Question

4. An ideal double-slit slide is illuminated by laser light with a wavelength of 750 nm. The slits are spaced 0.25 mm apart. The interference pattern is observed on a screen 2.0 m behind the slits. A. What is the bright fringe spacing on the screen? B. What is the smallest angle (with respect to the center of the screen) at which the light exiting the slide is perfectly destructive? C. What is the distance from the center of the screen to the first intensity minimum? D. What is the distance from the center of the screen to the first location the light intensity on the screen is exactly 44 the maximum intensity?

Answer 1

maxima is at bright fringes. condition for maxi d-sino = m.de bright - dark si 19 sino = 4 for small o. D do y mind, m= 0, 1, 2, - O y = t.D d © for first minima. y = 50x10 9 x 2 0.25x10 3 y = 6.08153m ole لاام (y=6mm] d-sino =) sino = 2 = 750x10-9 2xd 2x 0.25X103 sino=1.50x10 3 (0=1:5X105 radians ] center. distance from qi & ind Digino = d.o.sin 1.58103 3.onom y = 3x103 m @ 8 = 21 dsina 1 = 4 Lo Los2012 16 456 los 8/2 = 14 . Lost 02, 4 क = 2.6362

21. dosina sino =/ 0 - 21. dxy J.D .y 2.6362 2 TX 0.250 10 3 x 150x1592 ya 2. 82X10-3 (y = 2.52 mm] from central maxima.

Answer 2

given
lamda = 750 nm = 750*10^-9 m
d = 0.25 mm = 0.25*10^-3 m
R = 2 m

A) bright fringe spacing = lamda*R/d

= 750*10^-9*2/(0.25*10^-3)

= 0.006 m (or) 0.60 cm (or) 6.0 mm <<<<<<<<<<------------Answer

B) for 1st dark fringe, d*sin(theta) = lamda/2

sin(theta) = lamda/(2*d)

= 750*10^-9/(2*0.25*10^-3)

= 0.0015

theta = sin^-1(0.0015)

= 0.0859 degrees <<<<<<<<<<------------Answer

C) the distance from the center of the screen to the first minimum = (1/2)*lamda*R/d

= (1/2)*750*10^-9/(2*0.25*10^-3)

= 7.5*10^-4 m (or) 0.75 mm <<<<<<<<<<------------Answer

D) let phi is the phase diffrence

I = I_max*cos^2(phi/2)

(1/4)*I_max = cos^2(phi/2)

1/2 = cos(phi/2)

cos(60) = cos(phi/2)

==> phi = 120 degrees

= 2*pi/3 radians

now use,

2*d*sin(theta) = (lamda/(2*pi))*phi

2*d*sin(theta) = (lamda/(2*pi))*(2*pi/3)

2*d*sin(theta) = lamda/3

sin(theta) = lamda/(6*d)

= 750*10^-9/(6*0.25*10^-3)

= 0.0005

theta = sin^-1(0.0005)

= 0.0286 degrees

now use, tan(theta) = y/R

==> y = R*tan(theta)

= 2*tan(0.0286)

= 9.98*10^-4 m (or) 0.998 mm <<<<<<<<<<------------Answer