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Chapter 24: Problem 59

A slide projector, using slides of width \(5.08 \mathrm{cm},\) produces an image that is \(2.00 \mathrm{m}\) wide on a screen \(3.50 \mathrm{m}\) away. What is the focal length of the projector lens?

Short Answer

Expert verified
#tag_title# Step 2: Convert the units. #tag_content# To make all the units consistent, let's convert the image width and image distance to centimeters: - Image width: \(2.00 \mathrm{m} = 200 \mathrm{cm}\) - Image distance: \(3.50 \mathrm{m} = 350 \mathrm{cm}\) Now we have: - Object width (width of the slide) \(= 5.08 \mathrm{cm}\) - Image width \(= 200 \mathrm{cm}\) - Image distance \(= 350 \mathrm{cm}\)

Step by step solution

01

Write down the known values.

We are given the following information: - Object width (width of the slide) \(= 5.08 \mathrm{cm}\) - Image width \(= 2.00 \mathrm{m}\) - Distance between the projector lens and the screen (image distance) \(= 3.50 \mathrm{m}\)

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Most popular questions from this chapter

A converging lens with focal length \(3.00 \mathrm{cm}\) is placed $4.00 \mathrm{cm}$ to the right of an object. A diverging lens with focal length \(-5.00 \mathrm{cm}\) is placed \(17.0 \mathrm{cm}\) to the right of the converging lens. (a) At what location(s), if any, can you place a screen in order to display an image? (b) Repeat part (a) for the case where the lenses are separated by \(10.0 \mathrm{cm}.\)
A nearsighted man cannot clearly see objects more than \(2.0 \mathrm{m}\) away. The distance from the lens of the eye to the retina is \(2.0 \mathrm{cm},\) and the eye's power of accommodation is \(4.0 \mathrm{D}\) (the focal length of the cornea-lens system increases by a maximum of \(4.0 \mathrm{D}\) over its relaxed focal length when accommodating for nearby objects). (a) As an amateur optometrist, what corrective eyeglass lenses would you prescribe to enable him to clearly see distant objects? Assume the corrective lenses are $2.0 \mathrm{cm}$ from the eyes. (b) Find the nearest object he can see clearly with and without his glasses.
A slide projector has a lens of focal length \(12 \mathrm{cm} .\) Each slide is \(24 \mathrm{mm}\) by \(36 \mathrm{mm}\) (see the figure with Problem 16). The projector is used in a room where the screen is \(5.0 \mathrm{m}\) from the projector. How large must the screen be?
Unless the problem states otherwise, assume that the distance from the comea- lens system to the retina is \(2.0 \mathrm{cm}\) and the normal near point is \(25 \mathrm{cm}.\) If the distance from the lens system (cornea + lens) to the retina is $2.00 \mathrm{cm},$ show that the focal length of the lens system must vary between \(1.85 \mathrm{cm}\) and \(2.00 \mathrm{cm}\) to see objects from $25.0 \mathrm{cm}$ to infinity.
An object is located at \(x=0 .\) At \(x=2.50 \mathrm{cm}\) is a converging lens with a focal length of \(2.00 \mathrm{cm},\) at \(x=16.5 \mathrm{cm}\) is an unknown lens, and at \(x=19.8 \mathrm{cm}\) is another converging lens with focal length \(4.00 \mathrm{cm} .\) An upright image is formed at $x=39.8 \mathrm{cm} .$ For each lens, the object distance exceeds the focal length. The magnification of the system is \(6.84 .\) (a) Is the unknown lens diverging or converging? (b) What is the focal length of the unknown lens? (c) Draw a ray diagram to confirm your answer.
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