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3 years ago

14

Parallel rays from a distant object are traveling in air and then are incident on the concave end of a glass rod with a radius o

f curvature of 15.0 cm. The refractive index of the glass is 1.50. What is the distance between the vertex of the glass surface and the image formed by the refraction at the concave surface of the rod? Is the image in the air or in the glass?

1 answer:

skad [1K]3 years ago

8 0

Answer:

the distance of image from the vertex is 45 cm and the image formed is in the glass.

Explanation:

distance of object, u = - infinity

radius of curvature, R = - 15 cm

refractive index, n = 1.5

Let the distance of image is v.

Use the formula

$-\frac{n1}{u}+\frac{n2}{v}=\frac{n2- n1}{R}\\\\-\frac{1}{\infty }+\frac{1.5}{v}=\frac{1.5-1}{-15}\\\\v=45 cm$

The image is in the glass.

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To solve this problem we will apply the concepts related to resistance as a function of temperature, product of the relationship between the squared voltage and the power. Mathematically this is,

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v = Voltage

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Then we have,

R at 140°C (7 times room temperature),

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The relationship between normal temperature and increased temperature would then be given by,

$R(140\°C) = R(20\°C)(1 +\alpha (\Delta T))$

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Answer:

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b) The equivalent temperature of 2000 ºC is 4091.67 R.

c) The equivalent temperature of 2000 ºC is 2273.15 K.

d) The equivalent temperature of 2000 ºC is 1600 ºRe.

Explanation:

a) The equivalent temperature on the Fahrenheit scale is defined by the following formula:

$T_{F} = \frac{9}{5}\cdot T_{C}+32$ (Eq. 1)

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$T_{C}$ - Temperature, measured in degrees Celsius.

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If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

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$T_{F} = 3632\,^{\circ}F$

The equivalent temperature of 2000 ºC is 3632 ºF.

b) From result in a) we determine the equivalent temperature on the Rankine scale by using the following formula:

$T_{R} = T_{F}+459.67$ (Eq. 2)

Where:

$T_{F}$ - Temperature, measured in degrees Fahrenheit.

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If we know that $T_{F} = 3632\,^{\circ}F$ , then the temperature is:

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$T_{R} = 4091.67\,R$

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$T_{K} = T_{C}+273.15$ (Eq. 3)

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$T_{C}$ - Temperature, measured in degrees Celsius.

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$T_{K} = 2273.15\,K$

The equivalent temperature of 2000 ºC is 2273.15 K.

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Where:

$T_{C}$ - Temperature, measured in degrees Celsius.

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If we know that $T_{C} = 2000\,^{\circ}C$ , then the temperature is:

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