Light of wavelength 580 nm is incident on a slit having a width of `0.300` nm The viewing screen is 2.00 m from the slit. Find the positions of the fi

Question

Light of wavelength 580 nm is incident on a slit having a width of `0.300` nm The viewing screen is 2.00 m from the slit. Find the positions of the fi

Light of wavelength 580 nm is incident on a slit having a width of `0.300` nm The viewing screen is 2.00 m from the slit. Find the positions of the first dark fringe and the width of the central bright fringe.

Monjur Alahi

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2025-01-04 04:42

1 Answers

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Salim Ahmed Kawsar · Accepted Answer · 2023-02-05T15:29:48+06:00

The two dark fringes that flank the central bright fringe correspond to `n=pm1` in Eq. (i) hence we find that
`sintheta=pmlambda/a=pm(5.80xx10^(-7)m)/(0.300xx10^(-3)m)=pm1.93xx10^(-3)`
From the triangle in figure note that `tan theta=y_(1)//D` Because `theta` is very small, we can use the approximation `sinthetaapproxtantheta` thus `sin theta approxy_(1)//D` Therefore, the positions of the first minima measured from th central axis are given by
`y_(1)approxD sin theta=pmDlambda/a=pm3.87xx10^(-3)m`
The positive and negative signs correspond to the dark fringes on either side of the central bright fringe. Hence, the width of the central bright fringe is equal to
`2|y_(1)|=7.74xx10^(-3)m=7.74mm` Note that this value is much greater than the width of the slit. However, as the slit corresponding to smaller values of `theta` in fact for large values of a, the various maxima and minima are so closely spaced that only a large central bright area resembling the geometric image of the slit is observed.