- #1
mike2007
- 46
- 0
A positive lens has a focal length of 12cm. An object is located a distance of 3 cm from the lens.
How far is the lens from the image?
Is the image real or virtual, erect or inverted?
My attempt
f = +12cm
o = +3cm
i =?
1/o + 1/I = 1/f
1/i = 1/f – 1/o
= 1/12cm – 1/3cm
= 1/12 – 4/12
= - (3/12)
i = -4cm on the same side of the object.
Since the object lies in within the focal point, the image is virtual and erect.
How far is the lens from the image?
Is the image real or virtual, erect or inverted?
My attempt
f = +12cm
o = +3cm
i =?
1/o + 1/I = 1/f
1/i = 1/f – 1/o
= 1/12cm – 1/3cm
= 1/12 – 4/12
= - (3/12)
i = -4cm on the same side of the object.
Since the object lies in within the focal point, the image is virtual and erect.