Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.2m be

Question

Light from a helium-neon laser (? = 633 nm) is used to illuminate two narrow slits. The interference pattern is observed on a screen 3.2m behind the slits. Eleven bright fringes are seen, spanning a distance of 60mm .

What is the spacing (in mm) between the slits?

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Adela 5 years 2021-08-22T03:30:23+00:00 2 Answers 25 views 0

Answers ( )

  1. hongcuc2
    0
    2021-08-22T03:31:43+00:00
    Reply

    Answer:

    0.3376 mm

    Explanation:

    The computation of the spacing in mm between the slits is shown below:

    As we know that

    d = \frac{m\lambda L}{\Delta y}

    where,

    \lambda = wavelength

    L = distance from the scrren

    \Delta y = spanning distance

    As there are 11 bright fingers seen so m would be

    = 11 – 1

    = 10

    Now placing these values to the above formula

    So, the spacing is

    = \frac{(10)(633 \times 10^{-9})(3.2m)}{60 \times 10^{-3}}

    = 0.3376 mm

    We simply applied the above formula.

  2. Dulcie
    0
    2021-08-22T03:32:20+00:00
    Reply

    Answer:

    Explanation:

    Maximum occurs when the path difference is an integral multiple of wavelength

    Here \lambda – Wavelength, d- slit separation and m- Order of pattern

    Rearrange the equation for 

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\begin{aligned}d &=\frac{m \lambda}{\sin \theta} \\\text { Here, } \sin \theta &=\frac{y}{L} \quad\left(\begin{array}{l}\text { Here, } L-\text { separation between slit and screen } \\y-\text { Distance between respective fringe from center on screen }\end{array}\right)

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    d=\frac{m \lambda}{\left(\frac{y}{L}\right)} \\&=\frac{m \lambda L}{y}

    Here, order

    Due to the fact that there are 11 bright fringes seen, you take 11-1=10

    since starts from 0,1,2,3

    Substitute given values

    \begin{aligned}d &=\frac{(10)\left(633 \times 10^{-9} \mathrm{m}\right)(3.2 \mathrm{m})}{60 \times 10^{-3} \mathrm{m}} \\&=\left(3.376 \times 10^{-4} \mathrm{m}\right)\left(\frac{1 \mathrm{mm}}{10^{-3} \mathrm{m}}\right) \\&=0.3376 \mathrm{mm}\end{aligned}

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