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Zinaida [17]
3 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]3 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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Lewiston and Vernonville are 208 miles apart. A car leaves Lewiston traveling towards​ Vernonville, and another car leaves Verno
iragen [17]

Answer:

Average speed of the car A = 70 miles per hour

Average speed of the car B = 60 miles per hour

Explanation:

Average speed of the car A is v_{A} =\frac{x_{A} }{t_{A} } (Equation A) and Average speed of the car B is v_{B} =\frac{x_{B} }{t_{B} } (Equation B), where x_{A} and x_{B} are the distances and t_{A} and t_{B} are the times at which are travelling the cars A and B respectively.

We have to convert the time to the correct units:

1 hour and 36 minutes = 96 minutes

96 minutes . \frac{1 hour}{60 minutes} = 1.6 h

From the diagram (Please see the attachment), we can see that at the time they meet, we have:

v_{A} = \frac{208-x}{1.6h} + 10\frac{miles}{h} (Equation C)

v_{B} = \frac{208-x}{1.6h} (Equation D)

From Equation A and C, we have:

\frac{208-x}{1.6}+10 = \frac{x}{1.6}

208-x+16 = x

208 + 16 = 2x

x = \frac{224}{2}

x = 112 miles

Replacing x in Equation A:

v_{A}  = \frac{112miles}{1.6h}

v_{A} = 70 miles per hour

Replacing x in Equation B:

v_{B}  = \frac{208miles-112miles}{1.6h}

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v_{B}  = 60 miles per hour

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What is the radius of a tightly wound solenoid of circular cross-section that has 180turns if a change in its internal magnetic
Nata [24]

Answer:

Radius of cross section, r = 0.24 m

Explanation:

It is given that,

Number of turns, N = 180

Change in magnetic field, \dfrac{dB}{dt}=3\ T/s

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We need to find the radius of the solenoid (r). We know that emf is given by :

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A=0.19\ m^2

Area of circular cross section is, A=\pi r^2

r=\sqrt{\dfrac{A}{\pi}}

r=\sqrt{\dfrac{0.19}{\pi}}

r = 0.24 m

So, the  radius of a tightly wound solenoid of circular cross-section is 0.24 meters. Hence, this is the required solution.

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