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Chapter 2: Problem 54

A camera lens used for taking close-up photographs has a focal length of \(22.0 \mathrm{mm}\). The farthest it can be placed from the film is \(33.0 \mathrm{mm}\). (a) What is the closest object that can be photographed? (b) What is the magnification of this closest object?

Short Answer

Expert verified
The closest object distance that can be photographed is 66.0 mm, and the magnification of this closest object is -0.5, which indicates a reduction in size and an inversion of the image on the film plane.

Step by step solution

01

(a) Calculate the closest object distance (s)

First, let's find the closest object distance (s) using the lensmaker's equation: \(\frac{1}{f} = \frac{1}{s} + \frac{1}{i}\) Here, we are given the values f = 22.0 mm and i = 33.0 mm. Substitute these values into the equation: \(\frac{1}{22.0} = \frac{1}{s} + \frac{1}{33.0}\) To solve for s, subtract \(\frac{1}{33.0}\) from both sides: \(\frac{1}{s} = \frac{1}{22.0} - \frac{1}{33.0}\) Now we will find a common denominator and subtract the fractions: \(\frac{1}{s} = \frac{33 - 22}{(22)(33)} = \frac{11}{(22)(33)}\) Now we can find s by taking the reciprocal of the fraction: \(s = \frac{(22)(33)}{11} = \frac{726}{11} = 66.0\,\mathrm{mm}\) The closest object distance (s) is 66.0 mm.
02

(b) Calculate the magnification (m) of the closest object

Now let's find the magnification (m) using the magnification equation: \(m = -\frac{i}{s}\) We already know the values of i = 33.0 mm and s = 66.0 mm. Substitute these values into the equation: \(m = -\frac{33.0}{66.0}\) Simplify to find the magnification (m): \(m = -0.5\) The magnification of the closest object is -0.5, which indicates a reduction in size and an inversion of the image on the film plane.

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