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Chapter 24: Problem 17

A photographer wishes to take a photograph of the Eiffel Tower $(300 \mathrm{m}\( tall ) from across the Seine River, a distance of \)300 \mathrm{m}$ from the tower. What focal length lens should she use to get an image that is 20 mm high on the film?

Short Answer

Expert verified
Answer: The photographer should use a lens with a focal length of approximately 19.96 mm.

Step by step solution

01

Write down the given information

The height of the Eiffel Tower is 300 m, the distance from the photographer to the Tower (object distance) is 300 m, and we need an image 20 mm high on the film. We must find the focal length of the lens.
02

Determine the object distance (d_o) and image distance (d_i)

The object distance (d_o) is the distance from the lens to the object, which is 300 m.
03

Determine the magnification formula

The magnification formula is: magnification = (image height) / (object height) = (-d_i) / (d_o)
04

Solve for the image height

The image height is given as 20 mm. Since we're dealing with distances in meters, let's convert the image height to meters: 20 mm = 0.02 m.
05

Calculate the magnification

Now we will calculate the magnification using the formula. magnification = (0.02 m) / (300 m) = 1 / 15000
06

Determine the thin lens formula

The thin lens formula is: 1 / f = 1 / d_o + 1 / d_i
07

Use magnification and thin lens formulas to solve for f

Since magnification = (-d_i) / (d_o), we can rewrite this equation in terms of d_i: d_i = -d_o * magnification Now plug in the values for d_o and magnification: d_i = -300 m * (1 / 15000) = -0.02 m The image will be 0.02 m behind the lens. Now, use the thin lens formula to find the focal length: 1 / f = (1 / 300) + (1 / (-0.02))
08

Solve for the focal length

We will now solve for the focal length, f: 1 / f = (1 / 300) - (1 / 0.02) f = 1 / ((1 / 300) - (1 / 0.02)) f ≈ 0.01996 m or 19.96 mm The photographer should use a lens with a focal length of approximately 19.96 mm to get an image of the Eiffel Tower that is 20 mm high on the film.

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Most popular questions from this chapter

An astronomical telescope provides an angular magnification of 12. The two converging lenses are \(66 \mathrm{cm}\) apart. Find the focal length of each of the lenses.
The distance from the lens system (cornea + lens) of a particular eye to the retina is \(1.75 \mathrm{cm} .\) What is the focal length of the lens system when the eye produces a clear image of an object \(25.0 \mathrm{cm}\) away?
A microscope has an eyepiece of focal length \(2.00 \mathrm{cm}\) and an objective of focal length \(3.00 \mathrm{cm} .\) The eyepiece produces a virtual image at the viewer's near point \((25.0 \mathrm{cm}\) from the eye). (a) How far from the eyepiece is the image formed by the objective? (b) If the lenses are \(20.0 \mathrm{cm}\) apart, what is the distance from the objective lens to the object being viewed? (c) What is the angular magnification?
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?
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?
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