Discussion
In an Young's double slit experiment, the band width of the fringes observed is \(\beta\) , when light of wave length \(\lambda\) is used. With same experimental set up, to double the band width of the fringes, the wave length of light required is
Answer: C
The wavelength required is \(2\lambda\).
Fringe width is \(\beta = \frac{\lambda D}{d}\). With the same slit separation \(d\) and screen distance \(D\), \(\beta\) is directly proportional to \(\lambda\), so doubling \(\beta\) needs double the wavelength.
\(\frac{\lambda}{2}\) would halve the fringe width, and the smaller fractions would shrink it further.
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