During a double slit interference experiment, light with a wavelength of 5.0 × 10^ -7 m is used, and the separation between the slits is 1.0 mm . The screen is positioned 1.0 m from the slits. Determine how many bright

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Correct answer: C. 20

The fringe width in a double slit experiment is given by

\beta = \dfrac{\lambda D}{d}

Substituting the given values:

\lambda = 5.0 \times 10^{-7} \text{ m}

D = 1.0 \text{ m}

d = 1.0 \text{ mm} = 1.0 \times 10^{-3} \text{ m}

\beta = \dfrac{5.0 \times 10^{-7} \times 1.0}{1.0 \times 10^{-3}} = 5.0 \times 10^{-4} \text{ m}

The given width on the screen is

W = 1 \text{ cm} = 1.0 \times 10^{-2} \text{ m}

The number of bright fringes in this width is

n = \dfrac{W}{\beta}

n = \dfrac{1.0 \times 10^{-2}}{5.0 \times 10^{-4}} = 20

During a double slit interference experiment, light with a...

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