A certain digital camera having a lens with focal length 7.5 cm focuses on an object 1.85 m tall that is 4.25 m from the lens. (a) How far must the lens be from the photocells? (b) How tall is the image on the photocells? Is it erect or inverted? (c) A SLR digital camera often has pixels measuring 8.0 um by 8.0 um. How many such pixels does the height of this image cover?
I have part a and b, but I can't figure out c! help please!
To solve part (c) of the problem, we need to use the concept of image magnification. The image height is given by the ratio of the image distance to the object distance, multiplied by the object height.
Given:
Focal length (f) = 7.5 cm = 0.075 m
Object height (h) = 1.85 m
Object distance (d₀) = 4.25 m
Pixel size (ps) = 8.0 μm = 8.0 x 10^(-6) m
We have already solved part (a) and found the image distance (dᵢ) = 12.70 m.
Image magnification (M) is given by the formula:
M = -dᵢ / d₀
Substituting the values, we get:
M = -12.70 m / 4.25 m
M = -2.988
This means the image is inverted.
To find the height of the image on the photocells, we use the formula:
Image height (hᵢ) = M * h
Substituting the values, we get:
hᵢ = -2.988 * 1.85 m
hᵢ ≈ -3.48 m (approximated to two decimal places)
The height of the image on the photocells is approximately -3.48 m. However, since height cannot be negative, we take the absolute value of the result: |hᵢ| ≈ 3.48 m.
Finally, to find the number of pixels the height of this image covers, we divide the height of the image by the pixel size:
Number of pixels = |hᵢ| / ps
Substituting the values, we get:
Number of pixels ≈ 3.48 m / (8.0 x 10^(-6) m)
Number of pixels ≈ 435,000 pixels
Therefore, the height of the image on the photocells covers approximately 435,000 pixels.