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3 years ago

14

Research indicates that inattentional blindness often decreases when people work on tasks that require a great deal of attention

. Please select the best answer from the choices provided T F

2 answers:

Shalnov [3]3 years ago

7 0

Answer:

False

Explanation:

Inattentionla blindness is a condition that happens when you overview or do not notice an object that is not supposed to be in a certain place and that is strange for its context because you are paying attention to something else, this means that if you do a task that needs a great deal of attention you will actually increase the inattentional blindness.

QveST [7]3 years ago

6 0

The correct answer is F

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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

Contact [7]

<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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Explanation:

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