at what distance would a convex magnifing lens that has a focal length of 10 cm have to be held for an image to appear upright and 3 cm tall?
[6.67cm]
!!! The last words in given data have to be
“and 3 times as tall”
If so, then:
f = focal length of lens = 10cm
di = image distance from lens
do= object distance M =3 - magnification
Note that for the image to be upright it must be a virtual image, so di is negative.
di/do = -3
so di = -3do
1/10 = 1/do + 1/di
1/10 = 1/do - 1/(3•do) = 2/3•do
3•do = 20
do = 6.67cm
To determine the distance at which a convex magnifying lens with a focal length of 10 cm would have to be held for an upright and 3 cm tall image, we can use the lens formula:
1/f = 1/v - 1/u
where:
f is the focal length of the lens,
v is the image distance, and
u is the object distance.
Since the image is upright (which means it has positive (+) height), the magnification (M) will be positive (+). The magnification is given by the formula:
M = -v/u
Given that the height of the image (h') is 3 cm and the focal length (f) is 10 cm, we can use the magnification formula to find the image distance (v):
M = h'/h = -v/u
Rearranging the equation:
v = -M * u
Now, substituting the given values, we have:
3/(-10) = -v/u
Rearranging and simplifying:
v = -30/u
Since the image is 3 cm tall, its height (h') is equal to the negative of the object height (h):
h' = -h
Substituting the values, we have:
3 = -h
Therefore, the object height (h) = -3 cm.
Now we can solve for the object distance (u):
M = -v/u
-3/u = -30/u
Cross-multiplying:
-3u = -30
Simplifying:
u = -30 / -3
u = 10 cm
So, the object distance (u) is 10 cm.
Finally, we can find the image distance (v) by substituting the values into the equation:
v = -30/u
v = -30 / 10
v = -3 cm
Therefore, the distance at which the convex magnifying lens would have to be held for an upright and 3 cm tall image is 3 cm.
To find the distance at which an image appears upright and 3 cm tall through a convex magnifying lens with a focal length of 10 cm, we can use the magnification formula:
Magnification (M) = - (Image Height / Object Height) = -(h'/h)
Where:
M is the magnification,
h' is the image height, and
h is the object height.
In this case, the image height (h') is given as 3 cm. The object height (h) is not provided. However, we can still solve for the distance using the magnification formula and rearranging it as:
Distance (d) = f / (1 + 1/M)
Where:
d is the distance,
f is the focal length, and
M is the magnification.
Given that the focal length (f) is 10 cm, we can substitute these values into the formula and solve for the distance:
d = 10 cm / (1 + 1/M)
To find the value of M, we need to plug in the magnification formula:
M = - (h'/h) = - (3 cm / h)
Substituting this value back into the distance formula:
d = 10 cm / (1 + 1/(-3 cm / h))
Simplifying further:
d = 10 cm / (1 - 3 cm / h)
To simplify the equation, we can take the reciprocal of the denominator:
d = 10 cm / ((h - 3 cm) / h)
Now we can multiply the numerator and denominator by h:
d = 10 cm * (h / (h - 3 cm))
This equation allows us to calculate the distance (d) at which the image appears upright and 3 cm tall using the given object height (h).