Q: In a young’s experiment, one of the slits is covered with a transparent sheet of thickness 3.6 × 10^(-3) cm due to which position of central fringe to a position originally occupied by 30th fringe. If λ = 6000 A, then find the refractive index of the sheet.

Sol: The position of 30th bright fringe,

$\large y_{30} = \frac{30 \lambda D}{d} $

Now position shift of central fringe is

$\large y_0 = \frac{30 \lambda D}{d} $

But we know, $\large y_0 = \frac{D}{d} (\mu -1)t $

$\large \frac{30 \lambda D}{d} = \frac{D}{d} (\mu -1)t $

$\large (\mu -1) = \frac{30 \lambda}{t} = \frac{30 \times 6000 \times 10^{-10}}{3.6 \times 10^{-5}} $

$\large (\mu -1) = 0.5 $

∴ μ = 1.5.