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Chapter 24: Problem 56

A camera has a telephoto lens of 240 -mm focal length. The lens can be moved in and out a distance of \(16 \mathrm{mm}\) from the film plane by rotating the lens barrel. If the lens can focus objects at infinity, what is the closest object distance that can be focused?

Short Answer

Expert verified
Answer: The closest object distance that can be focused with these specifications is approximately 3360mm.

Step by step solution

01

Find the Image Distance when focusing on the closest object

When the lens is moved inwards by 16mm, the distance from the film plane is reduced, meaning the image distance is now 240mm - 16mm: \(d_i = 240\mathrm{mm} - 16\mathrm{mm} = 224\mathrm{mm}\) With this new value for \(d_i\), we can calculate the closest object distance.
02

Use the Thin Lens Equation to Calculate the Closest Object Distance

Using the thin lens equation, \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), and the values for \(f\) and \(d_i\) from the previous step, we can find the closest object distance \(d_o\). Substituting the values, we get: \(\frac{1}{240\mathrm{mm}} = \frac{1}{d_o} + \frac{1}{224\mathrm{mm}}\) Now, we need to solve for \(d_o\).
03

Solve for Closest Object Distance

To solve for \(d_o\), first subtract \(\frac{1}{224\mathrm{mm}}\) from both sides: \(\frac{1}{240\mathrm{mm}} - \frac{1}{224\mathrm{mm}} = \frac{1}{d_o}\) Now, find a common denominator: \(\frac{224 - 240}{240 \times 224} = \frac{1}{d_o}\) Simplifying the fraction: \(\frac{-16}{240 \times 224} = \frac{1}{d_o}\) Now, take the reciprocal of the fractions: \(d_o = -\frac{240 \times 224}{16}\) Finally, since object distance is positive, we can remove the negative sign: \(d_o = \frac{240 \times 224}{16}\) Now, calculate the value for \(d_o\): \(d_o \approx 3360\mathrm{mm}\) The closest object distance that can be focused is approximately 3360mm.

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Most popular questions from this chapter

A refracting telescope has an objective lens with a focal length of $2.20 \mathrm{m}\( and an eyepiece with a focal length of \)1.5 \mathrm{cm} .$ If you look through this telescope the wrong way, that is, with your eye placed at the objective lens, by what factor is the angular size of an observed object reduced?
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.
Two lenses, of focal lengths \(3.0 \mathrm{cm}\) and \(30.0 \mathrm{cm},\) are used to build a small telescope. (a) Which lens should be the objective? (b) What is the angular magnification? (c) How far apart are the two lenses in the telescope?
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?
The area occupied by one frame on 35 -mm film is \(24 \mathrm{mm}\) by $36 \mathrm{mm}-\( see the figure with Problem \)16 .$ The focal length of the camera lens is \(50.0 \mathrm{mm}\). A picture is taken of a person $182 \mathrm{cm}$ tall. What is the minimum distance from the camera for the person to stand so that the image fits on the film? Give two answers; one for each orientation of the camera.
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