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

Answer:

The required wavelength is 1.19 μm

Explanation:

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

$y_m = \dfrac{m \lambda D}{d}$

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:

$d = \dfrac{(1) \lambda D}{y_1}$

$d = \dfrac{ \lambda D}{y_1}$

replacing the value from the given question, we get:

$d = \dfrac{ (597 \ nm )\times (3.00 \ m)}{4.84 \ mm} \\ \\ d = \dfrac{ (597 \ nm \times (\dfrac{1 \ m}{10^9\ nm}) )\times (3.00 \ m)}{4.84 \ mm(\dfrac{1 \ m}{1000 \ mm })} \\ \\ d = 3.7 \times 10^{-4} \ m$

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

$y_m = (m+\dfrac{1}{2}) \dfrac{\lambda D}{d}$

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

∴

$y_0 = (0+\dfrac{1}{2}) \dfrac{\lambda D}{d}$

$y_0 = (\dfrac{1}{2}) \dfrac{\lambda D}{d}$

making λ the subject of the formula, we have:

$\lambda = \dfrac{2y_o d}{D} \\ \\ \lambda = \dfrac{2(4.84 \ mm) \times \dfrac{1 \ m}{1000 \ mm} (3.7 \times 10^{-4} \ m) }{3.00 \ m}$

$\lambda = 1.19 \times 10^{-6} \ m ( \dfrac{10^6 \mu m }{1\ m}) \\ \\ \lambda = 1.19 \mu m$