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laila [671]
3 years ago
13

Light rays in a material with index of refrection 1.35 1.35 can undergo total internal reflection when they strike the interface

with another material at a critical angle of incidence. Find the second material's index of refraction n n when the required critical angle is 75.1 ∘ .
Physics
1 answer:
nekit [7.7K]3 years ago
6 0

Answer:

1.30

Explanation:

To calculate the critical angle we have ti use the formula:

sin\theta_c=\frac{n_2}{n_1}

where theta_c is the critical angle, n1 is the index of refraction of the material where the light is totally reflected, and n2 is the refractive index of the other material.

By taking n_2 and replacing we obtain:

n_2=n_1sin\theta_c=(1.35)sin75.1\°=1.30

hope this helps!!

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otez555 [7]

Answer:

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Explanation:

From the question given above, the following data were obtained:

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Force (F) =?

NOTE: Electric constant (K) = 9×10⁹ Nm²/C²

The force exerted can be obtained as follow:

F = Kq₁q₂ / r²

F = 9×10⁹ × (1.6×10¯¹⁹)² / (1×10¯¹⁵)²

F = 9×10⁹ × 2.56×10¯³⁸ / 1×10¯³⁰

F = 2.304×10¯²⁸ / 1×10¯³⁰

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5 0
3 years ago
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tatuchka [14]

Answer:

1.8*10^8m/s

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Using

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V= 0.6*3E8m/s

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You prepare tea with 0.250 kg of water at 85.0 ºC and let it cool down to room temperature (20.0 ºC) before drinking it. Essenti
vladimir2022 [97]

Answer:

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Explanation:

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5 0
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user100 [1]
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6 0
3 years ago
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boyakko [2]

Answer:

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