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xz_007 [3.2K]
3 years ago
11

In a double-slit experiment, the slits are illuminated by a monochromatic, coherent light source having a wavelength of 609 nm.

An interference pattern is observed on the screen. The distance between the screen and the double-slit is 1.69 m and the distance between the two slits is 0.118 mm. A light wave propogates from each slit to the screen. What is the path length difference between the distance traveled by the waves for the sixth-order maximum (bright fringe) on the screen?
Physics
1 answer:
Ray Of Light [21]3 years ago
6 0

Answer:

\Delta x = 3.65 \mu m

Explanation:

As we know that the sixth order maximum will have path difference given as

\Delta x = N\lambda

here we know that

N = order of maximum

\lambda = 609 nm

now we have

N = 6

so we know that

\Delta x = 6(609 nm)

\Delta x = 3.65 \mu m

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Answer:

Explanation:

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A person walks from her home at position O to position F by taking a path that is comprised of three displacement vectors travel
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Answer:

Explanation:

A = 155 m at 25° North of west

B = 92 m due north

C = 64.7 m at an angle 50° East of north

Write the displacements in the vector form

\overrightarrow{A} = 155\left ( Cos25\widehat{i}-Sin25\widehat{j} \right )

\overrightarrow{A}=140.5\widehat{i}-65.5\widehat{j}

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\overrightarrow{C} = 64.7\left ( Cos50\widehat{i}+Sin50\widehat{j} \right )

\overrightarrow{C} = 41.6\widehat{i}+49.6\widehat{j}

(a)

\overrightarrow{A}.\overrightarrow{C}=\left ( 140.5\widehat{i}-65.5\widehat{j} \right ).\left ( 41.6\widehat{i}+49.5\widehat{j} \right )

\overrightarrow{A}.\overrightarrow{C} = 2602.55

(b)

\overrightarrow{A}\times \overrightarrow{C}=\left ( 140.5\widehat{i}-65.5\widehat{j} \right )\times \left ( 41.6\widehat{i}+49.5\widehat{j} \right )

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