given (young's double slit experiment the slit separation is 0.08mm and the slits distance to the screen is 1.6 m. the central maxima has intensity Imax and the light has a wavelength of 6nm.
the separation of the fringes was found to be 12mm near the center of the screen.
what is the intensity on the screen at a distance y= 32mm from the central maximum?
To calculate the intensity on the screen at a distance of 32mm from the central maximum, we can use the expression for the intensity in Young's double slit experiment, given by:
I = Imax * cos^2(π * d * sinθ / λ)
Where:
- I is the intensity at a particular point on the screen
- Imax is the intensity at the center maximum
- d is the slit separation
- θ is the angle between the central maximum and the point on the screen
- λ is the wavelength of the light
First, let's find the value of θ for the point on the screen at a distance of 32mm from the central maximum. We can use the small angle approximation, which states that sinθ ≈ θ for small angles. Therefore, we can say:
θ = y / D
Where:
- y is the distance of the point from the central maximum (32mm)
- D is the distance from the slits to the screen (1.6m)
Converting the values to meters, we have:
- y = 0.032m
- D = 1.6m
Now, let's substitute these values into the equation:
θ = 0.032 / 1.6
θ ≈ 0.02 radians
Now, substitute the values of Imax, d, θ, and λ into the equation for intensity:
I = Imax * cos^2(π * d * sinθ / λ)
I = Imax * cos^2(π * (0.08 * 10^-3) * sin(0.02) / 6 * 10^-9)
Simplifying further, using the wavelength in meters:
I = Imax * cos^2(π * (0.08 * 10^-3) * sin(0.02) / (6 * 10^-9))
Now, calculate the value of I using a scientific calculator or a programming language.