Question

Coherent light with wavelength 610 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?Express your answer in micrometers(not in nanometers).

Answer 1

Answer:

= 1220 nm

= 1.22 μm

Explanation:

given data:

wavelength $\lambda = 610 nm = 610\times 10 ^{-9} m$

distance of screen from slits D = 3 m

1st order bright fringe is 4.84 mm

condition for 1 st bright is

$d sin \theta =\lambda$     ---( 1)

and$tan \theta = \frac{y}{ D}$

$\theta = tan^{-1}\frac{(y }{D})$

= 0.0924 degrees

plug theta value in equation 1 we get

$d sin ( 0.0924) = 610 \times 10 ^{-9}$

$d = 3.78\times 10^{-4} m$

condition for 1 st dark fringe

$d sin \theta =\frac{λ'}{2}$

$\lambda '= 2 d sin\theta$

= 2λ    since from eq (1)

= 1220 nm

= 1.22 μm