You have found in the lab an old microscope, which has lost its eyepiece. It still has its objective lens, and markings indicate that its focal length is \(7.00 \mathrm{~mm}\). You can put in a new eyepiece, which goes in \(20.0 \mathrm{~cm}\) from the objective. You need a magnification of about 200. Assume you want the comfortable viewing distance for the final image to be \(25.0 \mathrm{~cm}\). You find in a drawer eyepieces marked \(2.00-, 4.00-,\) and \(8.00-\mathrm{cm}\) focal length. Which is your best choice?

We will use the lens equation to find the image distance (\(v\)) of the objective lens: \( \frac{1}{f} = \frac{1}{u} + \frac{1}{v} \) In this case, we know the focal length (\(f = 7 \mathrm{~mm}\)) and the object distance (\(u = 20 \mathrm{~cm} - 7 \mathrm{~mm} = 19.3 \mathrm{~cm}\)), so we can solve for \(v\): \( \frac{1}{v} = \frac{1}{7 \mathrm{~mm}} - \frac{1}{19.3 \mathrm{~cm}} \) Keep all values in the same unit (centimeters), and solve for \(v\): \( \frac{1}{v} = \frac{1}{0.7 \mathrm{~cm}} - \frac{1}{19.3 \mathrm{~cm}}\) \( \frac{1}{v} ≈ 1.429 - 0.052\) \( \frac{1}{v} ≈ 1.377 \) \(v ≈ \frac{1}{1.377} ≈ 0.726 \mathrm{~cm}\)

To find the magnification produced by the objective lens alone, we can use the formula: \( M_1 = - \frac{v}{u} \) \( M_1 = - \frac{0.726 \mathrm{~cm}}{19.3 \mathrm{~cm}} ≈ -0.0376 \)

Since we need a total magnification of about 200, let the magnification required by the eyepiece be \(M_2\). To find \(M_2\), we can use the formula: \( M_2 = \frac{M_{total}}{M_1} \) \( M_2 = \frac{200}{-0.0376} ≈ -5324 \)

Now, we want to find the best eyepiece lens that produces a magnification of about -5324. We can use the lens equation once again to find out which lens best fits the requirement. For each of the eyepieces, calculate their magnification using the lens equation: For \(f = 2 \mathrm{~cm}\): \( \frac{1}{25 \mathrm{~cm} - 2 \mathrm{~cm}} - \frac{1}{2 \mathrm{~cm}} = \frac{1}{v_2} \) By calculating we get: \( v_2 ≈ -7.06 \mathrm{~cm}\) The magnification produced by this lens, \( M_2 = - \frac{v_2}{25-2} ≈ -0.352 \) For \(f = 4 \mathrm{~cm}\): \( \frac{1}{25 \mathrm{~cm} - 4 \mathrm{~cm}} - \frac{1}{4 \mathrm{~cm}} = \frac{1}{v_3} \) By calculating we get: \( v_3 ≈ -18.5 \mathrm{~cm}\) The magnification produced by this lens, \( M_2 = - \frac{v_3}{25-4} ≈ -1 \) For \(f = 8 \mathrm{~cm}\): \( \frac{1}{25 \mathrm{~cm} - 8 \mathrm{~cm}} - \frac{1}{8 \mathrm{~cm}} = \frac{1}{v_4} \) By calculating we get: \( v_4 ≈ -83.7 \mathrm{~cm}\) The magnification produced by this lens, \( M_2 = - \frac{v_4}{25-8} ≈ -5.518 \) None of the lenses produce the exact magnification we want. However, the \(8.00 \mathrm{~cm}\) focal length lens has the closest approximate magnification of \(-5.518\) to the required value of \(-5324\). Therefore, the best choice is the \(8.00 \mathrm{~cm}\) focal length eyepiece lens.