Two slits are separated by 0215 mm An interference pattern i

Two slits are separated by 0.215 mm. An interference pattern is formed on a screen 51.0 cm away by 656.3-nm light. Calculate the fraction of the maximum intensity a distance 0.600 cm away from the central maximum.

Solution

d = 0.215 mm = 0.215*10^-3 m

R = 51 cm = 0.51 m

lamda = 656.3 nm = 656.3*10^-9 m

y = 0.6 cm = 0.006 m

path diffrence = d*sin(theta)

= 0.215*10^-3*0.006/sqrt(0.51^2 + 0.006^2)

= 2.52*10^-6 m

we know,

phase difference, phi = (2*pi/lamda)*path diffrence

= (2*pi/(656.3*10^-6))*2.52*10^-6

= 0.0241 rad

we know,

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

= Imax*cos^2(0.0241/2)

= 0.9995*Imax