A source of light of wavelength 5000 mathringA is placed at one end of a table 2 m long and 5 mm above its flat well polished top. The fringe width of the interference bands seen on a screen located at the end of the table is

Question

A source of light of wavelength 5000 mathringA is placed at one end of a table 2 m long and 5 mm above its flat well polished top. The fringe width of the interference bands seen on a screen located at the end of the table is (A) 2 × 10-5 : m  (B) 2 × 10-4 : m  (C) 2 × 10-3 : m  (D) 2 × 10-2 : m

Tardigrade Admin · Accepted Answer

Here , λ = 5000 mathring A = 5000 × 10-10 m = 5 × 10-7 m D - 2 m, d=5 mm = 5 × 10-3 m Fringe width, β = (λ D/d) = ((5 × 10-7m)(2m)/(5× 10-3 m)) = 2× 10-4 m

Q. A source of light of wavelength $5000 \overset{˚}{A}$ is placed at one end of a table $2 m$ long and $5 mm$ above its flat well polished top. The fringe width of the interference bands seen on a screen located at the end of the table is

$2 \times 1 0^{- 5} m$

$2 \times 1 0^{- 4} m$

$2 \times 1 0^{- 3} m$

$2 \times 1 0^{- 2} m$

Here ,
$λ = 5000 \overset{˚}{A} = 5000 \times 1 0^{- 10} m = 5 \times 1 0^{- 7} m$
$D - 2 m, d = 5 mm = 5 \times 1 0^{- 3} m$
Fringe width,
$β = \frac{λ D}{d} = \frac{( 5 \times 1 0 ^{- 7} m ) ( 2 m )}{( 5 \times 1 0 ^{- 3} m )} = 2 \times 1 0^{- 4} m$

Q. A source of light of wavelength 5000A˚ is placed at one end of a table 2m long and 5mm above its flat well polished top. The fringe width of the interference bands seen on a screen located at the end of the table is

2×10−5m

2×10−4m

2×10−3m

2×10−2m