In a Young's double slit experiment, the separation between consecutive dark fringes is 1.0 mm . The screen is situated at a distance of 2.5 m from the slits, while the separation between the slits is 1.0 mm . Determine

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Quantrex Academy · Accepted Answer

Correct answer: C. 400 nm

The separation between consecutive dark fringes is the fringe width

\beta

.

The formula for fringe width is given by:

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

Rearranging for wavelength

\lambda

:

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

Given values are:

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

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

D = 2.5 \text{ m}

Substituting the values into the formula:

\lambda = \dfrac{(1.0 \times 10^{-3}) \times (1.0 \times 10^{-3})}{2.5}

\lambda = \dfrac{10^{-6}}{2.5} \text{ m}

\lambda = 0.4 \times 10^{-6} \text{ m}

\lambda = 400 \times 10^{-9} \text{ m} = 400 \text{ nm}

In a Young's double slit experiment, the separation between...

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