Light intensity through a slit

[ghostops]

Homework Statement

After a laser beam passes through two thin parallel slits, the first completely dark fringes occur at ±14.0∘ with the original direction of the beam, as viewed on a screen far from the slits.

A)What is the ratio of the distance between the slits to the wavelength of the light illuminating the slits?
d/lambda = 2.07

B)What is the smallest positive angle, relative to the original direction of the laser beam, at which the intensity of the light is 110 the maximum intensity on the screen?

Θ=

Homework Equations

I = I_ocos²(ø/2)

ø = 2∏dsinθ/λ

The Attempt at a Solution

so for this I solved for phi and then substituted that into the second equation and for this sinθ is ≈θ I just solved for theta in the 2nd equation

for ø= 2cos^-1(√(1/10))
ø=2.498

then ø/(2∏(d/λ)) = Θ

for Θ I got .192 but I can't figure out what I did wrong

thank you in advance

 

[Simon Bridge]

I'm getting the same as you.
Are you expected to give your answer in degrees or radiens?

 

[ghostops]

radians is what we were told to do for all of these problems

 

[Simon Bridge]

What leads you to believe that you've done anything wrong then?

 

[HalfLostAndSearching]

for any help

I would like to clarify that the question is incomplete as it does not mention the value of the distance between the slits (d) or the wavelength of the light (λ). Without these values, it is not possible to accurately solve for the smallest positive angle at which the intensity of light is 110% of the maximum intensity on the screen.

However, assuming that the value of d/λ is 2.07 as given in part A, we can calculate the value of θ as follows:

ø = 2cos^-1(√(1/10)) = 2.498 radians

Substituting this value into the equation ø = 2πdsinθ/λ, we get:

2.498 = 2π(2.07)sinθ/λ

Solving for θ, we get:

θ = sin^-1(2.498λ/4.14d)

Without knowing the values of d and λ, we cannot accurately solve for θ. Therefore, it is not possible to provide a specific answer to part B of the question.