Double Slit Intensity Problem

[navm1]

Homework Statement

Coherent light with wavelength 585 nm passes through two very narrow slits that are
0.320 mm apart. The screen is 0.700 m away from the slits. The intensity is I0 at the
centre of the central maximum (theta=0)

(a) What is the phase difference (in radians) in the light from the two slits at an angle
of 23degrees from the centre line?

(b) What is the distance on the screen from the centre of the central maximum to the
point where the intensity has dropped to I0/2?

Homework Equations

Φ=2π/λ*dθ
I=I_0*cos²Φ/2

The Attempt at a Solution

a)
(2pi*3.2x10^-4)/(5.85x10^-7)*23pi/180=439pi radians

b) I=I_0cos^2(Φ/2)

cos^2((pi/2)/2)=1/2

So i figured Φ=2pi but the equation i wanted to use to solve is

Φ=(2*pi*d)/(λ)*θ

but i don't have a theta. For part a it was 23 but i figured there will be a completely new angle for this part of the question.

Thanks

 

[mfb]

but i don't have a theta

Well, you have to calculate it with that formula.
You can get Φ with first equation you have in (b). It is not 2 pi (plug it in and see if the result would be right).

 

[navm1]

Well, you have to calculate it with that formula.
You can get Φ with first equation you have in (b). It is not 2 pi (plug it in and see if the result would be right).

i meant to put pi sorry.

d is slit separation right? i think i was getting mixed up and thinking it was distance from screen to centre

 

[mfb]

Looks like slit separation in your formula.

Pi is still wrong. Please stop guessing and calculate the value.

 

[HalfLostAndSearching]

for your help!

I would approach this problem by first understanding the concept of the double slit experiment and how it relates to the interference of light. The intensity of light at a point on the screen is determined by the superposition of the waves from the two slits. The phase difference between the waves from the two slits is crucial in determining the intensity at a particular point on the screen.

(a) To find the phase difference, we can use the equation Φ=2π/λ*dθ, where λ is the wavelength of light, d is the distance between the slits, and θ is the angle from the center line. Plugging in the given values, we get Φ=2π/5.85x10^-7 * 0.320x10^-3 * 23π/180 = 0.439π radians.

(b) The intensity at a point on the screen is given by I=I_0*cos^2(Φ/2). To find the distance at which the intensity drops to I_0/2, we can set I=I_0/2 and solve for Φ. This gives us Φ=π/2 radians. Plugging this into the equation Φ=2π/λ*dθ and solving for θ, we get θ=λ/2d=5.85x10^-7/(2*0.320x10^-3)=0.915 degrees. This means that the distance on the screen from the center of the central maximum to the point where the intensity drops to I_0/2 is 0.915 degrees.

Overall, the key to solving this problem is understanding the concept of interference and using the appropriate equations to calculate the phase difference and intensity at different points on the screen.