A camera is supplied with two interchangeable lenses, whose focal lengths are 35.0 and 150.0 mm. A woman whose height is 1.80 m stands 7.00 m in front of the camera. What is the height (including sign) of her image on the film, as produced by
(a) the 35.0-mm lens
(b) the 150.0-mm lens?
You can use the lens equation to find (given fo, objectdistance) image distance.
Then, use the magnification equation to find image height (di/do=hi/di)
http://glenbrook.k12.il.us/GBSSCI/PHYS/Class/refrn/u14l5f.html
To find the height of the woman's image on the film, we can use the thin lens equation:
1/f = 1/do + 1/di
Where:
- f is the focal length of the lens
- do is the object distance (7 m in this case)
- di is the image distance (the distance from the lens to the film plane)
Let's calculate the height of the woman's image for each lens:
(a) 35.0-mm lens:
- f = 35.0 mm = 0.035 m
- do = 7 m
The formula above can be rearranged to solve for di:
di = 1 / (1/f - 1/do)
di = 1 / (1/0.035 - 1/7)
di ≈ 0.036 m
Now, we need to calculate the height of the image using similar triangles:
hi/ho = di/do
Where:
- hi is the image height
- ho is the object height (1.80 m in this case)
Substituting the values:
hi/1.80 = 0.036/7
hi = (0.036 * 1.80) / 7
hi ≈ 0.009 m
The height of the woman's image produced by the 35.0-mm lens is approximately 0.009 m.
(b) 150.0-mm lens:
- f = 150.0 mm = 0.150 m
- do = 7 m
Using the thin lens equation:
di = 1 / (1/0.150 - 1/7)
di ≈ 0.143 m
Using similar triangles to find the image height:
hi/1.80 = 0.143/7
hi = (0.143 * 1.80) / 7
hi ≈ 0.037 m
The height of the woman's image produced by the 150.0-mm lens is approximately 0.037 m.
To find the height of the woman's image on the film produced by each lens, we can use the lens formula:
1/f = 1/d₀ + 1/dᵢ
Where:
f is the focal length of the lens,
d₀ is the object distance (distance between the woman and the lens), and
dᵢ is the image distance (distance between the lens and the image).
Let's calculate the height of the woman's image for each lens:
(a) 35.0-mm lens:
Given:
f = 35.0 mm
d₀ = 7.00 m
To convert the object distance to the same unit as the focal length (mm), multiply by 1000:
d₀ = 7.00 m * 1000 = 7000 mm
Let's substitute the values into the lens formula:
1/35.0 = 1/7000 + 1/dᵢ
To find dᵢ, rearrange the equation:
1/dᵢ = 1/35.0 - 1/7000
Now calculate dᵢ:
1/dᵢ = (1 - 1/7000)/35.0
Taking the reciprocal of both sides:
dᵢ = 35.0/(1 - 1/7000)
Now we can calculate the height of the image using the magnification formula:
m = -dᵢ/d₀
The negative sign indicates that the image is upside down relative to the object.
Substitute the values:
m = -dᵢ/d₀ = -(35.0/(1 - 1/7000))/7000
Now we can find the height of the image by multiplying the magnification by the height of the object:
Height of Image = m * Height of Object = m * 1.80 m
(b) 150.0-mm lens:
Repeat the same steps as above, but this time with the given focal length of 150.0 mm:
f = 150.0 mm
Follow the same process to calculate the image distance, magnification, and finally, the height of the image.
This explanation outlines the steps to solve the problem. To find the actual numerical values, you can substitute the given values into the equations and calculate them using a calculator or a computer program.