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xz_007 [3.2K]
2 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]2 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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A formula for the normal systolic blood pressure for a man age a , measured in mmhg, is given as p=0. 006a2−0. 02a 120. Find the
djyliett [7]

The age of a man whose normal blood pressure measures 123 mm of hg

9 years

<h3>What is Quadratic equation ?</h3>

A quadratic equation as an equation of degree 2, meaning that the highest exponent of this function is 2. The standard form of a quadratic equation is y = ax^{2} + bx + c, where a, b, and c are numbers and a cannot be 0

P(A) = 0.006 a^{2} - 0.02a + 120

123 = 0.006- 0.02a + 120

0=0.006 a^{2} - 0.02a - 3

you can use the quadratic equation  formula to solve for the man's age.

A = (-b ± (\sqrt{b^{2} - 4*a*c})  ) / (2a)

A = (0.02 ±  \sqrt{(-0.02)^{2} - 4*0.006*(-3)}/ (2*0.006)

A = (0.02 ± \sqrt{0.0076}) / 0.012

A = 9 , -5.67

Age of the man will be 9 years

To learn more about quadratic equation  here

brainly.com/question/17177510?referrer=searchResults

#SPJ4

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