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andrew-mc [135]
3 years ago
9

Red light of wavelength 633 nmnm from a helium-neon laser passes through a slit 0.360 mmmm wide. The diffraction pattern is obse

rved on a screen 2.90 mm away. Define the width of a bright fringe as the distance between the minima on either side.
Physics
1 answer:
777dan777 [17]3 years ago
8 0

Answer:

      Δy= 5,075 10⁻⁶ m

Explanation:

The expression that describes the interference phenomenon is

      d sin θ = (m + ½) λ

As the observation is on a distant screen

     tan θ = y / x

     tan θ= sin θ/cos θ

As in ethanes I will experience the separation of the vines is small and the distance to the big screen

          tan θ = sin θ

Let's replace

     d y / x = (m + ½) λ

The width of a bright stripe at the difference in distance  

     y₁ = (m + ½) λ x / d

     m = 1

      y₁ = 3/2 λ x / d

Let's use m = 1, we look for the following interference,

             m = 2

             y₂ = (2+ ½) λ x / d

The distance to the screen is constant x₁ = x₂ = x₀

The width of the bright stripe is

           Δy = λ x / d (5/2 -3/2)

           Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)

           Δy= 5,075 10⁻⁶ m

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Answer:

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Explanation:

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height, h = 10 m

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Total energy of the system

T E = 50 J + 50 J

T E = 100 J

now,

A h = 10 m

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mg = 5 ..............(1)

at the top most Point the only Potential energy will be acting on the body.

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 100 = m g h

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9.8 Newton

Explanation:

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4 0
2 years ago
A 1-kilogram ball has a kinetic energy of 50 joules the speed of the ball is
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Answer:

10 m/s

Explanation:

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Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th
Kipish [7]

Answer:

Probability of tunneling is 10^{- 1.17\times 10^{32}}

Solution:

As per the question:

Velocity of the tennis ball, v = 120 mph = 54 m/s

Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = 2.0\times 10^{- 3}\ m

Max velocity of the tennis ball, v_{m} = 200\ mph = 89 m/s

Now,

The maximum kinetic energy of the tennis ball is given by:

KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J

Kinetic energy of the tennis ball, KE' = \frac{1}{2}mv^{2} = 0.5\times 0.1\times 54^{2} = 154.8\ m/s

Now, the distance the ball can penetrate to is given by:

\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}

\bar{h} = \frac{h}{2\pi} = \frac{6.626\times 10^{- 34}}{2\pi} = 1.0545\times 10^{- 34}\ Js

Thus

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}

\eta = 1.52\times 10^{-35}\ m

Now,

We can calculate the tunneling probability as:

P(t) = e^{\frac{- 2t}{\eta}}

P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}

P(t) = e^{-2.63\times 10^{32}}

Taking log on both the sides:

logP(t) = -2.63\times 10^{32} loge

P(t) = 10^{- 1.17\times 10^{32}}

6 0
3 years ago
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