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LenaWriter [7]
3 years ago
7

A piece of curved glass has a radius of curvature of r = 10.8 m and is used to form Newton's rings, as in the drawing. Not count

ing the dark spot at the center of the pattern, there are one hundred dark fringes, the last one being at the outer edge of the curved piece of glass. The light being used has a wavelength of 652 nm in vacuum. What is the radius R of the outermost dark ring in the pattern? (Hint: Note that r is much greater than R, and you may assume that tan(θ) = θ for small angles, where θ must be expressed in radians.)
Physics
2 answers:
Sergeeva-Olga [200]3 years ago
8 0

Answer:

The Radius is 2.65cm

Explanation:

MAVERICK [17]3 years ago
7 0

Answer:

Radius of the outer most dark fringe is 2.65 cm

Solution:

As per the question:

Radius of curvature of the glass, r = 10.8 m

No. of dark fringes, n = 100

Wavelength of light, \lambda = 652\ nm = 652\times 10^{- 9}\ m

Now,

To calculate the radius R of the outermost ring:

Radius of the dark fringe of nth order is given by:

R^{2} = nr\lambda = 100\times 10.8\times 652\times 10^{- 9} = 7.042\times 10^{- 4}

R = \sqrt{7.042\times 10^{- 4}} = 0.0265\ m = 2.65\ cm

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Answer:

Height.

Explanation:

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Hence, the property of the object (having a mass of 5 kilograms) which must differ to have different gravitational potential energies is the height from which they are falling from.

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2 years ago
Explain Sound level intensity with mathematical steps?
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Answer:

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8 0
2 years ago
A resistor is connected in series with an AC source that provides a sinusoidal voltage of v of t is equal to V times cosine of b
nekit [7.7K]
<h2>Answer:</h2>

In circuits, the average power is defined as the average of the instantaneous power  over one period. The instantaneous power can be found as:

p(t)=v(t)i(t)

So the average power is:

P=\frac{1}{T}\intop_{0}^{T}p(t)dt

But:

v(t)=v_{m}cos(\omega t) \\ \\ i(t)=i_{m}cos(\omega t)

So:

P=\frac{1}{T}\intop_{0}^{T}v_{m}cos(\omega t)i_{m}cos(\omega t)dt \\ \\ P=\frac{v_{m}i_{m}}{T}\intop_{0}^{T}cos^{2}(\omega t)dt \\ \\ But: cos^{2}(\omega t)=\frac{1+cos(2\omega t)}{2}

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