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In a YDSE, a total of 241 fringe can be seen on the screen. The set up is completely immersed in water. What will be the number of fringes now seen on the screen? |
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Answer» Solution :The total number of fringes is given by `(2D)/(lambda)+1` assuming `(2d)/(lambda)` to be an integer. `THEREFORE (2d)/(lambda)+1= 241 implies (2d)/(lambda)= 240 implies (2d)/(lambda)= 240` Now, let `lambda^(1)` be the changed WAVELENGTH when the YDSE set up is submerged in water. Then the total number of fringes will be `(2d)/(lambda)+1= (2 d mu)/(lambda)+1= 240(4/3)+1` `(therefore mu" for water "= 4"/"3)= 321`. |
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