In a Young’s double slit experiment set up, the two slits are kept 0.4mm apart and screen is placed at 1m from slits. If a thin transparent sheet of thickness is introduced in front of one of the slits then the bright fringe sh

Question

In a Young’s double slit experiment set up, the two slits are kept 0.4 mm apart and screen is placed at 1 m from slits. If a thin transparent sheet of thickness $20 μ m$ is introduced in front of one of the slits then centre bright fringe shifts by 20 mm on the screen. The refractive index of transparent sheet is given by $α / 10$ where $α$ is [2026]

Smart Achievers · Accepted Answer

(14)

$y_{shift} = \frac{(μ - 1) t D}{d}$

$209 \times 10^{- 3} = \frac{(μ - 1) \times 20 \times 10^{- 6} \times 1}{0.4 \times 10^{- 3}}$

$(μ - 1) = 0.4$

$μ = 1.4$

$\frac{α}{10} = 1.4, α = 14$

Solution

Ans.
(14)

$y_{shift} = \frac{(μ - 1) t D}{d}$

$209 \times 10^{- 3} = \frac{(μ - 1) \times 20 \times 10^{- 6} \times 1}{0.4 \times 10^{- 3}}$

$(μ - 1) = 0.4$

$μ = 1.4$

$\frac{α}{10} = 1.4, α = 14$

Want Us to Email you About Special Offers & Updates?

Solution

Ans. (14) yshift=(μ-1)tDd 209×10-3=(μ-1)×20×10-6×10.4×10-3 (μ-1)=0.4 μ=1.4 α10=1.4, α=14

Want Us to Email you About Special Offers & Updates?

Ans.
(14)

$y_{shift} = \frac{(μ - 1) t D}{d}$

$209 \times 10^{- 3} = \frac{(μ - 1) \times 20 \times 10^{- 6} \times 1}{0.4 \times 10^{- 3}}$

$(μ - 1) = 0.4$

$μ = 1.4$

$\frac{α}{10} = 1.4, α = 14$