A monochromatic light of wavelength 500 nm is incident normally on a single slit of width 0.2 mm to produce a diffraction pattern. Find the angular width of the central maximum obtained on the screen.

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Hassan Watkins · Accepted Answer

The wavelength of incident light is given as   λ  =  500  ×      10          −      9           mThe width of a single slit is given as   a  =  0.2  ×      10          −      3           mThe angular width of central maxima is,  w  =            2      λ        a      =            2      ×      500      ×              10                  −          9                            0.2      ×              10                  −          3                        =  5  ×      10          −      3        mThus, the angular width of the central maxima obtained on the screen is 5 mm.Given the width of fringe in the young double-slit experiment is   d  =  0.5  ×      10          −      3        mLet there are n fringes made by a double-slit in the central maxima of the diffraction,Then,  n  =            B      ′        B  where       B    ′    =            2      λ      D        a   is the width of central maxima in the diffraction and   B  =            λ      D        d   is the fringe width by the double slit.Substituting the known values,  n  =            (                        2          λ          D                a            )              (                        λ          D                d            )          =            2      λ      D        a    ×      d          λ      D          =  2      d    a      =            2      ×      0.5      ×              10                  −          3                            0.2      ×              10                  −          3                        =  5Thus, the number of fringes obtained is 5.

Explain two features to distinguish between the interference pattern in Young double slit experiment

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