Question

When performing optical spectroscopy (for example, photoluminescence or Raman spectroscopy), a laser beam is focused on the sample to be investigated by means of a lens having a focal distance \(f\). Assume that the laser beam exits a pupil \(D_{\mathrm{g}}\) in diameter that is located at a distance \(d_{0}\) from the focusing lens. For the case when the image of the exit pupil forms on the sample, calculate a) at what distance \(d_{\mathrm{i}}\) from the lens is the sample located. b) the diameter \(D_{i}\) of the laser spot (image of the exit pupil) on the sample. c) the numerical results for: \(f=10.0 \mathrm{~cm}, D_{\mathrm{o}}=2.00 \mathrm{~mm}, d_{\mathrm{o}}=1.50 \mathrm{~m}\).

Short answer

Question: Calculate the distance between the lens and the sample, and the diameter of the laser spot on the sample in an optical spectroscopy setup using a lens with a focal length of 10 cm, an object distance of 1.50 m, and a 2.00 mm wide exit pupil. Answer: The distance between the lens and the sample is 0.1034 m, and the diameter of the laser spot on the sample is 0.1378 mm.

Step-by-step solution

Apply the Thin-Lens Formula

We will be using the thin-lens formula to find the image distance di. The thin-lens formula is \(\frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i}\), where f is the focal distance, do is the object distance, and di is the image distance.

Solve for Image Distance \(d_{i}\)

Rearrange the thin-lens formula to obtain di, by using the values from the problem: \(d_i = \frac{1}{\frac{1}{f} - \frac{1}{d_o}}\) Plug in the given values: \(d_i = \frac{1}{\frac{1}{10.0\,\mathrm{cm}} - \frac{1}{1.50\,\mathrm{m}}}\) Now, calculate the value of di: \(d_i = 0.1034\,\mathrm{m}\) This is the distance between the lens and the sample.

Apply the Magnification Formula

To find the diameter of the laser spot on the sample (also known as the "image pupil"), we can use the magnification formula: \(M = \frac{d_i}{d_o} = \frac{D_i}{D_g}\) Here, M is the magnification, Di is the diameter of the image pupil (laser spot), and Dg is the diameter of the exit pupil.

Solve for Image Pupil Diameter \(D_{i}\)

Rearrange the magnification formula to solve for Di: \(D_i = MD_g\) Substitute the values found previously: \(D_i = \frac{0.1034\,\mathrm{m}}{1.50\,\mathrm{m}} \times 2.00\,\mathrm{mm}\) Now, calculate the value of Di: \(D_i = 0.1378\,\mathrm{mm}\) This is the size of the laser spot on the sample (image of the exit pupil).

Provide the Numerical Results

The distance between the lens and the sample is \(d_i = 0.1034\,\mathrm{m}\), and the diameter of the laser spot on the sample (image of the exit pupil) is \(D_i = 0.1378\,\mathrm{mm}\).