Question

1

(a) What is the minimum angular spread (in rad) of a 694 nm wavelength ruby laser beam that is originally 1.28 mm in diameter? ___ rad

(b) If this laser is aimed at a mountain cliff 14.5 km away, how big will the illuminated spot be? (Give the diameter of the spot in meters.) ____ m

(c) How big a spot would be illuminated on the Moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round-trip time and, hence, distance. Assume the distance from the surface of the Earth to the surface of the Moon is 3.82 ✕ 10⁸ m. Give the diameter of the spot in kilometers.) ______ km

2

a) Find the largest wavelength (in nm) of light falling on double slits separated by 5.50 µm for which there is a second-order maximum.

____nm

Answer 1

X VOXD @using the relation = 60614 x 164 x 3.82 X ☺ ~ 1.22 2x 108 252,694.8 Substitute guen veles 9 021-22 x 694 X10 1.28X103 O 0.0006614 252-65 x 10 m thus final diameter of light be amt neglecting the thickneus Duis 22 - 505 km @ ou 6:614710"rad low N 6.604 Xio Yadane -4 6 19.1806 km D=14.9 km Sowse of ligha beam & mountain blow distance us D final diameter would be 22 + D 505 km Tano = opp adi Tano XD Tanlo') = 0 radians (small) 2 = OXD 360614X109 x 14.5 X 10% 2998905903 m 2x + D = 2 (9.5903) + 16103 neglecting D as it is much smaller than b distance 92 19.1806 m final diameter of light beam on mountain is 19.1806m