An object is 30.0 cm to the left of a convex lens with a focal length of +8.0 cm. Calculate the distance from the lens. Would it be upright or inverted?
Use 1/f=1/di + 1/do
and
Magnification M=hi/ho=-di/do
Use metres throughout
do=0.30
f=+0.08 m (converging lens)
di=1/(1/f-1/do)
=0.1091 (image distance)
So image is real (di >0) and on the right-hand (emerging) side of the lens.
M=-di/do=-0.364
(image smller, inverted, M<0)
To calculate the distance from the lens, we can make use of the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the image distance from the lens, and
u is the object distance from the lens.
In this case:
f = +8.0 cm (given)
u = -30.0 cm (since the object is to the left of the lens and we take distances to the left as negative)
Using the lens formula and rearranging to solve for v, we get:
1/v = 1/f - 1/u
Substituting the values, we have:
1/v = 1/8.0 cm - 1/(-30.0 cm)
1/v = (30 - 8)/(8 * 30) = 22/240 = 11/120
Now, we can solve for v by taking the reciprocal:
v = 120/11 cm
Therefore, the image distance from the lens is approximately 10.9 cm.
To determine whether the image is upright or inverted, we can use the magnification formula:
m = -v/u
where:
m is the magnification,
v is the image distance from the lens, and
u is the object distance from the lens.
In this case, substituting the values, we have:
m = -(120/11 cm) / (-30.0 cm)
m ≈ 4.0/1
Since the magnification is positive (+4.0), the image is upright.