Question

In YDSE experiment, the fringes are displaced by a distance ‘x’ when a glass plate of refractive index 1.5 is introduced in the path of one of the beams. When this plate is replaced by another plate of same thickness the shift of fringes is ‘1.5x’ The refractive index of second plate is
1. 1.75
2. 1.40
3. 1.25
4. 1.67

Answer 1

Correct Answer - Option 1 : 1.75

The shift in fringe width due to plate of thickness ‘t’ and refractive index ‘u’ is

\(shift = \frac{{\left( {\mu - 1} \right)t{\rm{D}}}}{d}\)

For the first case

\(x = \frac{{\left( {1.5 - 1} \right)t{\rm{D}}}}{d}\)     ….(1)

For the second case

\(\frac{{3x}}{2} = \frac{{\left( {u - 1} \right)t{\rm{D}}}}{d}\)     ….(2)

Dividing (1) by (2)

\(\frac{2}{3} = \frac{{0.5}}{{\left( {u - 1} \right)}}\)

2u – 2 = 1.5

μ = 1.75