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Romashka-Z-Leto [24]
3 years ago
14

Structures on a bird feather act like a diffraction grating having 8000 8000 lines per centimeter. What is the angle of the firs

t-order maximum for 452 nm 452 nm light shone through a feather?
Physics
1 answer:
sammy [17]3 years ago
6 0

Answer:

Angle of first order maximum, \theta=21.19^{\circ}

Explanation:

Given that,

Wavelength of the light, \lambda=452\ nm=452\times 10^{-9}\ m

Number of lines, N = 8000 per cm

The relation between the number of lines and the slit width is given by :

d=\dfrac{1}{N}

d=\dfrac{1}{8000/cm}  

d=0.000125\ cm=1.25\times 10^{-6}\ m

The equation of grating is given by :

d\ sin\theta=n\lambda

n = 1

sin\theta=\dfrac{\lambda}{d}

sin\theta=\dfrac{452\times 10^{-9}}{1.25\times 10^{-6}}

\theta=21.19^{\circ}

So, the angle of the first-order maximum is 21.19 degrees. Hence, this is the required solution.

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arsen [322]

Answer:

40

Explanation:

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The balance between incoming solar energy and out going energy radiated into space is called...
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A metallic sheet has a large number of slits, 5.0 mm wide and 20 cm apart, and is used as a diffraction grating for microwaves.
san4es73 [151]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and constructive interference, that is to say everything that refers to an overlap of two or more equal frequency waves, which when interfering create a new pattern of waves of greater intensity (amplitude) whose cusp is the antinode.

Mathematically its definition can be given as:

d sin\theta =m\lambda

Where

d = Width of the slit

\theta = Angle between the beam and the source

m = Order (any integer) which represent the number of repetition of the spectrum, at this case 1 (maximum respect the wavelength)

Since the point of the theta angle for which the diffraction becomes maximum will be when it is worth one then we have to:

\lambda = d

\lambda = 20cm = 20*10^{-2}m

Applying the given relation of frequency, speed and wavelength then we will have that the frequency would be:

f = \frac{v}{\lambda}

Here the velocity is equal to the speed of light and the wavelength to the value previously found.

f = \frac{3*10^8}{0.2}

f = 1.5Ghz

Therefore the smallest microwave frequency for which only the central maximum occurs is 1.5Ghz

7 0
3 years ago
A source produces 20 crests and 20 troughs in 4 seconds. The second crest is 3 cm away from the first crest.Calculate :
sineoko [7]

Answer:

Solution given:

No of waves[N] =20crests & 20 troughs

=20waves

Time[T]=4seconds

distance[d]=3cm=0.03m

Now

<u>Wave</u><u> </u><u>length</u><u>=</u>3cm=3 × {10}^{ - 2}m

<u>Frequency</u>=\frac{No of waves}{time}

=\frac{20}{4}=5Hertz

and

Wave speed:wave length×frequency=3 × {10}^{ - 2}m×5=1.5 × {10}^{ - 1} \tt{ {ms}^{ - 1}}.

3 0
2 years ago
An Atwood machine consists of two masses, mA = 6.8 kg and mB = 8.0 kg , connected by a cord that passes over a pulley free to ro
Lisa [10]

To solve this problem it is necessary to apply the concewptos related to Torque, kinetic movement and Newton's second Law.

By definition Newton's second law is described as

F= ma

Where,

m= mass

a = Acceleration

Part A) According to the information (and as can be seen in the attached graph) a sum of forces is carried out in mass B, it is obtained that,

\sum F = m_b a

m_Bg-T_B = m_Ba

T_B = m_Bg-m_Ba

In the case of mass A,

\sum F = m_A a

T_A = m_Ag-m_Aa

Making summation of Torques in the Pulley we have to

\sum\tau = I\alpha

T_BR_0-T_AR_0=I\alpha

T_B-T_A=I\frac{a}{R^2_0}

Replacing the values previously found,

(m_Bg-m_Ba )-(m_Ag-m_Aa )=I\frac{a}{R^2_0}

(m_B-m_A)g-(m_B+m_A)a=I\frac{a}{R^2_0}

a = \frac{(m_B-m_A)g}{\frac{I}{R_0^2}+(m_B+m_A)}

a = \frac{(m_B-m_A)g}{\frac{MR^2_0^2/2}{R_0^2}+(m_B+m_A)}

a =\frac{(m_B-m_A)g}{\frac{M}{2}+(m_B+m_A)}

Replacing with our values

a =\frac{(8-6.8)(9.8)}{\frac{0.8}{2}+(8+6.8)}

a=0.7736m/s^2

PART B) Ignoring the moment of inertia the acceleration would be given by

a' =\frac{(m_B-m_A)g}{(m_B+m_A)}

a' =\frac{(8-6.8)(9.8)}{(8+6.8)}

a' = 0.7945

Therefore the error would be,

\%error = \frac{a'-a}{a}*100

\%error = \frac{0.7945-0.7736}{0.7736}*100

\%error = 2.7%

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