Problem

$\cdot$ The dimensions of the picture on a 35 $\m…

05:09

Problem 12 Medium Difficulty

An An LCD projector (see Sec. 25.2$)$ has a projection lens with $f$ -number of 1.8 and a diameter of 46 $\mathrm{mm}$ . The LCD array measures 3.30 $\mathrm{cm} \times 3.30 \mathrm{cm}$ and will be projected on a screen 8.00 m from the lens. If the array is $800 \times 600$ pixels, what will be the dimensions of a single pixel on the screen?

Answer

$=\left(4.13 \times 10^{-3} \mathrm{cm}\right) \times\left(5.5 \times 10^{-3} \mathrm{cm}\right)$

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Video Transcript

read it about. We know that for the a rare not the screen, but the array, the wit of a pixel. So it's calling W some air is equal to, Ah, the size of the is equal to the width of the rare over the member of pixels. So that would be 3.3 centimeters with the width of the array over 800 uh, pixels. Great. So that would be point Oh, for 0 1 to 5 millimetres her pixel. So the width per pixel is 0.4125 millimeters. Similarly, in the array, the height history 0.3 centimeters over 800 over 600 pixels. Because remember wit times height is 800 times 600. So per pixel, we have 0.5 millimetres, all right, And then we're given the F member and the diameter. And so we know that after number after an ratio is equal to F, this is the focal length over diameter. And so the focal length it is his equal to to that number times the diameter. And so that would be 1.8 times 46 millimeters are 4.6 centimeters on DSO. That's so the focal length is 8.28 centimeters. Okay, next to a given that the screen is eight meters away. So as prime image distance his eight meters and that is 800 centimetres on DSO. We know that one of her imaged one of her and so we we want object distance and we know that one of her object distance is equal to one over focal length minus one over image system. Therefore, one of her object distance is equal to one over 8.28 centimeters that we just found minus one over 800 centimetres. Therefore s will work out to be a 0.37 centimeters or 83.7 millimeters. All right, so the next step is to find magnification, and so m is just negative as prime over s. So that's negative. 800 over a 8000.37 both in centimeters. So em magnification is negative. 15. 95.6. That's an inverted image within the magnification of over 95. Therefore, Aunt, I'm going to write it but blew on the screen and we're gonna we're gonna used these values the array of the pixel wit in the Iran, the pics, all height in the array, Tom times the magnification. So on the screen, the whip of a pixel is called W sub s is just the absolute value of the magnification. We don't really care about the sign here, because we just want to know how big it is. Um, Times point. Oh, for one, 25 millimetres. And so I'm sorry to be switching between millimeters and centimetres, by the way, but it comes out to be 0.39 centimeters, which is 2.9 millimeters for the wit and the height of a pixel on the screen is per pixel. Um, and so the height is again magnification sometimes 0.5 the height of the array. And so that comes out to me 0.52 centimeters or five point to me Ah, millimeters per pixel. And that's it.

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