Light of wavelength 5.0 × 10^ -7 m is emitted in phase from two narrow slits. An interference pattern is formed on a screen placed 1.0 m away. Determine the separation between the sources required to yield a separation

Question

Quantrex Academy · Accepted Answer

Correct answer: A. 0.50 mm

The separation between consecutive maxima is the fringe width, given by the formula:

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

Rearranging for the separation between the sources

d

:

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

Substituting the given values

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

,

D = 1.0 \text{ m}

, and

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

:

d = \dfrac{5.0 \times 10^{-7} \times 1.0}{1.0 \times 10^{-3}}

d = 5.0 \times 10^{-4} \text{ m}

Converting to millimeters:

d = 0.50 \text{ mm}

Light of wavelength 5.0 × 10^ -7 m is emitted in phase from two...

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