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 find the intensity on the screen at a distance of y = 32mm from the central maximum, we can use the formula for the intensity of the double-slit interference pattern:
I = Imax * cos^2(theta)
where Imax is the intensity at the central maximum, and theta is the angle between the line from the central maximum to the point of interest and the perpendicular to the screen.
To find theta, we can use the small angle approximation, which states that for small angles, sin(theta) ≈ tan(theta) ≈ theta (in radians).
Let's calculate theta first:
Given that the separation of the fringes (d) is 12mm, and the distance between the slits and the screen (D) is 1.6m:
tan(theta) = y / D
= 32mm / 1.6m
≈ 0.02 rad
Now, we can calculate the intensity (I) at y = 32mm using the value of theta:
I = Imax * cos^2(theta)
= Imax * cos^2(0.02 rad)
Since we are not provided with the value of Imax, we cannot give an exact numerical value for the intensity. However, we can explain how to calculate it once the value of Imax is known.
If you have the value of Imax, substitute it into the equation above along with the value of theta to find the intensity at the point of interest.