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
Step-by-step explanation:
a) Width of the slit, d = 0.1 mm = 0.0001 m
Wavelength of the light,
![\lambda = 461.9 nm = 461.9 * 10^(-9) m](https://img.qammunity.org/2021/formulas/physics/college/54svh2re38ufdekn6u9t9uin3lr8l15xan.png)
The distance of the screen, R = 1 m
![d sin \theta = \lambda](https://img.qammunity.org/2021/formulas/physics/college/w20rn4t1aw2syz3voidrg1237ro3td1p2t.png)
where
is the angle at which first minima is visible
![0.0001 sin \theta = 461.9 * 10^(-9) \\sin \theta =(461.9 * 10^(-9))/(0.0001) \\sin \theta = 0.004619\\\theta = sin^(-1) 0.004619\\\theta = 0.2647^(0)](https://img.qammunity.org/2021/formulas/physics/college/tpu4t2a55o6rhq73nuogn0ib1poq8ntmg7.png)
The angular width of the central maximum =
![2 \theta](https://img.qammunity.org/2021/formulas/physics/college/ln1t3pvmu3w9e78nnsgt13wfiuhp9u6lrs.png)
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
![\lambda = 600 nm = 600 * 10^(-9) m](https://img.qammunity.org/2021/formulas/physics/college/i67zzke7wik91piuucu97onn0cmw7wurel.png)
R = 40 cm = 0.4 m
The distance from central bright band to third dark band, is given by the formula:
![y = (3 \lambda R)/(d) \\y = (3 * 600 * 10^(-9)*0.4 )/(0.0001) \\y = 0.0072 m](https://img.qammunity.org/2021/formulas/physics/college/f0rqno8uwsvbqhoo5kgtl1wu0mb9sfwsn8.png)
y = 0.72 cm