Question

If we block one of the two slits of a double slit experiment, the intensity coming through the other slit is 1.0 mW.   If the slit spacing, d, is 147 µm, and the wavelength is 700 nm, what is the intensity at 14 degrees if both slits are uncovered? (Note: You may assume that the slit width a≪λ.)

Answer is 2.66 mW....... please show how we got this. Thank you

Answer 1

given

Io = 1 mW

d = 147 micro m

lamda = 700 nm

theta = 14 degrees

we know,
On the screen Intensity at angle theta,

I = 4*Io*cos^2(phi/2) (here phi is phase diffrence)

we know rlation between phase diffrence and path diffrence

path diffrence = (2*pi/lamda)*path difference

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

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

= pi*147*10^-6*sin(14)/(700*10^-9)

= 159.6 radians

so,

I = 4*Io*cos^2(phi/2)

= 4*1*cos^2(159.6) (here use calculatror in radian mode)

= 2.66 mW