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Vesna [10]
3 years ago
15

The indices of refraction for violet light (λ = 400 nm) and red light (λ = 700 nm) in diamond are 2.46 and 2.41, respectively. A

ray of light traveling through air strikes the diamond surface at an angle of 51.0 ∘ to the normal.
Calculate the angular separation between these two colors of light in the refracted ray.
Physics
1 answer:
JulijaS [17]3 years ago
6 0

Answer:

0.42°

Explanation:

Using Snell's law of refraction which states that the ratio of the angle of sin of incidence to angle of sine of refraction is equal to a constant for a given pair of media. Mathematically,

Sin(i)/sin(r) = n

n is the refractive index of the medium

FOR VIOLET LIGHT:

n = 2.46

i = 51°

r = ?

To get r, we will use the Snell's law formula.

2.46 = sin51°/sinr

Sinr = sin51°/2.46

Sinr = 0.316

r = sin^-1(0.316)

rv = 18.42°

FOR RED LIGHT:

n = 2.41

i = 51°

r = ?

To get r, we will use the Snell's law formula.

2.41 = sin51°/sinr

Sinr = sin51°/2.41

Sinr = 0.323

r = sin^-1(0.323)

rd = 18.84°

The angular separation between these two colors of light in the refracted ray will be the difference between there angle of refraction.

Angular separation = rd - rv

= 18.84° - 18.42°

= 0.42°

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The car C has KE = 100, PE = 0

Explanation:

The principle of conservation of energy states that although energy can be transformed from one form to another, the total energy of the given system remains unchanged.

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3 years ago
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Answer:

a)  Q = 397.57 pC , Q = 3.18 104 pC , b) C = 1.157 10⁻¹⁰ F ,  V = 3.4375 V ,

c)  U = 54.7 nJ ,  d) ΔU = 54 nJ,

Explanation:

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        Q = C V

         

can also be calculated using geometry consideration

        C = e or A / d

         

we reduce to the SI system

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       d = 1.53 cm = 1.53 10⁻² m

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         Q = eo A / d V

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         Q = 3.9757 10⁻¹⁰ C

         

let's reduce to pC

         Q = 3.9757 10⁻¹⁰ C (10¹² pC / 1 C)

          Q = 397.57 pC

when the capacitor is introduced into the water the dielectric constant is different

           Q = k Q₀

           Q = 80 397.57

           Q = 3.18 104 pC

b) Find capacitance and voltage after submerged in water

           C = k C₀

           C = 80 8.85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²

           C = 1.157 10⁻¹⁰ F

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c) The stored energy is

             U = ½ C V²

              U = ½, 85 10⁻¹² 25 10⁻⁴ / 1.53 10⁻²     275²

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             U = ½ (kCo) (Vo / k) 2

             U = ½ Co Vo2 / k

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the energy change is

         ΔU = U₀ -U

          ΔU = 54.7 - 0.687375

           

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