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SCORPION-xisa [38]
3 years ago
7

Find the angle of incidence at which a light beam traveling from a medium having the index of refraction n1 = 2.7 to another med

ium having the index of refraction n2 = 1.6 suffers a total internal reflection.
Physics
1 answer:
luda_lava [24]3 years ago
7 0

Answer:

The angle of incidence at which total internal reflection takes place will be equal to \Theta _c=36.34^{\circ}

Explanation:

We have given that light is traveling from a medium  of refractive index x 2.7 to a medium of refractive index 1.6

So refractive index of first medium n_1=2.7 and n_2=1.6

We have to find the angle of incidence for total internal reflection

For total internal reflection

n_1sin\Theta _c=n_2sin90^{\circ}

\Theta _c=sin^{-1}\frac{n_2}{n_1}

\Theta _c=sin^{-1}\frac{1.6}{2.7}

\Theta _c=36.34^{\circ}

So the angle of incidence at which total internal reflection takes place will be equal to \Theta _c=36.34^{\circ}

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chubhunter [2.5K]

Answer:

 v_{1fy} = - 0.4549 m / s

Explanation:

This is an exercise of conservation of the momentum, for this we must define a system formed by the two balls, so that the forces during the collision have internal and the momentum is conserved

initial. Before the crash

      p₀ = m v₁₀

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      p_{f} = m v_{1f} + m v_{2f}

Recall that velocities are a vector so it has x and y components

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we write this equation for each axis

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       0 = -m v_{1fy} + m v_{2fy}

the exercise tells us the initial velocity v₁₀ = 1.83 m / s, the final velocity v_{2f} = 1.15, let's use trigonometry to find its components

      sin 23.3 = v_{2fy} / v_{2f}

      cos 23.3 = v_{2fx} / v_{2f}

      v_{2fy} = v_{2f} sin 23.3

      v_{2fx} = v_{2f} cos 23.3

we substitute in the momentum conservation equation

       m v₁₀ = m v_{1f} cos θ + m v_{2f} cos 23.3

       0 = - m v_{1f} sin θ + m v_{2f} sin 23.3

      1.83 = v_{1f} cos θ + 1.15 cos 23.3

       0 = - v_{1f} sin θ + 1.15 sin 23.3

      1.83 = v_{1f} cos θ + 1.0562

        0 = - v_{1f} sin θ + 0.4549

     v_{1f} sin θ = 0.4549

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we divide these two equations

      tan θ = - 0.5878

      θ = tan-1 (-0.5878)

       θ = -30.45º

we substitute in one of the two and find the final velocity of the incident ball

        v_{1f} cos (-30.45) = - 0.7738

        v_{1f} = -0.7738 / cos 30.45

        v_{1f} = -0.8976 m / s

the component and this speed is

       v_{1fy} = v1f sin θ

       v_{1fy} = 0.8976 sin (30.45)

       v_{1fy} = - 0.4549 m / s

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Answer: The correct option is (c.).

Explanation:

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Velocity of Cart A = 1.4 m/s towards right

Mass of the cart B = 1.0 kg

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Momentum (P)= Mass × Velocity

P_A=1.5 kg\times 1.4 m/s=2.1 kg m/s

P_B=1.0 kg\times (-1.4m/s)=-1.4 kg m/s

(Negative sign means velocity of the cart is in opposite direction of that of the cart A)

Total Momentum =P_A+P_B=2.10 kg m/s-1.40 kg m/s=0.70 kg m/s

Hence, the correct option is (c.).

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