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faust18 [17]
3 years ago
15

A beam of light strikes a sheet of glass at an angle of 57.0° with the normal in air. You observe that red light makes an angle

of 38.1 degrees with the normal in the glass, while violet light makes a 36.7 degree angle.
1.What are the indexes of refraction of this glass for these colors of light?
Answer in the order indicated. Separate your answers with a comma.

2.What are the speeds of red and violet light in the glass?
Answer in the order indicated. Separate your answers with a comma.
Physics
1 answer:
Contact [7]3 years ago
3 0
<h2>Answers: </h2>

1) 1.359, 1.403

2) 2.207(10)^{8}m/s,  2.138(10)^{8}m/s    

Explanation:

The described situation is known as Refraction.  

Refraction is a phenomenon in which a wave (the light in this case) bends or changes it direction when passing through a medium with a refractive index different from the other medium.  

In this context, the Refractive index n is a number that describes how fast light propagates through a medium or material, and is defined as the relation between the speed of light in vacuum (c=3(10)^{8}m/s) and the speed of light v in the second medium:

n=\frac{c}{v}   (1)

On the other hand we have the Snell’s Law:  

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

Where:  

n_{1} is the first medium refractive index . We are told is the air, hence n_{1}\approx 1

n_{2} is the second medium refractive index  

\theta_{1} is the angle of the incident ray  

\theta_{2} is the angle of the refracted ray  

Knowing this, let's begin with the answers:

<h2><u>1) Indexes of refraction for red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Using equation (2) according to Snell's Law and \theta_{1}=57.0\º   \theta_{2}=38.1\º:

(1)sin(57.0\º)=n_{2}sin(38.1\º)  

Finding n_{2}:

n_{2}=\frac{sin(57.0\º)}{sin(38.1\º)}  

n_{2}=1.359   (3)>>>Index of Refraction for red light

<h2>1b) Violet light</h2>

Again, using equation (2) according to Snell's Law and \theta_{1}=57.0\º   \theta_{2}=36.7\º:

(1)sin(57.0\º)=n_{2}sin(36.7\º)  

Finding n_{2}:

n_{2}=\frac{sin(57.0\º)}{sin(36.7\º)}  

n_{2}=1.403   (4) >>>Index of Refraction for violet light

<h2><u>2) Speeds of red and violet light</u></h2><h2 /><h2>1a) Red light</h2>

Here we are going to use equation (1):

n_{red}=\frac{c}{v_{red}}

v_{red}=\frac{c}{n_{red}}

Substituting (3) in this equation:

v_{red}=\frac{3(10)^{8}m/s}{1.359}

v_{red}=2.207(10)^{8}m/s >>>>Speed of red light

<h2>1a) Violet light</h2>

Using again equation (1):

n_{violet}=\frac{c}{v_{violet}}

v_{violet}=\frac{c}{n_{violet}}

Substituting (4) in this equation:

v_{violet}=\frac{3(10)^{8}m/s}{1.403}

v_{red}=2.138(10)^{8}m/s >>>>Speed of violet light

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Force exerted on the car driver by the seatbelt = 8139.4 N = 8.14 kN

Force exerted on the truck driver by the seatbelt = 1628.2 N = 1.63 kN

It is evident that the driver of the smaller vehicle has it worse. The car driver is in way more danger in this perfectly inelastic head-on collision with a bigger vehicle (the truck).

Explanation:

First of, we calculate the velocity of the vehicles after collision using the law of conservation of Momentum

Momentum before collision = Momentum after collision

Since the collision of the two vehicles was described as a head-on collision, for the sake of consistent convention, we will take the direction of the velocity of the bigger vehicle (the truck) as the positive direction and the direction of the car's velocity automatically is the negative direction.

Velocity of the truck before collision = 6.80 m/s

Velocity of the car before collision = -6.80 m/s

Let the velocity of the inelastic unit of vehicles after collision be v

Momentum before collision = (4000)(6.80) + (800)(-6.80) = 27200 - 5440 = 21,760 kgm/s

Momentum after collision = (4000 + 800)(v) = (4800v) kgm/s

Momentum before collision = Momentum after collision

21760 = 4800v

v = (21760/4800)

v = 4.533 m/s (in the direction of the big vehicle (the truck)

So, we then apply Newton's second law of motion which explains that the magnitude change in momentum is equal to the magnitude of impulse.

|Impulse| = |Change in momentum|

But Impulse = (Force exerted on each driver by the seatbelt) × (collision time) = (F×t)

Change in momentum = (Momentum after collision) - (Momentum before collision)

So, for the driver of the truck

Initial velocity = 6.80 m/s (the driver moves with the velocity of the truck)

Final velocity = 4.533 m/s

Change in momentum of the truck driver = (79)(6.80) - (79)(4.533) = 179.1 kgm/s

(F×t) = 179.1

F × 0.110 = 179.1

F = (179.1/0.11)

F = 1628.2 N = 1.63 kN

So, for the driver of the car

Initial velocity = -6.80 m/s (the driver moves with the velocity of the car)

Final velocity = 4.533 m/s

Change in momentum of the car driver = (79)(-6.80) - (79)(4.533) = -895.3 kgm/s

(F×t) = |-895.3|

F × 0.110 = 895.3

F = (895.3/0.11)

F = 8139.4 N = 8.14 kN

Hope this Helps!!!

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