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Two slits are illuminated by a 363 nm light. The angle between the zeroth-order bright band at the center of the screen and the fourth-order bright band is 14.9 ◦ . If the screen is 170 cm from the double-slit, how far apart is this bright band from the central peak

Respuesta :

Answer:

y = 0.44 m

Explanation:

As we know that path difference on the screen is given as

[tex]\Delta x = \frac{yd}{L}[/tex]

now for constructive interference we know that

path difference = integral multiple of wavelength

so we have

[tex]N\lambda = \frac{yd}{L}[/tex]

now for 4th maximum on the screen we can say

[tex]y = \frac{N\lambda L}{d}[/tex]

here N = 4

L = 170 cm = 1.70 m

[tex]\lambda = 363 nm[/tex]

also we know

path difference on screen = d[tex]sin\theta[/tex]

[tex]4\lambda = dsin14.9[/tex]

[tex]d = \frac{4\lambda}{sin14.9}[/tex]

now we have

[tex]y = \frac{4(\lambda)(1.70)}{\frac{4\lambda}{sin14.9}}[/tex]

[tex]y = 0.44 m[/tex]

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