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kykrilka [37]
3 years ago
15

The index of refraction for silicate flint glass is 1.66 for violet light that has a wavelength in air equal to 400 nm and 1.61

for red light that has a wavelength in air equal to 700nm. A ray of 700-nm wavelength red light and a ray of 400-nm wavelength violet light both have angles of refraction equal to 30 degrees upon entering the glass from the air.
(a) Which is greater, the angle of incidence of the ray of red light or the angle of incidence of the ray of violet light? Explain your answer.

(b) What is the difference betweeen the angles of incidence of the two rays?
Physics
1 answer:
nikitadnepr [17]3 years ago
4 0

Answer:

(a) Angle of incidence for violet is more than the angle of incidence for red

(b) 2.4°

Explanation:

refractive index for violet , v = 1.66

refractive index for red, nR = 1.61

wavelength for violet, λv = 400 nm

wavelength for red, λR = 700 nm

Angle of refraction, r = 30°

(a) Let iv be the angle of incidence for violet.

Use Snell,s law

nv = Sin iv / Sin r

1.66 = Sin iv / Sin 30

Sin iv = 0.83

iv = 56°

Use Snell's law for red

nR  = Sin iR / Sin r  

where, iR be the angle of incidence for red

1.61 = Sin iR / Sin 30

Sin iR = 0.805

iR = 53.6°

So, the angle of incidence for violet is more than red.

(b) iv - iR = 56° - 53.6° = 2.4°

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A paperweight consists of a 8.55-cm-thick plastic cube. Within the plastic a thin sheet of paper is embedded, parallel to opposi
ella [17]

Answer:

\mu=1.5322

Explanation:

Given:

Thickness of the paperweight cube, x=8.55\ cm

apparent depth from one side of the inbuilt paper in the plastic cube, i=4.02\ cm

apparent depth from the other side of the inbuilt paper in the plastic cube, i'=1.56\ cm

Now as we know that refractive index is given as:

\rm \mu=\frac{real\ depth}{apparent\ depth}

  • Let the real depth form first side of the slab be, d
  • Then the depth from the second side of the slab will be, d'=x-d=8.55-d

Since refractive index for an amorphous solid is an isotropic quantity so it remains same in all the direction for this plastic.

\mu=\mu'

\frac{d}{i} =\frac{d'}{i'}

\frac{d}{4.02}=\frac{8.55-d}{1.56}

d=6.1597\ cm

Now the refractive index:

\mu=\frac{d}{i}

\mu=\frac{6.1596}{4.02}

\mu=1.5322

7 0
3 years ago
What component of a longitudinal sound wave is analogous to a trough of a transverse wave?
anzhelika [568]

Explanation:

There are two components of a longitudinal sound wave which are compression and rarefaction. Similarly, there are two components of the transverse wave, the crest, and trough.

The crest of a wave is defined as the part that has a maximum value of displacement while the trough is defined as the part which corresponds to minimum displacement.

While compression is that space where the particles are close together while the rarefaction is that space where the particles are far apart from each other.

So, the refraction or the rarefied part of a longitudinal sound wave is analogous to a trough of a transverse wave.  

6 0
3 years ago
A puddle of water has a thin film of gasoline floating on it. A beam of light is shining perpendicular on the film. If the wavel
g100num [7]

Answer:

option A

Explanation:

Given,

wavelength of light,\lambda = 560\ nm

refractive index of gasoline, n₁ = 1.40

Refractive index of water, n₂ = 1.33

thickness of the film, t = ?

Condition of constructive interference is given by

2 n t = (m+\dfrac{1}{2})\lambda

For minimum thickness of the film m = 0

From the question we can clearly observe that phase change from gasoline to air

so, n = 1.4

2 \times 1.4 \times t = \dfrac{560}{2}

t = 100\ nm

Hence, the correct answer is option A

7 0
4 years ago
A 60 [Hz] high-voltagepower line carries a current of 1000 [A]. The power line is at a height of 50 [m] above the earth. What is
omeli [17]

Answer:

The magnitude of the magneticfield B[T] at a point on the surface of the earth directly below the power line is B=1.496x10^(-6) T.

Explanation:

The distance from the wire to a point in the surface is the heigth of the wire.

The formula for the magnetic field on any point at distance R from a wire conducting alternating current is:

B=\frac{\mu_0I}{2\pi R} =\frac{(4\pi\cdot 10^{-7}T\cdot m/A)(1000\,A)}{2\pi \cdot 50\,m} =1.496\cdot10^{-6}\,T

4 0
3 years ago
An ideal spring obeys hooke's law: f = −bikx. a mass of m = 0.3 kg hung vertically from this spring stretches the spring 0.13 m.
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The force = 0.3 * 9.8 = 2.94
x = 0.13

k = 2.94/0.13
k = 22.6 N/m
5 0
3 years ago
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