1)An object is 2.40 mm high, is 8.6 cm to the left of a convex lens of 5.9 cm focal length. Find image location and height.
2) A dentist uses a small mirror of radius 42 mm to locate a cavity in a patient's tooth. If the mirror is concave and is held 18 mm from the tooth. What is the magnification of the image?
To solve these problems, we can use the lens/mirror equation:
1) For the first problem:
Given:
Object height (h₁) = 2.40 mm = 0.0024 m
Object distance (d₁) = -8.6 cm = -0.086 m (negative because the object is to the left of the lens)
Focal length (f) = 5.9 cm = 0.059 m
Using the lens equation:
1/f = 1/d₀ + 1/d₁
Where:
d₀ is the image distance
To find the image distance (d₀):
1/f = 1/d₀ + 1/d₁
1/0.059 = 1/d₀ + 1/-0.086
Solving for d₀:
1/d₀ = 1/0.059 - 1/-0.086
d₀ = 0.0416 m
Now, to find the image height (h₀):
According to the magnification equation:
h₀/h₁ = -d₀/d₁
Solving for h₀:
h₀ = h₁ * (-d₀/d₁)
h₀ = 0.0024 * (-0.0416/-0.086)
Answer:
The image is located 0.0416 m to the right of the lens, and the image height is 0.0012 m.
2) For the second problem:
Given:
Radius of curvature (r) = 42 mm = 0.042 m
Object distance (d₁) = -18 mm = -0.018 m (negative because the object is to the left of the mirror)
Using the mirror equation:
1/f = 1/d₀ + 1/d₁
Where:
d₀ is the image distance
To find the image distance (d₀):
1/f = 1/d₀ + 1/d₁
1/0.042 = 1/d₀ + 1/-0.018
Solving for d₀:
1/d₀ = 1/0.042 - 1/-0.018
d₀ = 0.025 m
Now, to find the magnification of the image (M):
According to the magnification equation:
M = -d₀/d₁
Substituting the values:
M = -0.025/-0.018
Answer:
The magnification of the image is 1.39.
Note: Negative magnification indicates an inverted image.
To solve both of these problems, we can use the lens/mirror formula, which is
1/f = 1/do + 1/di,
where "f" is the focal length of the lens/mirror, "do" is the object distance, and "di" is the image distance.
We also need to understand the magnification formula, which is
magnification = -di/do,
where the negative sign indicates a virtual or inverted image.
Let's solve each problem step by step:
1) Object distance (do) = 8.6 cm = 86 mm
Focal length (f) = 5.9 cm = 59 mm
Using the lens formula:
1/59 = 1/86 + 1/di
Simplifying the equation:
1/di = 1/59 - 1/86
1/di = (86 - 59)/(86 * 59)
di = 86 * 59 / (86 - 59)
di ≈ 135.44 mm
The image distance is approximately 135.44 mm.
To find the image height, we can use the magnification formula:
magnification = -di/do
Substituting the given values:
magnification = -135.44/86
magnification ≈ -1.57
Since the magnification is negative, it indicates an inverted image. The height of the image can be found by multiplying the object height by the magnification:
Image height = (2.40 mm) * (-1.57)
Image height ≈ -3.77 mm
The approximate image location is 135.44 mm to the right of the lens, and the height is approximately -3.77 mm.
2) Radius of the mirror (f) = 42 mm
Object distance (do) = -18 mm (negative because the object is on the same side as the mirror)
Using the mirror formula:
1/42 = 1/-18 + 1/di
Simplifying the equation:
1/di = 1/42 + 1/18
1/di = (18 + 42)/(42 * 18)
di = 42 * 18 / (18 + 42)
di ≈ 24.00 mm
The image distance is approximately 24.00 mm.
To find the magnification:
magnification = -di/do
Substituting the given values:
magnification = -24.00/-18
magnification ≈ 1.33
The magnification is positive, indicating an upright image. The magnification of the image is approximately 1.33.