in young doble slit expt the bandwitdh of the fringe obtained with red light in 2mm. the apprutus is immersed in water then the resutant bandwidth

The resultant fringewidth orbandwidth will decrease by putting the entire setup in water. This is bcoz u know that lambda(wavelength) decreases in water than air keeping frequency constant as speed changes. It becomes (lamda)/u where u is refractive index of water.Thats why fringe width given by (lamda)D /d gets changed to (lambda)D/ud. where `u` is refractive index of water. so your answer will be 2/(4/3) which is 3/2. Hope u got it.