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Pachacha [2.7K]

3 years ago

13

Determine the index of refraction of glass that is struck by unpolarized light at 53.8 degrees and resulting in a fully polarize

d reflected beam. (Brewster angle).

1 answer:

nignag [31]3 years ago

5 0

Answer:

The refractive index of glass, $\mu_{g} = 1.367$

Solution:

Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.

To determine the refractive index of glass:

$tan\theta_{P} = \frac{\mu_{g}}{\mu_{a}}$ (1)

where

$\mu_{a}$ = refractive index of glass

$\mu_{g}$ = refractive index of glass

Now, using eqn (1)

$tan{53.8^{\circ}} = \frac{\mu_{g}}{1}$

$\mu_{g} = tan53.8^{\circ}$

$\mu_{g} = 1.367$

You might be interested in

A uniformly charged ball of radius a and charge –Q is at the center of a hollowmetal shell with inner radius b and outer radius

vlabodo [156]

Answer:

<u>r < a:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}$

<u>r = a:</u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>a < r < b:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>r = b:</u>

$E = \frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$

<u>b < r < c:</u>

E = 0

<u>r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

<u>r < c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

Explanation:

Gauss' Law will be applied to each region to find the E-field.

$\int \vec{E}d\vec{a} = \frac{Q_{encl}}{\epsilon_0}$

An imaginary sphere is drawn with radius r, which is equal to the point where the E-field is asked. The area of this imaginary sphere is multiplied by E, and this is equal to the charge enclosed by this imaginary surface divided by ε0.

<u>r<a:</u>

Since the ball is uniformly charged and not hollow, then the enclosed charge can be found by the following method: If the total ball has a charge -Q and volume V, then the enclosed part of the ball has a charge Q_enc and volume V_enc. Then;

$\frac{Q}{V} = \frac{Q_{encl}}{V_{encl}}\\\frac{Q}{\frac{4}{3}\pi a^3} = \frac{Q_{encl}}{\frac{4}{3}\pi r^3}\\Q_{encl} = \frac{Qr^3}{a^3}$

Applying Gauss' Law:

$E4\pi r^2 = \frac{-Qr^3}{\epsilon_0 a^3}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Qr}{a^3}\\E = \frac{r}{4\pi a^3}\frac{Q}{\epsilon_0}$

The minus sign determines the direction of the field, which is towards the center.

<u>At r = a: </u>

$E = \frac{1}{4\pi a^2}\frac{Q}{\epsilon_0}$

<u>At a < r < b:</u>

The imaginary surface is drawn between the inner surface of the metal sphere and the smaller ball. In this case the enclosed charge is equal to the total charge of the ball, -Q.

<u /> $E4\pi r^2 = \frac{-Q}{\epsilon_0}\\E = -\frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$ <u />

<u>At r = b:</u>

<u /> $E = -\frac{1}{4\pi b^2}\frac{Q}{\epsilon_0}$ <u />

Again, the minus sign indicates the direction of the field towards the center.

<u>At b < r < c:</u>

The hollow metal sphere has a net charge of +2Q. Since the sphere is a conductor, all of its charges are distributed across its surface. No charge is present within the sphere. The smaller ball has a net charge of -Q, so the inner surface of the metal sphere must possess a net charge of +Q. Since the net charge of the metal sphere is +2Q, then the outer surface of the metal should possess +Q.

Now, the imaginary surface is drawn inside the metal sphere. The total enclosed charge in this region is zero, since the total charge of the inner surface (+Q) and the smaller ball (-Q) is zero. Therefore, the Electric region in this region is zero.

E = 0.

<u>At r < c:</u>

The imaginary surface is drawn outside of the metal sphere. In this case, the enclosed charge is +Q (The metal (+2Q) plus the smaller ball (-Q)).

$E4\pi r^2 = \frac{Q}{\epsilon_0}\\E = \frac{1}{4\pi \epsilon_0}\frac{Q}{r^2}$

<u>At r = c:</u>

$E = \frac{1}{4\pi \epsilon_0}\frac{Q}{c^2}$

3 0

3 years ago

2. An athlete of average size is hanging from the end of a 20 m long rope, which has a mass of 4 kg and is attached to a hook in

a_sh-v [17]

Answer:

t = 0.319 s

Explanation:

With the sudden movement of the athlete a pulse is formed that takes time to move along the rope, the speed of the rope is given by

v = √T/λ

Linear density is

λ = m / L

λ = 4/20

λ = 0.2 kg / m

The tension in the rope is equal to the athlete's weight, suppose it has a mass of m = 80 kg

T = W = mg

T = 80 9.8

T = 784 N

The pulse rate is

v = √(784 / 0.2)

v = 62.6 m / s

The time it takes to reach the hook can be searched with kinematics

v = x / t

t = x / v

t = 20 / 62.6

t = 0.319 s

7 0

3 years ago

Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

worty [1.4K]

Answer:Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

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D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

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B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

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A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

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B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

A. Reducing its mass to one-half of its original value

B. Increasing its velocitato twice its original value

C. Reducing its velocity to one-half of its original value O

D. Increasing its mass to twice its original value Questlon 20 of 20 Which change to an object would quadruple its kinetic energy?

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Explanation:

6 0

3 years ago

Read 2 more answers

A ball is attached to a vertical spring. The ball is initially supported at a height y so that the spring is neither stretched n

scoundrel [369]

Answer:

All the three quantities will have non zero joules.

Explanation:

At the initial position of rest the system will have only gravitational potential energy while the other 2 quantities will be zero.

when the system reaches a height (y-h) only kinetic energy will be zero while the other 2 quantities will be non zero

At the position of (y-h/2) all the 3 quantities will be non zero.

3 0

3 years ago

A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t

Marrrta [24]

Answer:

a) $U = 652.545\,kJ$ , b) $v \approx 659.568\,\frac{m}{s}$

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

$Q = U$

The internal energy for a monoatomic ideal gas is:

$U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T$

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

$T = \frac{P\cdot V}{n\cdot R_{u}}$

$T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}$

$T = 52.325\,K$

The thermal energy contained by the gas is:

$U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)$

$U = 652.545\,kJ$

b) The physical model for the cat is constructed from Work-Energy Theorem:

$U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}$

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

$v = \sqrt{\frac{2 \cdot U}{m_{cat}}}$

$v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }$

$v \approx 659.568\,\frac{m}{s}$

3 0

3 years ago