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jok3333 [9.3K]
3 years ago
14

The index of refraction for red light in a certain liquid is 1.303; the index of refraction for violet light in the same liquid

is 1.326. part a find the dispersion θv−θr for red and violet light when both are incident on the flat surface of the liquid at an angle of 45.00 ∘ to the normal.
Physics
1 answer:
kotegsom [21]3 years ago
3 0
Snell's law states: n1/n2 = Sin Ф2/Sin Ф1

But n2/n1 = 1.303 and 1.326 for red light and violet light respectively (for this case) and Ф1 = 45

Therefore,
Фr = Sin ^-1{1/1.303 *Sin 45} = 32.87°
Фv = Sin ^-1 {1/1.326* Sin 45} = 32.23°

Then,
Dispersion, Фv - Фr = 32.23 - 32.87 = -0.64°
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Varvara68 [4.7K]

Answer:

-10.8°, or 10.8° below the +x axis

Explanation:

The x component of the resultant vector is:

x = 3.14 cos(30.0°) + 2.71 cos(-60.0°)

x = 4.07

The y component of the resultant vector is:

y = 3.14 sin(30.0°) + 2.71 sin(-60.0°)

y = -0.777

Therefore, the angle between the resultant vector and the +x axis is:

θ = atan(y / x)

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θ = -10.8°

The angle is -10.8°, or 10.8° below the +x axis.

3 0
3 years ago
The current that charges a capacitor transfers energy that is stored in the capacitor’s electric field. Consider a 2.0 μF capaci
lapo4ka [179]

Answer:

the capacitor voltage is V = 20 V

Explanation:

Given,

Capacitance of the capacitor =  2.0 μF

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time (t) =2.0 μs

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\dfrac{dE}{dt}=200 W

E =200\times 2 \times 10^{-6}

E =400 \times 10^{-6}\ J

we know,

E = \dfrac{1}{2}CV^2

V =\sqrt{\dfrac{2E}{C}}

V =\sqrt{\dfrac{2\times 400 \times 10^{-6}}{2\times 10^{-6}}}

V =\sqrt{400}

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A 1.50 cm high diamond ring is placed 20.0 cm from a concave mirror with radius of curvature 30.00 cm. The magnification is ____
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Answer:

Magnification, m = -0.42

Explanation:

It is given that,

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Focal length of mirror, f = R/2 = -15 cm (focal length is negative for concave mirror)

Using mirror's formula :

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\dfrac{1}{v}=\dfrac{1}{-15}+\dfrac{1}{-20}

v = -8.57 cm

The magnification of a mirror is given by,

m=\dfrac{-v}{u}

m=\dfrac{-(-8.57)}{-20}

m = -0.42

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