Question 6 Let Y be the distance from the center of the central fringe to the fi

Question

Question 6 Let Y be the distance from the center of the central fringe to the first minimum and let W be the slit width. If the central fringe is X times wider than the slit, then 2y = XW. Let L be the distance from the slit to the screen. If the screen is z times further from the slit than the slit is wide, then L = ZW . In a single-slit diffraction pattern, the central fringe is 350 times as wide as the slit. The screen is 13,000 times farther from the slit than the slit is wide. What is the ratio lambda/W, where lambda is the wavelength of the light shining through the slit and W is the width of the slit? Assume that the angle that locates a dark fringe on the screen is small, so that sin theta is approximately equal to tan theta. the tolerance is +/-3%

Explanation / Answer

the central fringe is 350 times as wide as the slit

so 2y = 350W

or y = 175W --------(1)

the distance between slit & screen

    L = 13000W -----------(2)

    now we have

      tan@ = y/L

from (1) & (2) we have

    tan@ = 175W / 13000W  -------(3)

and sin@ = ?/W -----(4)

since the angle @ is very small , so sin @ = tan@

    from (3) & (4) we get

      so

      175W / 13000W   =  ?/W

   or ?/W = 0.0135

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