Answer:
a) Angular width of the central maximum = 0.5294°
b) The distance of the third dark band from the central bright band = 0.72 cm
Explanation:
a) Width of the slit, d = 0.1 mm = 0.0001 m
Wavelength of the light, [tex]\lambda = 461.9 nm = 461.9 * 10^{-9} m[/tex]
The distance of the screen, R = 1 m
[tex]d sin \theta = \lambda[/tex]
where [tex]\theta[/tex] is the angle at which first minima is visible
[tex]0.0001 sin \theta = 461.9 * 10^{-9} \\sin \theta =\frac{461.9 * 10^{-9}}{0.0001} \\sin \theta = 0.004619\\\theta = sin^{-1} 0.004619\\\theta = 0.2647^{0}[/tex]
The angular width of the central maximum = [tex]2 \theta[/tex]
Angular width of the central maximum = 2 * 0.2647
Angular width of the central maximum = 0.5294°
b) d = 0.1 mm = 0.0001 m
[tex]\lambda = 600 nm = 600 * 10^{-9} m[/tex]
R = 40 cm = 0.4 m
The distance from central bright band to third dark band, is given by the formula:
[tex]y = \frac{3 \lambda R}{d} \\y = \frac{3 * 600 * 10^{-9}*0.4 }{0.0001} \\y = 0.0072 m[/tex]
y = 0.72 cm
