Solving Film Projection & Telescope Questions

AI Thread Summary
To determine the focal length of a projection lens for a film projection apparatus, the calculations show that a focal length of approximately 456 ft is required to project a 12 ft wide image onto a screen 50 ft away. For the astronomical telescope, the eyepiece focal length is calculated to be about 1.6 in, leading to an objective lens focal length of approximately 25.6 in when considering the telescope's total length of 24 in. These calculations utilize magnification formulas relevant to both scenarios. The provided solutions clarify the necessary focal lengths for effective image projection and telescope functionality. This guidance is crucial for completing the lab report accurately.
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URGENT Can you please help!

Hello,

I have just finished what seems like an infinite amount of questions for this lab report. 2 of them I am just not sure how to proceed. Any help would be appreciated.

Question 1:
In a film projection apparatus, it is desired to produce pictures 12 ft wide on a screen 50 ft from the lens. The size of the picture on the film is 1 in. wide. What must be the focal length of the projection lens used?

Question 2:
The magnifying power of an astronomical telescope 24 in. long is equal to 15. Determine the focal length of the objective lens and that of the eyepiece. Assume the telescope is focused on a distant object.

Thank you.
 
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Hello there,

I can definitely assist you with these questions. For question 1, we can use the formula: M = d/D = f/F, where M is the magnification, d is the distance between the lens and the film, D is the distance between the lens and the screen, f is the focal length of the lens, and F is the distance between the film and the screen. We know that M = 12 ft/1 in. = 144 in., D = 50 ft, and F = 50 ft - 12 ft = 38 ft. Plugging these values into the formula, we get: 144 = f/38, which gives us a focal length of 5472 in. or 456 ft. So, the focal length of the projection lens should be approximately 456 ft.

For question 2, we can use the formula: M = f/fe, where M is the magnification, f is the focal length of the objective lens, and fe is the focal length of the eyepiece. We know that M = 15 and f = 24 in. Plugging these values into the formula, we get: 15 = 24/fe, which gives us a focal length of the eyepiece of 1.6 in. To find the focal length of the objective lens, we can use the formula: f = fe + d, where d is the distance between the two lenses. We know that d = 24 in. and fe = 1.6 in. Plugging these values into the formula, we get: f = 24 + 1.6 = 25.6 in. So, the focal length of the objective lens should be approximately 25.6 in.

I hope this helps! Let me know if you have any further questions or need clarification on any steps. Good luck with your lab report!
 
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