In this simulation the thin strip at the top shows a representation of the brigh

Question

In this simulation the thin strip at the top shows a representation of the bright and dark fringes you would observe on a screen a certain distance away from a single or double slit illuminated by monochromatic light. The graph shows another representation of what you would observe, showing a sketch of the intensity of the light striking a particular part of the screen relative to the intensity of the light striking the center of the screen.

(a) Set the slit-to-screen distance to be 0.6 m. With the simulation in single-slit mode, consider the following four cases:
Case 1: Wavelength = 400 nm; Slit width = 0.40 mm.
Case 2: Wavelength = 400 nm; Slit width = 0.60 mm.
Case 3: Wavelength = 600 nm; Slit width = 0.40 mm.
Case 4: Wavelength = 600 nm; Slit width = 0.60 mm.

Rank the four cases according to the width of the central maximum on the screen, from largest to smallest. Use only > and/or = signs in your answer (e.g., 2>1>3=4). Use principles of physics to predict the answer, and then check your answer using the simulation.

(B)For a particular combination of settings you draw lines from the single slit to the screen, and you find that the angle between the lines corresponding to the peak of the central maximum and the first miminum to the right of the central maximum is 8.00 degrees. What is the angle between the peak of the central maximum and the second minimum to the right of the central maximum?

Explanation / Answer

a)

we know, widht of central maximum, w = 2*lamda*R/d

here lamda is wavelength

R is distance between slit to screein

d is slit size.

case 1) W = 2*400*10^-9*0.6/(0.4*10^-3) = 0.0012 m

case 2) W = 2*400*10^-9*0.6/(0.6*10^-3) = 0.0008 m

case 3) W = 2*600*10^-9*0.6/(0.4*10^-3) = 0.0018 m

case 4) W = 2*600*10^-9*0.6/(0.6*10^-3) = 0.0012 m

so, 3 > 1 = 4 > 2

b)

let d is the slit size

path difference for first minimum, d*sin(theta1) = 1*lamda ---(1)

path difference for second minimum, d*sin(theta2) = 2*lamda ---(2)

take equation(2)/equation(1)

sin(theta2)/sin(theta1) = 2

sin(theta2) = 2*sin(8)

sin(theta2) = 0.2783

theta2 = sin^-1(0.2783)

= 16.16 degrees

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