answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Marina86 [1]
2 years ago
13

Sunlight strikes a piece of crown glass at an angle of incidence of 38.0°. Calculate the difference in the angle of refraction b

etween a yellow (580 nm) and a green (550 nm) ray within the glass. The index of refraction is n=1.523 for yellow and n=1.526 for green light. B.) The ray now travels inside the glass. What is the minimum angle of incidence at which the yellow ray can hit the surface of the glass and become there totally internally reflected and not refracted?
Physics
1 answer:
zhuklara [117]2 years ago
5 0

Answer:

Difference in the angle of refraction = 0.3°

41.04° is the minimum angle of incidence.

Explanation:

Angle of incidence  = 38.0°

For yellow light :

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for yellow light which is 1.523

n₂ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.523}{1}

{sin\theta_2}=0.9377

Angle of refraction for yellow light = sin⁻¹ 0.9377 = 69.67°.

For green light :

Using Snell's law as:

\frac {sin\theta_2}{sin\theta_1}=\frac {n_1}{n_2}

Where,  

Θ₁ is the angle of incidence

Θ₂ is the angle of refraction

n₁ is the refractive index for green light which is 1.526

n₂ is the refractive index of air which is 1

So,  

\frac {sin\theta_2}{sin{38.0}^0}=\frac {1.526}{1}

{sin\theta_2}=0.9395

Angle of refraction for green light = sin⁻¹ 0.9395 = 69.97°.

The difference in the angle of refraction = 69.97° - 69.67° = 0.3°

Calculation of the critical angle for the yellow light for the total internal reflection to occur :

The formula for the critical angle is:

{sin\theta_{critical}}=\frac {n_r}{n_i}

Where,  

{\theta_{critical}} is the critical angle

n_r is the refractive index of the refractive medium.

n_i is the refractive index of the incident medium.

n₁ is the refractive index for yellow light which is 1.523 (incident medium)  

n₂ is the refractive index of air which is 1 (refractive medium)

Applying in the formula as:

{sin\theta_{critical}}=\frac {1}{1.523}

The critical angle is = sin⁻¹ 0.6566 = 41.04°

You might be interested in
a 1.2x10^3 kilogram car is accelerated uniformly from 10. meters per second to 20 meters per second in 5.0 seconds. what is the
irinina [24]
Force , F = ma

F =  m(v - u)/t               

Where m = mass in kg, v= final velocity in m/s, u = initial velocity in m/s
t = time, Force is in Newton.

m= 1.2*10³ kg,  u = 10 m/s,  v = 20 m/s, t = 5s

F =  1.2*10³(20 - 10)/5

F = 2.4*10³ N = 2400 N


7 0
1 year ago
Two students, sitting on frictionless carts, push against each other. Both are initially at rest and the mass of student 1 and t
Zepler [3.9K]

Answer:

  v₂ = v/1.5= 0.667 v

Explanation:

For this exercise we will use the conservation of the moment, for this we will define a system formed by the two students and the cars, for this isolated system the forces during the contact are internal, therefore the moment conserves.

Initial moment before pushing

    p₀ = 0

Final moment after they have been pushed

    p_{f} = m₁ v₁ + m₂ v₂

   p₀ =  p_{f}

   0 = m₁ v₁ + m₂ v₂

   m₁ v₁ = - m₂ v₂

Let's replace

   M (-v) = -1.5M v₂

   v₂ = v / 1.5

  v₂ = 0.667 v

6 0
1 year ago
An electron moving parallel to a uniform electric field increases its speed from 2.0 × 107 m/s to 4.0 × 107 m/s over a distance
jeka94

Answer:

1.8\times 105 N/C

Explanation:

We are given that

u=2\times 10^7 m/s

v=4\times 10^7 m/s

d=1.9 cm=\frac{1.9}{100}=0.019 m

Using 1m=100 cm

We have to find the electric field strength.

v^2-u^2=2as

Using the formula

(4\times 10^7)^2-(2\times 10^7)^2=2a(0.019)

16\times 10^{14}-4\times 10^{14}=0.038a

0.038a=12\times 10^{14}

a=\frac{12}{0.038}\times 10^{14}=3.16\times 10^{16}m/s^2

q=1.6\times 10^{-19} C

Mass of electron,m=9.1\times 10^{-31} kg

E=\frac{ma}{q}

Substitute the values

E=\frac{9.1\times 10^{-31}\times 3.16\times 10^{16}}{1.6\times 10^{-19}}

E=1.8\times 105 N/C

7 0
2 years ago
Two point charges totaling 8 μC exert a repulsive force of 0.15 N on one another when separated by 0.5 m. What is the charge on
nexus9112 [7]

Answer:

B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C

Explanation:

From Coulomb's Law the electrostatic repulsive force is given by the following formula:

F = kq₁q₂/r²

where,

F = Repulsive Force = 0.15 N

k = Coulomb's Constant = 9 x 10⁹ N.m²/C²

q₁ = Magnitude of 1st Charge = ?

q₂ = Magnitude of 2nd Charge = ?

r = Distance between Charges = 0.5 m

Therefore,

0.15 N = (9 x 10⁹ N.m²/C²)q₁q₂/(0.5 m)²

q₁q₂ = (0.15 N)(0.5 m)²/(9 x 10⁹ N.m²/C²)

q₁q₂ = 4.17 x 10⁻¹²

q₁ = (4.17 x 10⁻¹²)/q₂   -------------------- equation (1)

The sum of charges is given as:

q₁ + q₂ = 8 μC

q₁ + q₂ = 8 x 10⁻⁶

using equation (1):

(4.17 x 10⁻¹²)/q₂ + q₂ = 8 x 10⁻⁶

(4.17 x 10⁻¹²) + q₂² = 8 x 10⁻⁶ q₂

q₂² - (8 x 10⁻⁶) q₂ + (4.17 x 10⁻¹²) = 0

Solving this quadratic equation:

q₂ = 7.4 x 10⁻⁶ C   (OR)   q₂ = 0.56 x 10⁻⁶ C

q₂ = 7.4 μC   (OR)   q₂ = 0.6 μC

Therefore,

q₁ = (4.17 x 10⁻¹² C)/(7.4 x 10⁻⁶ C)

q₁ = 0.6μC

Now, if we solve with q₂ = 0.6 μC, we will get q₁ = 7.4 μC.

Therefore, the correct option will be:

<u>B. 7.4 x 10⁻⁶ C, 0.6 x 10⁻⁶ C</u>

8 0
1 year ago
A 1700kg rhino charges at a speed of 50.0km/h. what is the magnitude of the average force needed to bring the rhino to a stop in
klio [65]
First, let's convert the initial speed of the rhino:
v_i = 50 km/h = 13.9 m/s
The acceleration of the rhino is given by:
a= \frac{v_f-v_i}{t}= \frac{0-13.9 m/s}{0.5 s}=-27.8m/s^2
where the negative sign means it is a deceleration. We can neglect the negative sign since we are interested only in the magnitude of the force applied, which is given by Netwon's second law:
F=ma = (1700 kg)(27.8 m/s^2)=4.7 \cdot 10^4 N
7 0
1 year ago
Read 2 more answers
Other questions:
  • 568 muons were counted by a detector on the top of Mount Washington in a one hour period of time. Assuming moving muons keep tim
    8·2 answers
  • A sled sliding on a flat,icy surface with a constant velocity is best described by
    15·1 answer
  • A gas of helium atoms at 273 k is in a cubical container with 25.0 cm on a side. (a) what is the minimum uncertainty in momentum
    11·1 answer
  • Complete each statement about the sign of the work done on a baseball. Carlton catches a baseball and his hand moves backward as
    13·1 answer
  • A piston–cylinder device contains 0.15 kg of air initially at 2 MPa and 350°C. The air is first expanded isothermally to 500 kPa
    7·1 answer
  • A 10. g cube of copper at a temperature T1 is placed in an insulated cup containing 10. g of water at a temperature T2. If T1 &g
    12·1 answer
  • 8.4-1 Consider a magnetic field probe consisting of a flat circular loop of wire with radius 10 cm. The probe’s terminals corres
    12·1 answer
  • Un camión cargado, de masa 3000 kg, ha adquirido en la bajada de un puerto una velocidad excesiva de 108 km/h. Al ver una curva
    11·1 answer
  • A mass m slides down a frictionless ramp and approaches a frictionless loop with radius R. There is a section of the track with
    7·1 answer
  • For the first 10 seconds a squirrel runs 3 m/s to look for an acorn. The next 5 seconds he eats an acorn that he finds. Afterwar
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!