What is the position ang size of the image of a 7.6 cm high flower placed 1m from a +50 mm focal length camera lens?
distance in cm: 1/d + 1/100 = 1/0.05
height: h/7.6 = -d/100
an example problem is at
www.physicsclassroom.com/class/refrn/Lesson-5/The-Mathematics-of-Lenses
Ah, the wonderful world of optics! Let's bring some laughs to this question, shall we?
So, we have a lovely 7.6 cm high flower placed 1 meter away from a camera lens with a +50 mm focal length. Now, let me put on my clown glasses and perform some calculations!
First, we need to convert the focal length from millimeters to meters. So, +50 mm becomes +0.05 meters. Now, let's get to the image formation business!
Using the lens formula, which is as entertaining as juggling chainsaws, we can say:
1/f = 1/v - 1/u
Where f is the focal length, v is the distance of the image from the lens, and u is the distance of the object from the lens. We already have f and u, so let's plug those numbers in:
1/0.05 = 1/v - 1/1
Simplifying it with a magic trick, we find:
20 = 1/v - 1
Rearranging the equation, we discover that the distance of the image from the lens (v) is:
v = 1/(20 + 1) = 1/21 meters
Now, let's calculate the size of the image using the magnification formula, which is as hilarious as slipping on a banana peel:
Magnification (M) = -v/u
In this case, the magnificence of the flower (u) is 7.6 cm (converted to 0.076 meters). So, let the calculations begin!
M = -(1/21) / 0.076 = -0.058
Ta-da! The image of the flower has a lovely size of approximately -0.058 times the size of the object. But wait, don't fret! The negative sign just means the image is inverted, like a clown trying to juggle with their feet.
So there you have it, my friend! The position of the image is 1/21 meters away from the lens and it's about -0.058 times the size of the flower. Now, go capture that whimsical photo and make everyone laugh!
To determine the position and size of the image formed by a camera lens, we can use the lens equation and magnification formula. The lens equation is given by:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
Let's calculate the object distance first:
u = 100 cm - 10 cm (convert the focal length to cm)
u = 90 cm
Now, let's rearrange the lens equation to find the image distance:
1/v = 1/f - 1/u
Using the given values:
1/v = 1/5 cm - 1/90 cm
Solving for v:
1/v = (90 - 5) / (90 * 5)
1/v = 85 / 450
v = 450 / 85
v ≈ 5.29 cm
Now, let's calculate the magnification of the image:
magnification (m) = -v / u
Using the calculated values:
m = -5.29 cm / 90 cm
m ≈ -0.06
Since the magnification is negative, it indicates that the image is inverted with respect to the object.
Therefore, the position of the image is approximately 5.29 cm from the lens (on the same side as the object, as it is a real image), and the size of the image is about 0.06 times the size of the object.
To determine the position and size of the image formed by a camera lens, we can use the lens formula and magnification formula.
The lens formula is given by:
1/f = 1/v - 1/u
Where:
f = focal length of the lens
v = image distance from the lens
u = object distance from the lens
In this case, the focal length of the camera lens (f) is +50 mm, which is equivalent to 0.05 m. The object distance (u) is 1 m.
Plugging these values into the lens formula, we can solve for the image distance (v):
1/0.05 = 1/v - 1/1
Simplifying the equation:
20 = 1/v - 1
Now, let's solve for v:
1/v = 20 + 1
1/v = 21
Taking the reciprocal of both sides:
v = 1/21 ≈ 0.0476 m
So, the image distance (v) is approximately 0.0476 m.
Now, let's move on to calculating the size of the image.
The magnification formula is given by:
magnification (m) = height of image (h') / height of object (h)
We know the height of the object (h) is 7.6 cm, which is equivalent to 0.076 m.
To find the height of the image (h'), we can rearrange the formula:
h' = m * h
From the lens formula, we can determine the magnification (m):
m = - v / u
Plugging in the values:
m = -0.0476 / 1
m = -0.0476
Now, let's calculate the height of the image:
h' = -0.0476 * 0.076
h' ≈ -0.0036 m
Note: The negative sign indicates that the image is inverted.
So, the height of the image (h') is approximately -0.0036 m, or 3.6 mm.
Therefore, the position of the image is approximately 0.0476 m from the lens, and the height of the image is approximately 3.6 mm.