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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 鈮� 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

A microscope has an eyepicce that gives an angular magnification of 5.00 for a final image at infinity and an objective lens of focal length $15.0 \mathrm{mm}\(. The tube length of the microscope is \)16.0 \mathrm{cm} .$ (a) What is the transverse magnification due to the objective lens alone? (b) What is the angular magnification due to the microscope? (c) How far from the objective should the object be placed?
An object is located \(10.0 \mathrm{cm}\) in front of a converging lens with focal length \(12.0 \mathrm{cm} .\) To the right of the converging lens is a second converging lens, \(30.0 \mathrm{cm}\) from the first lens, of focal length \(10.0 \mathrm{cm} .\) Find the location of the final image by ray tracing and verify by using the lens equations.
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.
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.
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.
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