Respuesta :
Answer:
a) [tex]\theta_A = 20.49^{0}[/tex], [tex]\theta_B = 1.003^{0}[/tex]
b) y1 = 0.75 m, y2 = 0.035 m
Explanation:
Wavelength, [tex]\lambda = 700 nm = 700 * 10^{-9} m[/tex]
Distance between the slits and the viewing screen, L = 2m
Slit spacing for experiment A, [tex]d_{A} = 2 \mu m = 2 * 10^{-6} m[/tex]
Slit spacing for experiment B, [tex]d_{B} = 40 \mu m = 40 * 10^{-6} m[/tex]
For maximum light intensity, [tex]n \lambda = d sin \theta[/tex]
a) For experiment A, [tex]n \lambda = d_A sin \theta[/tex]
n = 1 ( first maximum )
[tex]700 * 10^{-9} = 2 * 10^{-6} sin \theta_A\\sin \theta_A = \frac{700 * 10^{-9}}{2 * 10^{-6}} \\sin \theta_A = 0.35\\ \theta_A = sin^{-1} 0.35\\\theta_A = 20.49^{0}[/tex]
For experiment B, [tex]n \lambda = d_B sin \theta[/tex]
n = 1 ( first maximum )
[tex]700 * 10^{-9} = 40 * 10^{-6} sin \theta_B\\sin \theta_B = \frac{700 * 10^{-9}}{40 * 10^{-6}} \\sin \theta_B = 0.0175\\ \theta_B = sin^{-1} 0.0175\\\theta_B = 1.003^{0}[/tex]
b) The location on the screen of the first maximum:
y = L tanθ
L = 2 m
For experiment A, the location on the screen of the first maximum is calculated as:
[tex]y_1 = L tan \theta_{A}[/tex], [tex]\theta_A = 20.49^{0}[/tex]
[tex]y_1 = 2 tan 20.49\\y_1 = 0.75 m[/tex]
For experiment B, the location on the screen of the first maximum is calculated as:
[tex]y_2 = L tan \theta_{B}[/tex], [tex]\theta_B = 1.003^{0}[/tex]
[tex]y_2 = 2 tan 1.003\\y_2 = 0.035 m[/tex]
