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EastWind [94]
2 years ago
15

A laser is shone through a double slit and a particular interference pattern is observed on a screen some distance away. If the

separation between the openings is increased, the distance between the fringes will
Physics
1 answer:
Grace [21]2 years ago
5 0

If the separation between the openings in a laser is increased, then the distance between the interference fringes decreases

<h3>What is Interference fringe ?</h3>

Interference fringe refers to bands caused by different lights which can be found in phase or not each other.

  • Distances between laser fringes are short which is due to light wavelength.
  • The interference fringes can be estimated by knowing slit separation and wavelength.

In conclusion, if the separation between the openings in a laser is increased, then the distance between the interference fringe decreases

Learn more about Interference fringe here:

brainly.com/question/14264436

#SPJ4

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Find the magnitude and direction of an electric field that exerts a 4.80×10−17N westward force on an electron. (b) What magnitud
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Explanation:

(a) E = F/q

E = 4.8×10^-17/1.6×10^-19

E = 300 N/C

(b) same magnitude of electric field is exerted on proton

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3 years ago
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2 years ago
Two astronauts (each with mass 100 kg) are drifting together through space. They are connected to each other by a rope 5 m in le
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Answer:

1000 kgm²/s, 400 J

1000 kgm²/s, 1000 J

600 J

Explanation:

m = Mass of astronauts = 100 kg

d = Diameter

r = Radius = \frac{d}{2}

v = Velocity of astronauts = 2 m/s

Angular momentum of the system is given by

L=mvr+mvr\\\Rightarrow L=2mvr\\\Rightarrow L=2\times 100\times 2\times 2.5\\\Rightarrow L=1000\ kgm^2/s

The angular momentum of the system is 1000 kgm²/s

Rotational energy is given by

K=I\omega^2\\\Rightarrow K=\frac{1}{2}(mr^2)\left(\frac{v}{r}\right)^2\\\Rightarrow K=mv^2\\\Rightarrow K=100\times 2^2\\\Rightarrow K=400\ J

The rotational energy of the system is 400 J

There no external toque present so the initial and final angular momentum will be equal to the initial angular momentum 1000 kgm²/s

L_i=L_f\\\Rightarrow 2mv_ir_i=2mv_fr_f\\\Rightarrow v_f=\frac{v_ir_i}{r_f}\\\Rightarrow v_f=\frac{2\times 2.5}{0.5}\\\Rightarrow v_f=10\ m/s

Energy

E_2=mv_f^2\\\Rightarrow E_2=100\times 10\\\Rightarrow E_2=1000\ J

The new energy will be 1000 J

Work done will be the change in the kinetic energy

W=E_2-E\\\Rightarrow W=1000-400\\\Rightarrow W=600\ J

The work done is 600 J

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3 years ago
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