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wel
1 year ago
9

In a double slit experiment, if the separation between the two slits is 0.050 mm and the distance from the slits to a screen is

2.5 m, find the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits. Group of answer choices
Physics
1 answer:
meriva1 year ago
8 0

The spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

<h3>What is double slit experiment?</h3>

The double-slit experiment serves as a proof in current physics that both light and matter may exhibit properties of classically defined waves and particles. It also illustrates the inherently probabilistic nature of quantum mechanical events. Thomas Young carried out the first experiment of this kind employing light in 1802, illustrating how light behaves like a wave. It was formerly believed that light was made up of either waves or particles. About a century later, with the advent of modern physics, it was discovered that light may in fact exhibit behaviour like that of both waves and particles. The identical behaviour of electrons was first shown by Davisson and Germer in 1927, and it was later extended to atoms and molecules.

The separation between the slits, d = 0.05mm = 5×10⁻⁵ m

The distance from the slits to a screen, D = 2.5 m

Let x is the spacing between the first-order and second-order bright fringes when coherent light of wavelength 600 nm illuminates the slits,

λ = 600nm = 6× 10⁻⁷ m  

We know that the bright fringe is given by :

y = nλD/d

So, the spacing between the first-order and second-order bright fringes is :

x = 2λD/d - λD/d

x =  λD/d

x = 6 × 10⁻⁷ × 2.5/5 × 10⁻⁵

x = 0.03 m

or

x = 3 cm

So, the spacing between the first-order and second-order bright fringes is 3 cm. Hence, this is the required solution.

to learn more about double slit experiment go to -

brainly.com/question/28108126

#SPJ4

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andre [41]

Answer:

d=1.84\ mm

Explanation:

<u>Capacitance</u>

A two parallel-plate capacitor has a capacitance of

\displaystyle C=\frac{\epsilon_o A}{d}

where

\epsilon_o=8.85\cdot 10^{-12}\ F/m

A = area of the plates = \pi r^2

d = separation of the plates

\displaystyle d=\frac{\epsilon_o A}{C}=\frac{\epsilon_o \pi r^2}{C}

We need to compute C. We'll use the circuit parameters for that. The reactance of a capacitor is given by

\displaystyle X_c=\frac{1}{wC}

where w is the angular frequency

w=2\pi f=2\pi \cdot 36000=226194.67\ rad/s

Solving for C

\displaystyle C=\frac{1}{wX_c}

The reactance can be found knowing the total impedance of the circuit:

Z^2=R^2+X_c^2

Where R is the resistance, R=15 K\Omega=15000\Omega. Solving for Xc

X_c^2=Z^2-R^2

The magnitude of the impedance is computed as the ratio of the rms voltage and rms current

\displaystyle Z=\frac{V}{I}

The rms current is the peak current Ip divided by \sqrt{2}, thus

\displaystyle Z=\frac{\sqrt{2}V}{I_p}

I_p=0.65\ mA/1000=0.00065\ A

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\displaystyle X_c^2=Z^2-R^2=\left(\frac{\sqrt{2}V}{I_p}\right)^2-R^2

Or, equivalently

\displaystyle X_c=\sqrt{\frac{2V^2}{I_p^2}-R^2}

\displaystyle X_c=\sqrt{\frac{2\cdot 18^2}{0.00065^2}-15000^2}

X_c=36176.34\ \Omega

The capacitance is now

\displaystyle C=\frac{1}{226194.67\cdot 36176.34}=1.22\cdot 10^{-10}\ F

The radius of the plates is

r=18\ cm/2=9 \ cm = 0.09 \ m

The separation between the plates is

\displaystyle d=\frac{8.85\cdot 10^{-12} \cdot \pi\cdot 0.09^2}{1.22\cdot 10^{-10}}

d=0.00184\ m

\boxed{d=1.84\ mm}

8 0
3 years ago
A 2190 kg car moving east at 10.5 m/s collides with a 3220 kg car moving east. The cars stick together and move east as a unit a
Bezzdna [24]

To solve this problem it is necessary to apply the concepts related to the conservation of the Momentum describing the inelastic collision of two bodies. By definition the collision between the two bodies is given as:

m_1v_1+m_2v_2 = (m_1+m_2)V_f

Where,

m_{1,2}= Mass of each object

v_{1,2}= Initial Velocity of Each object

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Our values are given as

m_1 = 2190Kg

v_1 =10.5m/s

m_2 = 3220kg

V_f = 4.74m/s

Replacing we have that

m_1v_1+m_2v_2 = (m_1+m_2)V_f

(2190)(10.5)+(3220)v_2 = (2190+3220)(4.74)

v_2 = 0.8224m/s

Therefore the the velocity of the 3220 kg car before the collision was 0.8224m/s

8 0
4 years ago
A business letter is informal correspondence and is sent to people within
jarptica [38.1K]

Answer:

FALSE!!!

Explanation:

Business letters are formal and normally given to someone of higher importance.

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3.86 m/s^2  is the value of gravity  on this large, but low-density, world.

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Kepler-12b

diameter= 1.7 times of Jupiter (R_Jupiter = 6.99 × 10^7 m),

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g = GM/r^2

g = 6.67×10^-11 × 0.43×1.9×10^27/( 1.7×6.99×10^7)^2

g = 3.859 ~ 3.86 m/s^2

Gravity, also referred to as gravitation, is the unchanging force of attraction that binds all matter together in mechanics. It is by far the weakest known force in nature, so it has no effect on determining the internal properties of common matter.

On Earth, everything has weight, or a gravitational pull that is imposed by the planet's mass and proportional to the object's mass. A measure of the force of gravity is the acceleration that freely falling objects experience. At the surface of the Earth, gravity accelerates at a rate of about 9.8 meters per second. As a result, an object's speed increases during free fall by about 9.8 meters per second. At the Moon's surface, a freely falling body accelerates to about 1.6 m/s2.

To know more about  gravity visit : brainly.com/question/14428640

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6 0
1 year ago
According to Newton's second law of motion, force can be calculated by multiplying an object's mass by its acceleration. If weig
madreJ [45]
You would do m*a so 65kg * 9.8 m/s^2 which equates to 637 Newton’s
4 0
3 years ago
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