A microscope has an objective lens of focal length \(5.00 \mathrm{mm} .\) The objective forms an image \(16.5 \mathrm{cm}\) from the lens. The focal length of the eyepiece is \(2.80 \mathrm{cm} .\) (a) What is the distance between the lenses? (b) What is the angular magnification? The near point is \(25.0 \mathrm{cm} .\) (c) How far from the objective should the object be placed?

: We know that for a lens, the lens formula is given by \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\) For the objective lens, \(d_o\) = distance of the object from the objective lens (unknown) \(d_i\) = distance of the image formed by the objective lens = \(0.165 \mathrm{m}\) \(f_{objective}\) = focal length of the objective lens = \(0.005 \mathrm{m}\) Plugging these values into the lens formula, we get \(\frac{1}{0.005} = \frac{1}{d_o} + \frac{1}{0.165}\) Solving for \(d_o\): \(d_o = \frac{1}{(\frac{1}{0.005} - \frac{1}{0.165})} = 0.00536 \mathrm{m}\) Now, let's consider the eyepiece which acts as a magnifying glass. For the eyepiece, object distance from eyepiece = image distance (formed by the objective lens) - distance between the lenses = \(0.165 - d_{lenses}\). We also know that for maximum magnification, the eyepiece should be adjusted to form the final image at the near point (\(0.25 \mathrm{m}\)). So, \(d_i^{(2)} = 0.25 \mathrm{m}\) \(f_{eyepiece} = 0.028 \mathrm{m}\) Applying the lens formula for the eyepiece, \(\frac{1}{0.028} = \frac{1}{0.165 - d_{lenses}} + \frac{1}{0.25}\) We now solve for the distance between the lenses \(d_{lenses}\): \(d_{lenses} = 0.165 - \frac{1}{(\frac{1}{0.028} - \frac{1}{0.25})} = 0.132 \mathrm{m}\)

: Angular magnification \(M\) is given by the formula: \(M = 1 + \frac{d_i^{(2)}}{f_{eyepiece}}\) Plugging in the values from our calculations: \(M = 1 + \frac{0.25}{0.028} = 9.93\) The angular magnification is approximately 9.93.

: We have already found the object distance from the objective lens, \(d_o\), in step (a): \(d_o = 0.00536 \mathrm{m}\) or \(5.36 \mathrm{mm}\) The object should be placed approximately \(5.36 \mathrm{mm}\) away from the objective lens.