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siniylev [52]
3 years ago
10

A two-slit pattern is viewed on a screen 1.26 m from the slits. If the two fourth-order maxima are 53.6 cm apart, what is the to

tal width of the central bright fringe?
Physics
1 answer:
Anettt [7]3 years ago
4 0

Answer:

= 6.55cm

Explanation:

Given that,

distance = 1.26 m

distance between  two fourth-order maxima = 53.6 cm

distance between central bright fringe and fourth order maxima

y = Y / 2

  =  53.6cm / 2

  = 26.8 cm

  =0.268 m

tan θ = y / d

         = 0.268 m /  1.26 m

         = 0.2127

       θ = 12°

4th maxima

d sinθ = 4λ

d / λ = 4 / sinθ

d / λ = 4 / sin 12°

d / λ = 19.239

for first (minimum)

d sinθ = λ / 2

sinθ =  λ / 2d

       =  1 / 2(19.239)

       = 1 / 38.478

       = 0.02599

    θ =  1.489°

tan θ = y / d

y = d tan θ

  = 1.26 tan 1.489°

  = 0.03275

the total width of the central bright fringe  

Y = 2y

  = 2(0.03275)

  = 0.0655m

  = 6.55cm

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