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When diffraction is performed in water then why will the fringes will be smaller in number than when it is performed in air?
Maxima for a double slit occur when n lambda = d sin (theta)so the angles between maxima (and hence the spacing) depends on sin(theta)Reaaranging: sin (theta) = n lambda/dWhen the apparatus is put in water, lambda changes. The velocity in water is less so mthe distance travelled durin on e period is less ie the wavelength (lambda) decreases. So sin(theta) decreases and hence the spacing between the maxima.
Maxima for a double slit occur when n lambda = d sin (theta)
so the angles between maxima (and hence the spacing) depends on sin(theta)
Reaaranging: sin (theta) = n lambda/d
When the apparatus is put in water, lambda changes. The velocity in water is less so mthe distance travelled durin on e period is less ie the wavelength (lambda) decreases. So sin(theta) decreases and hence the spacing between the maxima.
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