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3 years ago

A ray of red light in air is incident at an angle of 30. on a

Physics

1 answer:

Vlad [161]3 years ago

8 0

Answer:

20 degrees.

Explanation:

From Snell’s law of refraction:

sinθ1•n1 = sinθ2•n2

where θ1 is the incidence angle, θ2 is the refraction angle, n1 is the refraction index of light in medium1, and n2 is the refraction index for virgin olive oil. The incidence angle of the red light is θ1 = 30 degrees.

The red light is in air as medium1, so n1 (air) = 1.00029

So, to find θ2, the refracted angle:

sinθ1•1.00029 = sinθ2•1.464

sin(30)•1.00029 / 1.464 = sinθ2

0.5•1.00029 / 1.464 = sinθ2

sinθ2 = 0.3416291

θ2 = arcsin(0.3416291)

θ2 = 19.976 degrees

To the nearest degree,

θ2 = 20 degrees.

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The circumference of a sphere was measured to be

professor190 [17]

To solve this problem we will apply the concepts related to the calculation of the surface, volume and error through the differentiation of the formulas given for the calculation of these values in a circle. Our values given at the beginning are

$\phi = 76cm$

$Error (dr) = 0.5cm$

The radius then would be

$\phi = 2\pi r \\76cm = 2\pi r\\r = \frac{38}{\pi} cm$

And

$\frac{d\phi}{dr} = 2\pi \\d\phi = 2\pi dr \\0.5 = 2\pi dr$

PART A ) For the Surface Area we have that,

$A = 4\pi r^2 \\A = 4\pi (\frac{38}{\pi})^2\\A = \frac{5776}{\pi}$

Deriving we have that the change in the Area is equivalent to the maximum error, therefore

$\frac{dA}{dr} = 4\pi (2r) \\dA = 4r (2\pi dr)$

Maximum error:

$dA = 4(\frac{38}{\pi})(0.5)$

$dA = \frac{76}{\pi}cm^2$

The relative error is that between the value of the Area and the maximum error, therefore:

$\frac{dA}{A} = \frac{\frac{76}{\pi}}{\frac{5776}{\pi}}$

$\frac{dA}{A} = 0.01315 = 1.31\%$

PART B) For the volume we repeat the same process but now with the formula for the calculation of the volume in a sphere, so

$V = \frac{4}{3} \pi r^3$

$V = \frac{4}{3} \pi (\frac{38}{\pi})^3$

$V = \frac{219488}{3\pi^2}$

Therefore the Maximum Error would be,

$\frac{dV}{dr} = \frac{4}{3} 3\pi r^2$

$dV = 2r^2 (2\pi dr)$

$dV = 4r^2 (\pi dr)$

Replacing the value for the radius

$dV = 4(\frac{38}{\pi})^2(0.5)$

$dV = \frac{2888}{\pi^2} cm^3$

And the relative Error

$\frac{dV}{V} = \frac{ \frac{2888}{\pi^2}}{ \frac{219488}{3\pi^2} }$

$\frac{dV}{V} = 0.03947$

$\frac{dV}{V} = 3.947\%$

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3 years ago

which describes the relationship between the frequency, wavelength, and speed of a wave as the wave travels through diffrent med

Goryan [66]

As speed changes, wavelength changes, and frequency remains the same

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3 years ago

Cardiovascular fitness measures the health of which body system?

Aloiza [94]

Answer:

I believe it is the circulatory system.

Explanation:

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3 years ago

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3 years ago

Solar cells are given antireflection coatings to maximize their efficiency. Consider a silicon solar cell (n = 3.50) coated with

alisha [4.7K]

Answer:

t = 121 nm

Explanation:

Given:

- Silicon refractive index n_1 = 3.50

- Silicon dioxide refractive index n_2 = 1.45

- The wavelength of light in air λ_air = 700 nm

Find:

What is the minimum coating thickness that will minimize the reflection at the wavelength of 700 nm.

Solution:

- The film’s index of refraction (n_2 = 1.45) is less than that of solar cell (n_1 = 3.50) so there will be a reflective phase change at the first boundary (air–film), and at the second boundary (film–solar cell). The relationship for destructive interference for two reflective phase changes is as follows:

2*t = (m + 0.5)*(λ/n_2) m = 0, 1, 2, ....

- Solve for thickness t where m = 0 (for the thinnest film).

t = 0.25*(λ/n_2)

t = 0.25*(700/1.45)

t = 121 nm ... (rounded to 3 sig. fig)

- This coating technique is important to increase the efficiency of solar cells; If the light can’t reflect, then it must transmit into the solar cell material.

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3 years ago