Answer in Optics for Nicholas Legault #87950
The angular position “theta_m” of the first minimum is determined by the equation “sin theta_m = lambda / d”, where “lambda = 725 times 10^{-9}, text{m}” is the wavelength, and “d = 18.5 times 10^{-6}, text{m}” is the width of the slit. We obtain “sin theta_m = 0.725 / 18.5 approx 0.0392”. In view of the smallness of this number, we have “sin theta_m approx theta_m approx 0.0392”. The angular width of the central maximum is twice this angle: “theta = 2 theta_m approx 0.0784”. The angular width in degrees is then “360, theta / 2 pi approx 4.5^circ”. The width in centimetres is proportional to the distance to the screen “R = 125, text{cm}” and is given approximately by “theta R = 0.0784 times 125, text{cm} = 9.8, text{cm}”.
Answer: (a) “4.5^circ”, (b) “9.8, text{cm}”.
