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Zinaida [17]
3 years ago
13

If a soap bubble is 115 nm thick, what wavelength is most strongly reflected at the center of the outer surface when illuminated

normally by white light? Assume that n = 1.33. Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
sp2606 [1]3 years ago
7 0

Answer:

611.8 nm

Explanation:

n_{oil} = Index of refraction of soap bubble  = 1.33

t = thickness of the soap bubble = 115 nm = 115 x 10⁻⁹ m

\lambda = wavelength of light = ?

m = order = 0

For reflection , the necessary condition is

2 n_{oil} t = (m + 0.5) \lambda

2 (1.33)(115\times 10^{-9})= (0 + 0.5) \lambda

\lambda = 6.118\times 10^{-7}

\lambda = 611.8 nm

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What is the energy of a photon whose frequency is 6.0 x 10^20?
omeli [17]

Answer:

3.75 MeV

Explanation:

The energy of the photon can be given in terms of frequency as:

E = h * f

Where h = Planck's constant

The frequency of the photon is 6 * 10^20 Hz.

The energy (in Joules) is:

E = 6.63 x10^(-34) * 6 * 10^(20)

E = 39.78 * 10^(-14) J = 3.978 * 10^(-13) J

We are given that:

1 eV = 1.06 * 10^(-19) Joules

This means that 1 Joule will be:

1 J = 1 / (1.06 * 10^(-19)

1 J = 9.434 * 10^(18) eV

=> 3.978 * 10^(-13) J = 3.978 * 10^(-13) * 9.434 * 10^(18) = 3.75 * 10^(6) eV

This is the same as 3.75 MeV.

The correct answer is not in the options, but the closest to it is option C.

6 0
3 years ago
An open pipe of length 0.39 m vibrates in the third harmonic with a frequency of 1400 Hz what is the distance from the center of
Svetach [21]

Length of the pipe = 0.39 m

Number of harmonics = 3

Now there are 3 loops so here we can say

3\times \frac{\lambda}{2} = 0.39

\lambda = 0.26 m

now here at the center of the pipe it will form Node

we need to find the distance of nearest antinode

So distance between node and its nearest antinode will be

d = \frac{\lambda}{4}

d = \frac{0.26}{4} = 0.065 m = 6.5 cm

So the distance will be 6.5 cm

3 0
3 years ago
Which is capeble of housing astronaughts while they conduct reasearch
dusya [7]

Answer:

2

Explanation:

2

5 0
3 years ago
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In the high jump, the kinetic energy of an athlete is transformed into gravitational potential energy without the aid of a pole.
Fiesta28 [93]

Answer:

6.0 m/s

Explanation:

According to the law of conservation of energy, the total mechanical energy (potential, PE, + kinetic, KE) of the athlete must be conserved.

Therefore, we can write:

KE_i+PE_i =KE_f+PE_f

or

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where:

m is the mass of the athlete

u is the initial speed of the athlete (at the bottom)

0 is the initial potential energy of the athlete (at the bottom)

v = 0.80 m/s is the final speed of the athlete (at the top)

g=9.8 m/s^2 is the acceleration due to gravity

h = 1.80 m is the final height of the athlete (at the top)

Solving the equation for u, we find the initial speed at which the athlete must jump:

u=\sqrt{v^2+2gh}=\sqrt{0.80^2+2(9.8)(1.80)}=6.0 m/s

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Verdich [7]

Answer:

Wavelength?

Explanation:

7 0
3 years ago
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