PHYSICS HIGH SCHOOL

A thin layer of turpentine (n=1.472) is floating on water (n=1.333). Light of wavelength 589 nm initially traveling in air is incident on the turpentine at an angle of 24.8° measured with respect to the air-turpentine interface. What is the light's refraction angle in the turpentine?

Answers

Answer 1

Answer:

The angle of refraction of the light in turpentine will be $\boxed{38.1^\circ}$ .

Explanation:

Given:

The refractive index of the turpentine is $1.472$ .

The refractive index of the water is $1.333$ .

The angle that the light makes with the surface of the turpentine is $24.8^\circ$ .

Concept:

As the light is incident on the surface of the turpentine at the air-turpentine interface, the light gets refracted from the surface. The refraction of light occurs due to change in the refractive index at the air-turpentine surface.

The angle of the refracted ray after undergoing refraction is given by the Snell's law.

$\boxed{n_1sin\,i=n_2sin\,r}$

Here, $n_1$ is the refractive index of first medium, $n_2$ is the refractive index of second medium, $i$ is the angle of incidence and $r$ is the angle of refraction.

The angle of incidence is the angle made by the light from the normal to the surface. So, in this case, the angle of incidence will be $(90^\circ-24.8^\circ)$ i.e. $65.2^\circ$ .

Substitute the values in above expression.

$\begin{aligned}(1)sin\,(65.2^\circ)&=(1.472)sin\,(r)\nr&=sin^(-1)\left((1)/(1.472)sin\,(65.2^\circ)\right)\n&=sin^(-1)(0.616)\n&\approx38.1^\circ\end{aligned}$

Thus, The angle of refraction of the light in turpentine will be $\boxed{38.1^\circ}$ .

Learn More:

1. How does the reflection differ from refraction and diffraction brainly.com/question/3183125

2. The reason for the refraction of light at air water interface brainly.com/question/3095091

3. Approximate length of the unsharpened pencil brainly.com/question/6140057

Answer Details:

Grade: High School

Subject: Physics

Chapter: Refraction

Keywords:

thin layer, turpentine, Snell's law, refraction, air, interface, normal, incident, light, refracted, refraction, refraction angle.

Answer 2

Answer:

16.6 degrees

Explanation:

We can solve the problem by using Snell's law for refraction:

$n_1 sin \theta_1 = n_2 sin \theta_2$

where:

$n_1 = 1.00$ is the index of refraction of the first medium (air)

$\theta_1 = 24.8^(\circ)$ is the angle of incidence

$n_2 = 1.472$ is the index of refraction of the second medium (turpentine)

$\theta_2$ is the angle of refraction of light in turpentine

By re-arranging the equation and using the numbers, we find

$sin \theta_2 = (n_1)/(n_2) sin \theta_i = (1.0)/(1.472) sin 24.8^(\circ)=0.285\n\theta_2 = sin^(-1) (0.285)=16.6^(\circ)$

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Based on their locations on the periodic table, which two elements would youexpect to form positive ions with a +3 charge?
O A. Nitrogen, N
O B. Aluminum, Al
O C. Phosphorus, P
D. Boron, B

Answers

Final answer:

The elements that would form positive ions with a +3 charge are aluminum (Al) and boron (B).

Explanation:

Based on their locations on the periodic table, the two elements that would form positive ions with a +3 charge are aluminum (Al) and boron (B).

When an element gains or loses electrons to form an ion, it takes on a charge that corresponds to the number of electrons gained or lost. Aluminum, located in Group 13 of the periodic table, tends to lose its three valence electrons to form an ion with a +3 charge. Likewise, boron, located in Group 13 as well, will also lose its three valence electrons to achieve a +3 charge.

Learn more about positive ions with a +3 charge here:

brainly.com/question/36949752

#SPJ3

Answer:

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Explanation:

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What is the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 grams of water?

Answers

Considering the definition of percentage by mass, the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 grams of water is 6.25 %.

The percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$percent by mass= (mass of solute)/(mass of solution)x100$

In this case, you know:

mass of solute= 5 g
mass of water= 75 g
mass of solution= mass of solute + mass of water= 5 + 75 g= 80 g

Replacing:

$percent by mass= (5 g)/(80 g)x100$

Solving:

percent by mass= 6.25 %

Finally, the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 grams of water is 6.25 %.

Learn more:

Hello!

What the percent by mass of 5.0 g of iron (II) sulfate dissolved in 75.0 g of water?

We have the following data:

m1 (solute mass - iron II sulfate) = 5.0 g

m2 (solvent mass - water) = 75.0 g

m (solution mass) = m1 + m2 = 5.0 + 75.0 = 80.0 g

%m/m (percent mass by mass) = ?

We apply the data to the formula

$\%\:m/m = (m_1)/(m)*100$

$\%\:m/m = (5)/(80)*100$

$\%\:m/m = 0.0625*100$

$\boxed{\boxed{\%\:m/m = 6.25}}\Longleftarrow(solute\:in\:percent)\:\:\:\:\:\:\bf\blue{\checkmark}$

Answer:

The percent by mass of solute and mass solution of Iron II Sulfate is 6.25%

_______________________

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A student who has trouble hearing and must have a written copy of class lectures in order to participate in class is considered to have which of the following?

Answers

For this, the student has problems when participating in class discussions. I believe that this type of student has a participation restriction disability. This disability refers to an individual who cannot focus or is not capable of participating in social activities.

Answer: The answer is c

Explanation:

A block of mass m sits at rest on a rough inclined ramp that makes an angle θ with the horizontal. What must be true about force of static friction f on the block?A) f > mg
B) f = mgsin(?)
C) f > mgcos(?)
D) f = mgcos(?)
E) f > mgsin(?)

Answers

When the block is at rest, the static frictional force is equal to the horizontal component of the block's weight (F = mgsin(θ)).

The static frictionalforce on the body at rests is determined by applying Newton's second law of motion.

F = ma

where;

F is applied force on a body
m is the mass of the body
a is the acceleration of the body

If the block is at rest, then the net horizontal force on the block is zero.

$\Sigma F_x = 0\n\nmg sin(\theta ) - F_s = 0\n\nF_s = mg sin(\theta)$

Thus, when the block is at rest, the static frictional force is equal to the horizontal component of the block's weight.

Learn more here:brainly.com/question/13758352

Answer:

Option B

Explanation:

For a system of block on inclined ramp shown in the attached image. From the attached image, the Normal force N, weight mg and frictional force f act on the block. The sum of vertical forces should be zero just as sum of vertical forces should be zero when the system is in equilibrium condition.

Taking sum of forces along the inclined plane we deduce that

[tex]f=mgsin \theta[tex]

Therefore, option B is the correct option.

The blood pressure in human body is greater at the feet than the brain

Answers

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yes,
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