Question

In a double-slit arrangement of the figure below, a 0.15 mm, L-140cm.2-(643 nm and y 1.6 cm and y 1.8 cm. a) What is the path difference A for the two rays from the two slits arriving at P. b) Express the path difference c) Does P correspond toa maximum, a minimum or intermediate condition? d) How does the irradiance at the screen vary, if the contribution of one slit is l

Answer 1

X = $\lambda D/ d$

= 643 x 10^-9 x 1.4 / 0.15 x 10^-3

= 6x10^-3 m = 6 mm ... fringe width

as y = n X

n = 3 = odd number = dark fringe.

c) Minimum condition

d)

For distant screen assumption Assumption of infinite source distance gives plane wave at slit so that all amplitude elements are in phase. e' For D>> a this approaches a right angle and θ~ θ Condition for maximum mAD a - slit width

a)

path difference = 3/2 $\lambda$ = 964.5 nm

b)

3/2 $\lambda$ = dsin@

sin@ = 3$\lambda$/2d = 3x643x10^-9 / 2x0.15x10^-3 = 6.43x10^-3

= 0.368 deg