image (and therefore, also virtual)
• hi is  if the image an inverted image (and therefore, also real)