Two slits are illuminated with red light ($\lambda$= 650 nm).The slits are 0.25 mm apart and the distance to the screen is 1.25 m.What fraction of the maximum intensity on the screen is the intensity measured at a distance 3.0 mm from the central maximum?
A)0.94
B)0.92
C)0.96
D)0.98
E)0.99

Question

Quizplus · Accepted Answer

The formula for calculating the intensity of the interference pattern is:
I = I0cos^2((πd sinθ)/λ)
where I0 is the intensity of each slit, d is the distance between the slits, θ is the angle between the central maximum and the point on the screen being measured, and λ is the wavelength of the light.
At a point 3.0 mm from the central maximum, we can calculate the value of sinθ as:
sinθ = (3.0 mm) / (1.25 m)
sinθ = 0.0024
Substituting the values into the formula and simplifying, we get:
I = I0cos^2(π(0.25 mm)(0.0024)/(650 nm))
I = I0(0.94)^2
I = 0.88I0
Therefore, the fraction of the maximum intensity measured at a distance 3.0 mm from the central maximum is:
0.88 / I0
To find the fraction of the maximum intensity on the screen, we need to look at the intensity at the central maximum, which is given by:
Icentral = 4I0
Therefore, the fraction of the maximum intensity on the screen is:
0.88 / 4I0 = 0.22I0 / Icentral
0.22I0 / Icentral = 0.22 / 4
0.22 / 4 = 0.055
1 - 0.055 = 0.945
Therefore, the fraction of the maximum intensity on the screen is 0.945, or 0.94 to two significant figures. The answer is (A).

Two Slits Are Illuminated with Red Light λ\lambdaλ = 650 Nm)The Slits Are 0

Two Slits Are Illuminated with Red Light $\lambda$ = 650 Nm)The Slits Are 0