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Alex17521 [72]
3 years ago
11

Light travels from crown glass (n=1:52) into air (n=1.00). The angle of refraction in

Physics
1 answer:
RUDIKE [14]3 years ago
8 0

The angle of incidence in glass is 34.7^{\circ}

Explanation:

We can solve this problem by applying Snell's law of refraction, which states that:

n_1 sin \theta_1 = n_2 sin \theta_2

where

n_1, n_2 are the index of refraction of the first and second medium, respectively

\theta_1, \theta_2 are the angle of incidence and refraction, respectively

In this problem we have:

n_1 = 1.52 is the index of refraction of the first medium (glass)

n_2 = 1.00 is the index of refraction of the second medium (air)

\theta_2 =60^{\circ} is the angle of refraction in glass

Solving for \theta_i, we find the angle of incidence:

\theta_1 = sin^{-1} (\frac{n_2 sin \theta_2}{n_1})=\sin^{-1}(\frac{(1.00)(sin 60^{\circ})}{1.52})=34.7^{\circ}

Learn more about refraction:

brainly.com/question/3183125

brainly.com/question/12370040

#LearnwithBrainly

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What is the critical angle of a light beam passed from a medium (n=2) to a medium (n=1.2)?
sertanlavr [38]

Answer:

θ_c = 36.87°

Explanation:

Index of refraction for index medium; n_i = 2

Index of refraction for Refractive medium; n_r = 1.2

Formula to find the critical angle is given;

n_i(sin θ_c) = n_r(sin 90)

Where θ_c is critical angle.

Thus;

2 × (sin θ_c) = 1.2 × 1

(sin θ_c) = 1.2/2

(sin θ_c) = 0.6

θ_c = sin^(-1) 0.6

θ_c = 36.87°

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Which of the following statements about electromagnetic radiation it true? A.electromagnetic waves with long wavelength are more
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Answer:

Electromagnetic radiation is an electric and magnetic disturbance traveling through space at the speed of light (2.998 × 108 m/s). It contains neither mass nor charge but travels in packets of radiant energy called photons, or quanta.

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Which item is a pure substance?
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Answer:

Solid gold. 10 carat indicates gold purity is not 100%. For 24 karat gold is pure gold.

Explanation:

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3. What do we call the ONLY part of the electromagnetic spectrum that we can
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Answer:

Visible Light

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6 0
2 years ago
Two charged point-like objects are located on the x-axis. The point-like object with charge q1 = 4.60 µC is located at x1 = 1.25
mylen [45]

Answer:

a) the total electric potential is 2282000 V

b) the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

Explanation:

Given the data in the question and as illustrated in the image below;

a) Determine the total electric potential (in V) at the origin.

We know that; electric potential due to multiple charges is equal to sum of electric potentials due to individual charges

so

Electric potential at p in the diagram 1 below is;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we know that; Coulomb constant, k = 9 × 10⁹ C

q1 = 4.60 uC = 4.60 × 10⁻⁶ C

r1 = 1.25 cm = 0.0125 m

q2 = -2.06 uC = -2.06 × 10⁻⁶ C

location x2 = −1.80 cm; so r2 = 1.80 cm = 0.018 m

so we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0125 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.018 )

Vp = (3312000) + ( -1030000 )

Vp = 3312000 -1030000

Vp = 2282000 V

Therefore, the total electric potential is 2282000 V

b)

the total electric potential (in V) at the point with coordinates (0, 1.50 cm).

As illustrated in the second image;

r1² = 0.015² + 0.0125²

r1 = √[ 0.015² + 0.0125² ]

r1 = √0.00038125

r1 = 0.0195

Also

r2² = 0.015² + 0.018²

r2 = √[ 0.015² + 0.018² ]

r2 = √0.000549

r2 = 0.0234

Now, Electric Potential at P in the second image below will be;

Vp = V1 + V2

Vp = kq1/r1 + kq2/r2

we substitute

Vp = ( 9 × 10⁹ × 4.60 × 10⁻⁶/ 0.0195 ) + ( 9 × 10⁹ × -2.06 × 10⁻⁶ / 0.0234 )

Vp = 2123076.923 + ( -762962.962 )

Vp = 2123076.923 -792307.692

Vp =  1330769.23 V

Therefore, the total electric potential (in V) at the point with coordinates (0, 1.50 cm) is 1330769.23 V

4 0
2 years ago
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