Question

Coherent light with wavelength 597 nm passes through two very narrow slits, and theinterference pattern is observed on a screen a distance of 3.00{\rm m} from the slits. The first-order bright fringe is adistance of 4.84 {\rm mm} from the center of the central bright fringe.

For what wavelength of light will thefirst-order dark fringe be observed at this same point on thescreen?

Answer 1

The required wavelength is 1.19 μm

Step-by-step explanation:

In the double-slit study, the formula below determines the position of light fringes
`y_m` on-screen.

`y_m = (m \lambda D)/(d)`

where;

m = fringe order

d = slit separation

λ = wavelength

D = distance between screen to the source

For the first bright fringe, m = 1, and we make (d) the subject, we have:

`d = ((1) \lambda D)/(y_1)`

`d = ( \lambda D)/(y_1)`

replacing the value from the given question, we get:

`d = ( (597 \ nm )* (3.00 \ m))/(4.84 \ mm) \\ \\ d = ( (597 \ nm * ((1 \ m)/(10^9\ nm)) )* (3.00 \ m))/(4.84 \ mm((1 \ m)/(1000 \ mm ))) \\ \\ d = 3.7 * 10^(-4) \ m`

In the double-slit study, the formula which illustrates the position of dark fringes
`y_m` on-screen can be illustrated as:

`y_m = (m+(1)/(2)) (\lambda D)/(d)`

The value of m in the dark fringe first order = 0

∴

`y_0 = (0+(1)/(2)) (\lambda D)/(d)`

`y_0 = ((1)/(2)) (\lambda D)/(d)`

making λ the subject of the formula, we have:

`\lambda = (2y_o d)/(D) \\ \\ \lambda = (2(4.84 \ mm) * (1 \ m)/(1000 \ mm) (3.7 * 10^(-4) \ m) )/(3.00 \ m)`

`\lambda = 1.19 * 10^(-6) \ m ( (10^6 \mu m )/(1\ m)) \\ \\ \lambda = 1.19 \mu m`