An object with a height of 42 cm is placed 1.6 m in front of a concave mirror with a focal length of 0.45 m. Find the location and magnification of the image produced by the mirror, using the mirror and magnification equations.
First, you need to learn the equations. It's not that hard.
1/Do + 1/Di = 1/f
Do is the object distance from the mirror and Di is the distance to the image from the mirror. f is the focal length.
Use that equation to solve for Di.
1/1.6 + 1/Di = 1/0.45
1/Di = 1.597
Di = 0.6261 m
The magnification is M = Di/Do.
Well, well, well, let's solve this optical puzzle! We have a concave mirror with a focal length of 0.45 m and an object with a height of 42 cm.
Now, let's use the mirror equation:
1/f = 1/v - 1/u
Where f is the focal length, v is the image distance, and u is the object distance.
Plugging in the values:
1/0.45 = 1/v - 1/1.6
Now, let's solve this equation and find the image distance (v).
1/0.45 = 1/v - 1/1.6
2.22 = 1/v - 0.625
1/v = 2.22 + 0.625
1/v = 2.845
v = 1/2.845
v ≈ 0.351 m
So, the image distance (v) is approximately 0.351 m.
Now, let's find the magnification (m) using the magnification equation:
m = -v/u
Plugging in the values:
m = -0.351/1.6
m ≈ -0.219
Therefore, the location of the image produced by the mirror is approximately 0.351 m in front of the mirror and the magnification is about -0.219.
Hope that puts a smile on your face!
To find the location of the image produced by the mirror, we can use the mirror equation:
1/f = 1/d₀ + 1/dᵢ
Where:
f = focal length of the mirror
d₀ = object distance from the mirror
dᵢ = image distance from the mirror
Plugging in the given values, we get:
1/0.45 = 1/1.6 + 1/dᵢ
To solve for dᵢ, we can rearrange the equation:
1/dᵢ = 1/0.45 - 1/1.6
Combining the fractions, we get:
1/dᵢ = (1.6 - 0.45)/(0.45 * 1.6)
Simplifying further:
1/dᵢ = 1.15 / 0.72
Dividing both sides by 1.15:
1/dᵢ = 1.597
Taking the reciprocal of both sides:
dᵢ = 1/1.597 ≈ 0.626 m
Therefore, the location of the image produced by the mirror is approximately 0.626 meters in front of the mirror.
To find the magnification of the image, we can use the magnification equation:
magnification (m) = -dᵢ / d₀
Plugging in the values:
m = -0.626 / 1.6
Simplifying:
m ≈ -0.391
Therefore, the magnification of the image produced by the mirror is approximately -0.391.
To find the location and magnification of the image produced by the concave mirror, we can use two important equations: the mirror equation and the magnification equation.
1. The mirror equation:
The mirror equation relates the object distance (do), image distance (di), and the focal length (f) of the mirror. It is given by:
1/f = 1/do + 1/di
Where:
- f is the focal length of the mirror.
- do is the object distance from the mirror (positive for objects in front of the mirror).
- di is the image distance from the mirror (positive for real images on the same side as the object, but negative for virtual images).
2. The magnification equation:
The magnification equation relates the height of the object (ho), the height of the image (hi), and the magnification (M):
M = hi / ho = -di / do
Where:
- ho is the height of the object.
- hi is the height of the image.
- M is the magnitude of the magnification (positive for upright images, but negative for inverted images).
Let's solve the problem step by step using the given information:
Given:
- ho (height of the object) = 42 cm = 0.42 m (converted to meters).
- do (object distance) = 1.6 m.
- f (focal length of the mirror) = 0.45 m.
1. To find the image distance (di), we can use the mirror equation:
1/f = 1/do + 1/di
Substituting the given values:
1/0.45 = 1/1.6 + 1/di
Now solve for di:
1/di = 1/0.45 - 1/1.6
1/di = (1.6 - 0.45) / (0.45 * 1.6)
1/di = 1.15 / 0.72
di = 0.72 / 1.15
di = 0.626 m
2. To find the magnification (M), we can use the magnification equation:
M = -di / do
Substituting the given values:
M = -0.626 / 1.6
M ≈ -0.391
Therefore, the location of the image is approximately 0.626 m in front of the concave mirror, and the magnification of the image is approximately -0.391.