How far apart are an object and an image formed by a 75.0cm focal length converging lens if the
image is 2.5x larger than the object and is real.
Please, help me.
Use two equations to get di and do. First, 1/f=1/do+1/di (textbooks usually use p and q).
Also, M = -di/do.
Since the image is real, di is positive and M is negative (image is upside down). Using the second equation, -2.5 = -di/do. Solving for di, di=2.5*do.
We can substitute this into first equation to get, 1/75 = 1/do + 1/(2.5*do). Solving for do, do = 105.
Now we put this into either equation, for example, 1/75 = 1/105 + 1/di and solve for di: di = 262.5
So finally, di + do = 105 + 262.5 = 367.5 cm
Why did the object go on a vacation? Because it needed some space from the image! In this case, we can use the magnification formula: magnification = -image distance / object distance. Since the image is 2.5 times larger, the magnification is 2.5. Rearranging the formula, we get the object distance = -image distance / magnification. Since the image is real, the image distance will be positive. Plugging in the values, we get object distance = -image distance / 2.5. Now, we also know that the focal length of the lens is 75.0 cm. Using the lens formula: 1/focal length = 1/object distance - 1/image distance, we can substitute the values we've found to solve for the image distance. Once we have the image distance, we can find the object distance by rearranging the formula. So imagine the lens and the numbers stepping back a bit to create some distance, and let's solve the problem together!
To find the separation between the object and the image formed by a converging lens, we can use the lens formula:
1/f = 1/di - 1/do
Where:
f is the focal length of the lens
di is the image distance from the lens
do is the object distance from the lens
Given:
f = 75.0 cm
Object-to-image magnification (M) = 2.5
We can use the magnification formula:
M = -di/do
Rearranging this equation, we get:
do = -di/M
Substituting the given values into the above equation, we have:
do = -di/2.5
Now, let's substitute the value of do in the lens formula:
1/75.0 = 1/di - 2.5/di
Combining the fractions on the right side:
1/75.0 = (1 - 2.5)/di
Simplifying the equation further:
1/75.0 = -1.5/di
Cross-multiplying:
di = 75.0/-1.5
di = -50.0 cm
Since the image is real, the negative sign indicates that the image is formed on the opposite side of the lens from the object.
Now, let's find do using the magnification formula:
M = di/do
Substituting the given values:
2.5 = -50.0/do
Cross-multiplying:
do = -50.0/2.5
do = -20.0 cm
Again, the negative sign indicates that the object is on the opposite side of the lens from the image.
Therefore, the object and the image are 20.0 cm apart.
To solve this problem, we need to use the lens formula, which relates the object distance (u), image distance (v), and focal length (f) of a lens. The lens formula is given by:
1/f = 1/v - 1/u
In this case, we are given the focal length (f = 75.0 cm) and that the image formed is real. For a converging lens, a real image is formed when the object is placed beyond the focal point.
Since the image is 2.5 times larger than the object, we can say that the image height (h') is 2.5 times larger than the object height (h). This gives us the equation:
h' = 2.5h
Now, let's substitute these values into the magnification formula:
magnification (m) = h'/h = -v/u
Since the image is real, the magnification (m) is positive, so we can rewrite the equation as:
m = v/u
Solving for v:
v = mu
We know that the focal length (f) can be expressed as:
f = 1/(2πn(1/R1 - 1/R2))
Where n is the refractive index of the lens material and R1 and R2 are the radii of curvature of the lens surfaces.
To find v, we first need to find the value of u. We'll use the equation:
1/f = 1/v - 1/u
Since the image is formed on the same side as the object (real image), the value of v is negative. We can rewrite the equation as:
1/f = -1/v - 1/u
Substituting the given values:
1/0.75 = -1/v - 1/u
Simplifying, we get:
1/v + 1/u = -1/0.75
Now, let's substitute the value of v from the magnification equation:
1/u + 1/(mu) = -1/0.75
Simplifying further:
1/u + m/u = -1/0.75
Substituting the magnification value (m = 2.5):
1/u + (2.5/u) = -1/0.75
Combining like terms:
(3.5/u) = -1/0.75
Cross multiplying:
1/u = -(3.5)(0.75)
Simplifying:
1/u = -2.625
Taking the reciprocal of both sides:
u = -1/2.625
u ≈ -0.38 cm
Now that we have the value of u, we can substitute it back into the lens formula to find v:
1/f = 1/v - 1/u
Substituting the given values:
1/0.75 = 1/v + 1/(-0.38)
Simplifying:
1/v = 1/0.75 - 1/(-0.38)
Adding the fractions:
1/v = (2/3) + (2/5)
Finding a common denominator:
1/v = (10/15) + (6/15)
Combining the fractions:
1/v = 16/15
Taking the reciprocal of both sides:
v = 15/16
v ≈ 0.94 cm
Therefore, the object distance (u) is approximately -0.38 cm and the image distance (v) is approximately 0.94 cm. The two are approximately 1.32 cm apart.