Question

(a) Calculate the distance from the central maximum to the first bright region on either side of the central maximum.
 mm

(b) Calculate the distance between the first and second dark bands in the interference pattern.
 mm

Answer 1

a)
The first maxima after the central maxima occurs at d sin$\theta$ = $\lambda$
Where d is the separation between the slits and $\lambda$ is the wavelength.
sin$\theta$ = $\lambda$/d
= (0.546 x 10^-6) / (0.245 x 10^-3)
= 2.23 x 10^-3.
Since $\theta$ is very small, sin$\theta$ = tan$\theta$
tan$\theta$ = y/D
Where y is the distance from the central maxima to the first maxima and D is the distance to the screen.
y = D tan$\theta$
= 1.23 x 2.23 x 10^-3
= 2.74 x 10^-3 m
= 2.74 mm.

b)
For the first dark band, d sin$\theta$1 = $\lambda$/2
sin$\theta$1 = 0.5$\lambda$ / d
For the second dark band, d sin$\theta$2 = 3$\lambda$/2
sin$\theta$2 = 1.5$\lambda$ / d

Consider y2 and y1 be be distance to the second and first dark band respectively.
y2 - y1 = D [tan$\theta$2 - tan$\theta$1]
= D [sin$\theta$2 - sin$\theta$1]
= D/d[1.5$\lambda$ - 0.5 $\lambda$]
= D$\lambda$/d
= [1.23 x 546.1 x 10^-9] / [0.245 x 10^-3]
= 2.74 x 10^-3 m.
= 2.74 mm

Note: The distance between two successive bright fringe of dark fringe is the same.