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arsen [322]
3 years ago
15

The smallness of the critical angle θc for diamond means that light is easily "trapped" within a diamond and eventually emerges

from the many cut faces. This makes a diamond more brilliant than stones with smaller n and larger θc. Traveling inside a diamond, a light ray is incident on the interface between diamond and air. What is the critical angle for total internal reflection? The refraction index for diamond is 2.23 . Answer in uni
Physics
1 answer:
hoa [83]3 years ago
3 0

Answer:

26.6°

Explanation:

refractive index of diamond, n = 2.23

When a ray of light passes from denser medium to the rarer medium and refracts at an angle of 90 degree from the normal of the surface, such angle of incidence in the denser medium is called the critical angle.

By the Snell's law

\frac{Sini}{Sin r }= n

For critical angle, angle of incidence is critical angle, i = θc and angle of refraction, r = 90

So,

Sin θc / Sin 90 = 1 / 2.23

Sin θc = 0.448

θc = 26.6°

Thus, the critical angle is 26.6°.

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if there is a duck behind two ducks and a duck between two ducks and a duck in front of two ducks how many ducks are there? look
TiliK225 [7]
Five ducks I guess so
8 0
2 years ago
An atomic nucleus has a charge of +40e. an electron is 10-9 m from the nucleus. what is the force on the electron?
sergiy2304 [10]
The electron charge is equal to -e=-1.6\cdot 10^{-19}C. The atomic nucleus of the problem has a charge of +40 e=40\cdot (1.6\cdot 10^{-19}C)=6.4\cdot 10^{-18}C. The distance between the nucleus and the electron is r=10^{-9}m, so we can calculate the electrostatic (Coulomb) force between the two:
F=k_e  \frac{(-e)(+40e) }{r^2} =8.99\cdot 10^9 Nm^2C^{-2}  \frac{(-1.6\cdot 10^{-19}C)(6.4\cdot 10^{-18}C)}{(10^{-9}m)^2} =
=-9.2 \cdot 10^{-9} N
which is attractive, since the two charges have opposite sign.
4 0
3 years ago
A coil of wire with 50 turns lies in the plane of the page and has an initial area of 0.250 m2. The coil is now stretched to hav
Oxana [17]

Answer:

The induced emf in the coil is 170 volt and its direction is clockwise direction.

Explanation:

Given that,

Number of turns of a coil, N = 50

Initial area, A_i=0.25\ m^2

Final area, A_f=0

Time, t = 0.1 s

The magnetic field points into the page and has a strength of 1.36 T.

We need to find the magnitude and direction of the average value of the induced emf. Due to change in area of the coil and emf will be induced in it. The induced emf is given by :

\epsilon=-N\dfrac{d\phi}{dt}\\\\\text{since}\ \phi=BA\\\\\epsilon=-N\dfrac{d(BA)}{dt}\\\\\epsilon=-NB\dfrac{A_f-A_i}{dt}\\\\\epsilon=-50\times 1.36\dfrac{0-0.25}{0.1}\\\\\epsilon=170\ V

So, the induced emf in the coil is 170 volt and its direction is clockwise direction. Hence, this is the required solution.

3 0
2 years ago
How does light travel differently from sound
Serhud [2]
Answer: Light waves move faster than sound waves, which are longitudinal.
Light waves can also travel through space.

Sound waves, being longitudinal, are slower than light waves, and do not travel through space.
4 0
3 years ago
The Earth possesses an electric field of (average) magnitude near its surface. The field points radially inward. Calculate the n
IgorLugansk [536]

Answer:

Incomplete question

This is the complete question

The Earth possesses an electric field of (average) magnitude 150 near its surface. The field points radially inward. Calculate the net electric flux outward through a spherical surface surrounding, and just beyond, the Earth's surface.

Explanation:

Given that,

Electric field is 150N/C inward

i.e E = - 150 i

Then, electric flux (Φ) is given as

Φ=∮E.dA

Surface area of a sphere is

A=4πr²

dA=8πr dr

The area is outwardly i.e

dA=-8πr dr i

Φ=∮E.dA

Φ=∮-150 i • 8πr i dr from 0 to R

Φ= -1200π∮rdr. From 0 to R

Φ= -1200π r²/2. From 0 to R

Φ= -600π [R² -0²]

Φ= -600πR²

Where R is radius of earth which is 6,400,000m

Φ= -600πR²

Φ= -600π(6400000)²

Φ= -7.72×10^16 Nm²/C

7 0
3 years ago
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