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jeka94
2 years ago
8

Light of wavelength 510 nm is used to illuminate normally two glass plates 24.3 cm in length that touch at one end and are separ

ated at the other by a wire of radius 0.023 mm. How many bright fringes appear along the total length of the plates.
Physics
1 answer:
bulgar [2K]2 years ago
3 0

Answer:

m = 180 \ bright \ fringes

Explanation:

From the question we are told that

    The  wavelength is  \lambda  =  510 \ nm = 510 *10^{-9} \  m

    The length of the of the plate is  l  =  24.3 \ cm  =  0.243 \  m

    The radius of the wire  is r=  0.023 \ mm  =  2.3 *10^{-5} \ m

Generally the diameter of the wire which is the distance between the glass  plates is mathematically represented as

             d =  2 *   r

=>          d =  2 *   2.3 *10^{-5}

=>          d =  4.6 *10^{-5} \  m

 Generally the condition for constructive interference is mathematically represented as

           2 *d = m \lambda

=>        m = \frac{2d}{\lambda}

=>        m = \frac{2 * (4.6* 10^{-5})}{ 510 *10^{-9} }

=>        m = 180 \ bright \ fringes

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Explanation:

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then

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⇒   0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒  ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

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⇒  (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒  (ω₁ / ω₂) = 1.9079

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