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sergey [27]
2 years ago
15

Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed

on a screen 3.35 m away. Define the width of a bright fringe as the distance between the minima on either side.
(a) What is the width of the central bright fringe?

(b) What is the width of the first bright fringe on either side of the central one?
Physics
1 answer:
kogti [31]2 years ago
4 0

Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
  • D is the distance from the slit to the screen
  • a is the slit wide
  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

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Answer:

a) La magnitud del desplazamiento es de 5 m

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c) El ángulo entre la fuerza y el plano es 0 °

Explanation:

a) La magnitud del desplazamiento se encuentra por la relación;

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\left | X \right | = \sqrt{4^{2}+3^{2}} = 5 \ m

De manera similar, la magnitud de la fuerza, F, se encuentra como sigue;

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Lo que da;

\left | F \right | = \sqrt{16^{2}+12^{2}} = 20 \ N

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c) La dirección de la fuerza viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{F_y}{F_x} \right ) = tan^{-1} \left (\dfrac{12}{16} \right )  = 38.9^{\circ}

La dirección del plano viene dada por la siguiente fórmula;

tan^{-1} \left (\dfrac{X_y}{X_x} \right ) = tan^{-1} \left (\dfrac{3}{4} \right )  = 38.9^{\circ}

Por tanto, el ángulo entre la fuerza y el plano = 0 °

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