In YDSE, a thin film (μ=1.6) of thickness 0.01 mm is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the 10 text th bright fringe earlier. The wave length of wave is

Question

In YDSE, a thin film (μ=1.6) of thickness 0.01 mm is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the 10 text th bright fringe earlier. The wave length of wave is (A) 600 mathringA (B) 6000 mathringA (C) 60 mathringA (D) 660 mathringA

Tardigrade Admin · Accepted Answer

Shift of fringes is given by ((μ-1) t D/d) This is equal to position of 10 text th bright fringe (10 λ D/d) (0.6 × 1 × 10-5 D/d)=(10 λ D/d) λ=0.6 × 10-6 ∴ λ=6 × 10-7 4 or λ=6000 mathringA

Q. In YDSE, a thin film $(μ = 1.6)$ of thickness $0.01 mm$ is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the $1 0^{th}$ bright fringe earlier. The wave length of wave is

$600 \overset{˚}{A}$

$6000 \overset{˚}{A}$

$60 \overset{˚}{A}$

$660 \overset{˚}{A}$

Shift of fringes is given by $\frac{( μ - 1 ) t D}{d}$
This is equal to position of $1 0^{th}$ bright fringe $\frac{10 λ D}{d}$
$\frac{0.6 \times 1 \times 1 0 ^{- 5} D}{d} = \frac{10 λ D}{d}$
$λ = 0.6 \times 1 0^{- 6}$
$∴ λ = 6 \times 1 0^{- 7} 4$
or $λ = 6000 \overset{˚}{A}$

Q. In YDSE, a thin film (μ=1.6) of thickness 0.01mm is introduced in the path of one of the two interfering beams. The central fringe moves to a position occupied by the 10th bright fringe earlier. The wave length of wave is

600A˚

6000A˚

60A˚

660A˚