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Light of wavelength 687 nm is incident on a single slit 0.75 mm wide. At what distance from the slit should a screen be placed if the second dark fringe in the diffraction pattern is to be 1.7 mm from the center of the diffraction pattern?

Respuesta :

Answer:

L = 0.93 meter

Explanation:

As we know that the diffraction pattern will show the minimum on the screen at the position where it follows the equation

[tex]a sin\theta = N\lambda[/tex]

here we know that for second fringe position we have

[tex]sin\theta = \frac{2\lambda}{a}[/tex]

[tex]sin\theta = \frac{2(687 nm)}{0.75 mm}[/tex]

[tex]sin\theta = 1.832 \times 10^{-3}[/tex]

[tex]\theta = 0.105 degree[/tex]

now we know that the position of minimum is given at 1.7 mm from the center

so we have

[tex]tan\theta = \frac{y}{L}[/tex]

[tex]tan(0.105) = \frac{1.7\times 10^{-3}}{L}[/tex]

[tex]L = 0.93 meter[/tex]

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