1. Due to the wave nature of light, light shined on a single slit will produce a

Question

1. Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shined on a slit with width 0.440 mm.

(a) Find the width of the central maximum located 1.65 m from the slit.

(b) What is the width of the first order fringe?

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2.Working on your car, you spill oil (index of refraction = 1.55) on the ground into a puddle of water (n = 1.33). You notice a rainbow pattern appear across the oil slick. Recalling the lessons you learned in physics class, you realize that you can calculate where constructive and destructive interference occur based on the thickness of the oil slick. (Assume that the average wavelength is 536 nm.)

What are the first three thicknesses necessary for constructive interference? (Enter your answers from smallest to largest.)

_____nm

(b) What are the first three (nonzero) thicknesses necessary for destructive interference? (Enter your answers from smallest to largest.)

_____nm

_____nm _____nm

_____nm

(b) What are the first three (nonzero) thicknesses necessary for destructive interference? (Enter your answers from smallest to largest.)

_____nm _____nm

_____nm

1. Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shined on a slit with width 0.440 mm. (a) Find the width of the central maximum located 1.65 m from the slit. (b) What is the width of the first order fringe? 2.Working on your car, you spill oil (index of refraction = 1.55) on the ground into a puddle of water (n = 1.33). You notice a rainbow pattern appear across the oil slick. Recalling the lessons you learned in physics class, you realize that you can calculate where constructive and destructive interference occur based on the thickness of the oil slick. (Assume that the average wavelength is 536 nm.) What are the first three thicknesses necessary for constructive interference? (Enter your answers from smallest to largest.) (b) What are the first three (nonzero) thicknesses necessary for destructive interference? (Enter your answers from smallest to largest.)

Explanation / Answer

Number 1)

Part A)

For a single slit, the formula to apply is

y/L = m(wavelength)/d

y/1.65 = (1)(520 X 10-9)/(4.4 X 10-4)

y = 1.95 X 10-3 m

The central max is twice that length

2(1.95 X 10-3) = 3.9 X 10-3 m (which is 3.9 mm)

Part B)

The width of any other order fringe is half the central max. That means the answer is

1.95 X 10-3 m (1.95 mm)

Number 2)

Part A)

For constructive interference of this type, the formula to apply is

2nt = (m + .5)(wavelength)

For the first three thicknesses, we will use m = 0, 1, and 2

So...2(1.55)t = (.5)(536 X 10-9)

t = 8.65 X 10-8 m which is 86.5 nm

For the second...

So...2(1.55)t = (1.5)(536 X 10-9)

t = 2.59 X 10-7 m which is 259 nm

For the third

So...2(1.55)t = (2.5)(536 X 10-9)

t = 4.32 X 10-7 m which is 432 nm

So in summary for constructive, the thicknesses are 86.5 nm, 259 nm, and 432 nm

Part B)

The formula just changes to 2nt = m(wavelength) where m = 1, 2, and 3

So...2(1.55)t = (1)(536 X 10-9)

t = 1.73 X 10-9 m which is 173 nm

So...2(1.55)t = (2)(536 X 10-9)

t = 3.46 X 10-7 m which is 346 nm

So...2(1.55)t = (3)(536 X 10-9)

t = 5.19 X 10-7 m which is 519 nm

So in summary for destructive, the thicknesses are 173 nm, 346 nm, and 519 nm

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