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Zinaida [17]
4 years ago
14

A narrow beam of light containing red (660 nm) and blue (470 nm) wavelengths travels from air through a 1.00 cm thick flat piece

of crown glass and back to air again. The beam strikes at an incident angle of 30 degrees. (a) At what angles do the two colors emerge
Physics
1 answer:
miskamm [114]4 years ago
8 0

Answer:

The color blue emerges at 19.16° and the color red emerges at 19.32°.

Explanation:

The angle at which the two colors emerge can be calculated using the Snell's Law:

n_{1}sin(\theta_{1}) = n_{2}sin(\theta_{2})

Where:

n₁ is the refractive index of the incident medium (air) = 1.0003

n₂ is the refractive index of the refractive medium:

    blue light in crown glass = 1.524

    red light in crown glass = 1.512

θ₁ is the angle of the incident light = 30°

θ₂ is the angle of the refracted light                            

<u>For the red wavelengths we have:</u>

\theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.512}) = 19.32 ^{\circ}

<u>For the blue wavelengths we have</u>:

\theta_{2} = arcsin(\frac{n_{1}sin(\theta_{1})}{n_{2}}) = arcsin(\frac{1.0003*sin(30)}{1.524}) = 19.16 ^{\circ}

Therefore, the color blue emerges at 19.16° and the color red emerges at 19.32°.  

I hope it helps you!

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HELP ASAPPP which statement explains the first law of thermodynamics a)energy is created and destroyed or b) energy is transform
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Potential energy due to an objects compression or extension is called _______________________ energy.
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4 years ago
A wire 40cm long and of diameter 0.60mm has a resistance of 1.5 ohm what is the resistivity of the material of which it is made
zavuch27 [327]

Answer:

The resistivity of the wire is 1.05\times 10^{-6}\ \Omega-m

Explanation:

We have,

Length of a wire is 40 cm or 0.4 m

Diameter of a wire is 0.60 mm

Radius is 0.3 mm or 0.0003 m

Resistance of a wire is 1.5 ohm

Now we need to find the resistivity of the material of which it is made. The resistance of a wire in terms of its resistance, length and area is given by :

R=\rho \dfrac{l}{A}

\rho = resistivity

\rho=R \dfrac{\pi r^2}{l}\\\\\rho=1.5\times  \dfrac{3.14\times (0.0003 )^2}{0.4}\\\\\rho=1.05\times 10^{-6}\ \Omega-m

So, the resistivity of the wire is 1.05\times 10^{-6}\ \Omega-m.

3 0
3 years ago
Problem 4 A meteoroid is first observed approaching the earth when it is 402,000 km from the center of the earth with a true ano
Nikitich [7]

Answer:

Part a: The eccentricity is 1.086.

Part b: The altitude at closest approach is 5088 km

Part c: The velocity at perigee is 8.516 km/s

Part d: The turn angle is 134.08 while the aiming radius is 5641.28 km

Explanation:

<h2>Part a </h2>

Specific energy is given by

\epsilon=\frac{v^2}{2}-\frac{\mu}{r}

Here

  • ε is the specific energy
  • v is the velocity which is given as 2.23 km/s
  • μ is the gravitational constant whose value is 398600
  • r is the distance between earth and the meteorite which is 402,000 km

                         \epsilon=\frac{v^2}{2}-\frac{\mu}{r}\\\epsilon=\frac{2.2^2}{2}-\frac{398600}{402,000}\\\epsilon=1.495 km^2/s^2

Value of specific energy is also given as

\epsilon=\frac{\mu}{2a}\\a=\frac{\mu}{2\epsilon}\\a=\frac{398600}{2\times 1.495}\\a=13319 km

Orbit formula is given as

r=a(\frac{e^2-1}{1+ecos \theta})\\ae^2-recos\theta-(a+r)=0

Putting values in this equation and solving for e via the quadratic formula gives

ae^2-recos\theta-(a+r)=0\\(133319)e^2-(402000)(cos 150) e-(133319+402000)=0\\133319e^2+348142.21 e-535319=0\\\\e=\frac{-348142.21 \pm \sqrt{348142.21^2-4(133319)(535319)}}{2 (133319)}\\\\e=1.086 \, or \, -3.69

As the value of eccentricity cannot be negative so the eccentricity is 1.086.

<h2>Part b</h2>

The radius of trajectory at perigee is given as

r_p=a(e-1)\\

Substituting values gives

r_p=133319 (1.086-1)\\r_p=11465.4 km

Now for estimation of altitude z above earth is given as

z=r_p-R_E\\z=11465.4-6378\\z=5087.434\\z\approx 5088 km

So the altitude at closest approach is 5088 km

<h2>Part c</h2>

radius of perigee is also given as

r_p=\frac{h^2}{\mu}\frac{1}{1+e}

Rearranging this equation gives

h=\sqrt{r_p\mu(1+e)}\\h=\sqrt{11465.4 \times 3986000 \times (1+1.086)}\\h=97638.489 km^2/s

Now the velocity at perigee is given as

v_p=\frac{h}{r_p}\\v_p=\frac{97638.489}{11465.4}\\v_p=8.516 km/s\\

So the velocity at perigee is 8.516 km/s

<h2>Part d</h2>

Turn angle is given as

\delta =2 sin^{-1} (\frac{1}{e})

Substituting value in the equation gives

\delta =2 sin^{-1} (\frac{1}{e})\\\delta =2 sin^{-1} (\frac{1}{1.086})\\\delta =134.08

Aiming radius is given as

\Delta =a \sqrt{e^2-1}

Substituting value in the equation gives

\Delta =a \sqrt{e^2-1}\\\Delta =13319 \sqrt{1.086^2-1}\\\Delta=5641.28 km

So the turn angle is 134.08 while the aiming radius is 5641.28 km

3 0
3 years ago
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