Ahe-ne laser (light amplification by stimulated emission radiation) is created by filling a glass cylinder with he-ne gas and exciting it with a current (like a neon tube) which then emits light at a single wavelength: λ = 633 nm for our he-ne laser. the two caps of the cylinder are mirrored, one completely reflective and the other only partially so that light can escape, thus acting as an aperture for the laser beam. the cylinder is called a resonant cavity because the reflection of light from the ends of the tube establishes a standing wave along the length of the tube, and only resonant frequencies are amplified and escape the tube. a) in actuality, the stimulated emission has a very small bandwidth: λ = 633.000 ± 0.001 nm. because of this, for an arbitrary length of the resonant cavity there will in fact be several resonant frequencies. i. suppose the length of the resonant cavity was 60 cm. how many resonant frequencies are there? ii. will shortening or lengthening the tube reduce the number of resonant frequencies? explain. iii. is there an optimal length the resonant cavity should be so that only one resonant frequency exists? if so what is it? b) as the beam exits the aperture, it is not perfectly parallel due to diffraction. i. if the aperture has a diameter of 2 mm, what is the angular deflection of the beam, i. e. the angle from parallel to the first intensity minimum? what if the aperture size is 2 µm? does this increase or decrease by decreasing the size of the aperture? explain. ii. suppose the laser can create a beam with an output power of 1 mw. note that reducing the size of the aperture would increase the intensity of the emitted beam as it exits the laser. why? iii. suppose we want to create a dot on a screen 5 m away from the laser, and we want the dot to have maximal intensity. considering your answer to the previous question, explain why there is a minimum aperture size that will achieve maximum intensity of the dot on the screen. in particular, argue that the optimal aperture size occurs when the size of the aperture is equal to the size of the dot on the screen. you may want to use ray diagrams in your explanation. iv. what is the optimal size of the aperture such that the intensity of the dot on a screen 5 m away is maximum?