In a two-slit experiment, with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10-2 m towards the slits, the change in fringe width is 10-3 m. Then the wavelength of light used is (given that distance between the slits is 0.03 mm )

Question

In a two-slit experiment, with monochromatic light, fringes are obtained on a screen placed at some distance from the slits. If the screen is moved by 5 × 10-2 m towards the slits, the change in fringe width is 10-3 m. Then the wavelength of light used is (given that distance between the slits is 0.03 mm ) (A) 4500 mathringA (B) 5000 mathringA (C) 5500 mathringA (D) 6000 mathringA

Tardigrade Admin · Accepted Answer

Fringe width β=(λ D/d) ∴ Δ β=(λ Δ D / d ) or, λ=(Δ β d /Δ D )=(10-3 × 0.03 × 10-3/5 × 10-2)=(10-3 × 3 × 10-5/5 × 10-2) =6 × 10-7 m =6000 mathringA

$4500 \overset{˚}{A}$

$5000 \overset{˚}{A}$

$5500 \overset{˚}{A}$

$6000 \overset{˚}{A}$

Fringe width $β = \frac{λ D}{d}$
$∴ Δ β = \frac{λ Δ D}{d}$
or, $λ = \frac{Δ β d}{Δ D} = \frac{1 0 ^{- 3} \times 0.03 \times 1 0 ^{- 3}}{5 \times 1 0 ^{- 2}} = \frac{1 0 ^{- 3} \times 3 \times 1 0 ^{- 5}}{5 \times 1 0 ^{- 2}}$ $= 6 \times 1 0^{- 7} m = 6000 \overset{˚}{A}$

4500A˚

5000A˚

5500A˚

6000A˚

Fringe width β=dλD​ ∴Δβ=dλΔD​ or, λ=ΔDΔβd​=5×10−210−3×0.03×10−3​=5×10−210−3×3×10−5​ =6×10−7m=6000A˚

$4500 \overset{˚}{A}$

$5000 \overset{˚}{A}$

$5500 \overset{˚}{A}$

$6000 \overset{˚}{A}$

Fringe width $β = \frac{λ D}{d}$
$∴ Δ β = \frac{λ Δ D}{d}$
or, $λ = \frac{Δ β d}{Δ D} = \frac{1 0 ^{- 3} \times 0.03 \times 1 0 ^{- 3}}{5 \times 1 0 ^{- 2}} = \frac{1 0 ^{- 3} \times 3 \times 1 0 ^{- 5}}{5 \times 1 0 ^{- 2}}$ $= 6 \times 1 0^{- 7} m = 6000 \overset{˚}{A}$