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(a) The screen is 3.20m from the split.
(b) The closest minima for green, distance Δy = 0.45 cm.
When a wave hits a barrier or opening, numerous events are referred to as diffraction. It is described as the interference or bending of waves via an aperture or around the corners of an obstruction into the area that forms the geometric shadow of the obstruction or aperture.
(a)Equation of minima = sinθ = mλ/α
Given, m = 3, λ = 6.33X10⁻⁷, α = 0.00015
Putting the values in formula to get θ.
θ = sin⁻¹ ( ) = 0.01266 rad
triangle need to be drawn to find relationship between θ, y$ and L
tan(θ) = y/L where; y = 4.05 cm
L = y/tan(θ) = 3.20
Hence, the screen is 3.20m from the split.
(b) Find the closest minima for green
minima equation is sinθ = mλ/α where, m = 4 (minima with smallest distance)
sinθ = 4λ/α
θ = sin⁻¹ () = 0.01688 rad
Calculate L using
tanθ = y/L
L = 4.5 cm
From equation subtract y₃ from y:
4.50 cm - 4.05 cm = 0.45 cm
Hence, distance Δy = 0.45 cm.
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I understand that the question you are looking for is "Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the ot
Question
Two lasers, one red (with wavelength 633.0 nm) and the other green (with wavelength 532.0 nm), are mounted behind a 0.150-mm slit. On the other side of the slit is a white screen. When the red laser is turned on, it creates a diffraction pattern on the screen.
a. The distance y3,red from the center of the pattern to the location of the third diffraction minimum of the red laser is 4.05 cm. How far L is the screen from the slit? Express this distance L in meters to three significant figures.
b. With both lasers turned on, the screen shows two overlapping diffraction patterns. The central maxima of the two patterns are at the same position. What is the distance Δy between the third minimum in the diffraction pattern of the red laser (from Part A) and the nearest minimum in the diffraction pattern of the green laser?
Answer:
D = -4/7 = - 0.57
C = 17/7 = 2.43
Explanation:
We have the following two equations:
First, we isolate C from equation (2):
using this value of C from equation (3) in equation (1):
<u>D = - 0.57</u>
Put this value in equation (3), we get:
<u>C = 2.43</u>