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3 years ago

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Light rays in a material with index of refraction 1.31 can undergo total internal reflection when they strike the interface with

another material at a critical angle of incidence. Find this material\'s index of refraction when the required critical angle is 78.3°.

1 answer:

3241004551 [841]3 years ago

3 0

Answer:

The refractive index of the material is 1.28.

Explanation:

It is given that,

Refractive index of medium 1, n₁ = 1.31

Critical angle, $\theta_c=78.3$

At critical angle rays will reflects at 90 degrees. Using Snell's law as:

$n_1sin\theta_1=n_2sin\theta_2$

At critical angle, $n_1sin\ \theta_c=n_2\ sin90$

$n_1sin\ \theta_c=n_2$

$n_2=1.31\times sin(78.3)$

$n_2=1.28$

So, the refractive index of the material is 1.28. Hence, this is the required solution.

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3 years ago

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<h2>B) Newton's 2nd law</h2>

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A meter stick balances at the 50.0-cm mark. If a mass of 50.0 g is placed at the 90.0-cm mark, the stick balances at the 61.3-cm

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Answer:

H = 45 m

Explanation:

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v₀ = √(v₀ₓ² + v₀y²)

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Therefore,

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v₀ = 33.54 m/s

Now, we find the launch angle of the ball by using the following formula:

θ = tan⁻¹ (v₀y/v₀ₓ)

θ = tan⁻¹ (30/15)

θ = tan⁻¹ (2)

θ = 63.43°

Now, the maximum height attained by the ball is given by the formula:

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3 years ago

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Explanation:

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k=constant

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$0.5=e^{-kt^n}$

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3 years ago