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ozzi
2 years ago
5

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

a layer of silicon dioxide (n = 1.45). What is the minimum coating thickness that will minimize the reflection at the wavelength of 700 nm, where solar cells are most efficient?
Physics
1 answer:
alisha [4.7K]2 years ago
8 0

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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Law Incorporation [45]
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3 0
2 years ago
What conclusion can be derived by comparing the central tendencies of the two data sets?
zhannawk [14.2K]
The answer is B. I don’t think I need to explain this,
Mean is average, Mode is the most common number, and Median is the middle number when you put the numbers is numerical order from least to greatest
3 0
2 years ago
Odległość między kolejnymi grzbietami fal na morzu wynosi 20 m. Łódź opada z grzbietu fali, unosi się i osiąga ponownie najwyższ
Veronika [31]

Answer:

Explanation:

The distance between successive wave crests at sea is 20 m. The boat descends from the crest of the wave, rises and reaches the highest position again within 5 s. Calculate the wave propagation speed.

Given that,

The distance between two successive crest is 20m

Wavelength is the distance between two successive crest or trough

Then, it's wavelength is λ = 20m

The time to reached the maximum height is 5seconds, then it will take (5×4) to complete one period

Then,

Period T = 20seconds

From wave equation

v = fλ

Where

v is speed

f is frequency and

λ is wavelength

The frequency is related to the period

f =  1 / T

Then,

v = λ / T

So, v = 20 / 20

v = 1 m/s

The speed of propagation of the wave is 1m/s

To Polish

Jeśli się uwzględni,

Odległość między dwoma kolejnymi grzebieniami wynosi 20 m

Długość fali to odległość między dwoma kolejnymi grzebieniami lub dolinami

Zatem jego długość fali wynosi λ = 20 m

Czas do osiągnięcia maksymalnej wysokości wynosi 5 sekund, a następnie ukończenie jednego okresu zajmie (5 × 4)

Następnie,

Okres T = 20 sekund

Z równania falowego

v = fλ

Gdzie

v to prędkość

f oznacza częstotliwość, a

λ jest długością fali

Częstotliwość jest związana z okresem

f = 1 / T

Następnie,

v = λ / T

Zatem v = 20/20

v = 1 m / s

Prędkość propagacji fali wynosi 1m/s

6 0
2 years ago
The wind blows a jay bird south with a force of 300 Newtons. The
adoni [48]

Answer:

F = 316.22 N

Explanation:

Given that,

The wind blows a jay bird south with a force of 300 Newtons.

The  jay bird flies north, against the wind, with a force of 100  newtons.

Both the forces are acting perpendicular to each other. The net force is given by the resultant of forces as follows :

F=\sqrt{300^2+100^2} \\\\F=316.22\ N

Hence, the net force on the jay bird is 316.22 N.

6 0
2 years ago
A ball weighing 1 lb is attached to a string 2 feet long and is whirled in a vertical circle at a constant speed of 10 ft/sec.
fredd [130]

Explanation:

It is given that,

Mass of the ball, m = 1 lb

Length of the string, l = r = 2 ft

Speed of motion, v = 10 ft/s

(a) The net tension in the string when the ball is at the top of the circle is given by :

F=\dfrac{mv^2}{r}-mg

F=m(\dfrac{v^2}{r}-g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}-1\ lb\times 32\ ft/s^2)

F = 18 N

(b) The net tension in the string when the ball is at the bottom of the circle is given by :

F=\dfrac{mv^2}{r}+mg

F=m(\dfrac{v^2}{r}+g)

F=1\ lb\times (\dfrac{(10\ ft/s)^2}{2}+1\ lb\times 32\ ft/s^2)

F = 82 N

(c) Let h is the height where the ball at certain time from the top. So,

T=mg(\dfrac{r-h}{r})+\dfrac{mv^2}{r}

T=\dfrac{m}{r}(g(r-h)+v^2)

Since, v^2=u^2-2gh

T=\dfrac{m}{r}(u^2-3gh+gr)

Hence, this is the required solution.

6 0
2 years ago
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