He–Ne laser cavity has a spacing of 15 cm between the mirrors, and the optical

mode in the cavity has a diameter of 3 mm (a) Determine the frequency differ

ence between adjacent laser modes. (b) Determine the frequency difference be

tween all possible cavity modes contained within the laser cavity volume. (c) The

He–Ne gas mixture provides optical gain over a frequency width of 1.5 GHz.

Compare the number of laser modes that are within this width to the total number

of cavity modes within this width.

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Ans  a) For finding the frequency difference between two adjacent nodes, we know the relation that: -     L=\frac{c}{\Delta \nu}     where \Delta \nu = frequency difference L = length of the cavity = 15 cm = 0.15 m c = speed of the light = 3 x 108 m/s       0.15=\frac{3 \times 10^{8}}{\Delta \nu}     \Delta \nu=20 \times 10^{8} \mathrm{~Hz}       b) If the frequency difference between adjacent nodes are = 20 x 108 Hz, then the frequency difference between all nodes will be: -     \Delta \nu=\frac{c}{2 L}     \Delta \nu=\frac{3 \times 10^{8}}{2 \times 0.15}     \Delta \nu=1 \times 10^{9} \mathrm{~Hz}         c)   To find the no. of laser modes, we have a simple formula: -     \Delta \nu_{n}=\frac{n c}{2 L}     where \Delta \nu_{n} = frequency width = 1.5 GHz   n = no. of laser modes L = length of the cavity = 15 cm       1.5 \times 10^{9}=\frac{n \times 3 \times 10^{8}}{2 \times 0.15}     n=1.5       ...