Coherent light with wavelength = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed on a screen t

Question

Coherent light with wavelength = 600 nm falls on two very narrow closely spaced slits and the interference pattern is observed on a screen that is 4 m from the slits. Near the center of the secreen the separation between adjacent maxima is 2 mm. What is the distance between the two slits?

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Farah 5 years 2021-08-22T23:53:33+00:00 1 Answers 5 views 0

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  1. thuhuong
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    2021-08-22T23:55:25+00:00
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    Answer:

    The distance between the two slits is 1.2mm.    

    Explanation:

    The physicist Thomas Young establishes, through its double slit experiment, a relationship between the interference (constructive or destructive) of a wave, the separation between the slits, the distance between the two slits to the screen and the wavelength.

    \Lambda x = L\frac{\lambda}{d}  (1)

    Where \Lambda x is the distance between two adjacent maxima, L is the distance of the screen from the slits, \lambda is the wavelength and d is the separation between the slits.  

    If light pass through two slits a diffraction pattern in a screen will be gotten, at which each bright region corresponds to a crest, a dark region to a trough, as consequence of constructive interference and destructive interference in different points of its propagation to the screen.  

    Therefore, d can be isolated from equation 1.

    d = L\frac{\lambda}{\Lambda x}  (2)

    Notice that it is necessary to express L and \lambda in units of millimeters.

    L = 4m \cdot \frac{1000mm}{1m} ⇒ 4000mm

    \lambda = 600nm \cdot \frac{1mm}{1x10^{6}nm} ⇒ 0.0006mm

    d = (4000mm)\frac{0.0006mm}{2mm}

    d = 1.2mm

    Hence, the distance between the two slits is 1.2mm.

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