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o-na [289]
3 years ago
10

White light, with uniform intensity across the visible wavelength range of 400 nm - 690 nm, is perpendicularly incident on a wat

er film, of index of refraction 1.33 and thickness 300 nm. At what wavelength (in nanometers) is the light reflected by the film brightest to an observer
Physics
1 answer:
solniwko [45]3 years ago
6 0

Answer:

\lambda=532nm

Explanation:

From the question we are told that:

Wavelength range of white light  400 nm - 690 nm

Index of refraction n=1.33

Thickness T=300*10^{-9}m

Generally Constructive interference is mathematically given by

2nd cos\alpha=(m-\frac{1}{2})\lambda\\\\2nd cos\alpha=2nT\\\\Where \alpha=0\textdegree

Therefore

For m=12*1.33*300nm=(1-\frac{1}{2})\\\\\lambda=\frac{2*1.33*300nm}{(m-\frac{1}{2})}\\\\\lambda=1596nm\\\\For m=2\\\\2*1.33*300nm=(2-\frac{1}{2})\\\\

\lambda=532nm

Therefore the wavelength (in nanometers)  the light reflected by the film brightest to an observer is

\lambda=532nm

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3 years ago
Light enters an equilateral prism with an incident angle of 35° to the normal of the surface. Calculate the angle at which the
julia-pushkina [17]

Answer:

65.9°

Explanation:

When light goes through air to glass

angle of incidence, i = 35°

refractive index, n = 1.5

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1.5=\frac{Sin35}{Sinr}

Sin r = 0.382

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Now the ray is incident on the glass surface.

A = r + r'

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\frac{1}{n}=\frac{Sinr'}{Sini'}

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4 0
3 years ago
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You’ve been given the challenge of balancing a uniform, rigid meter-stick with mass M = 95 g on a pivot. Stacked on the 0-cm end
Mariulka [41]

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Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

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d = 4750n/3.1+95n

6 0
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