If in a Young's double slit experiment, the slit width is 3 cm, the separation between slits and screen is 7 cm and wavelength of light is 1000 mathringA, then fringe width will be (α=1°, μ=1.5) :

Question

If in a Young's double slit experiment, the slit width is 3 cm, the separation between slits and screen is 7 cm and wavelength of light is 1000 mathringA, then fringe width will be (α=1°, μ=1.5) : (A) 2 × 10-5 m (B) 2 × 10-3 m (C) 0.2 × 10-5 m (D) none of these

Tardigrade Admin · Accepted Answer

Fringe width β=(λ D/d) ...(1) Given: λ=1000 mathringA=10-5 cm , d=3 cm, D=7 cm Putting given value in eq (i) ∴ β =(10-5 × 7/3) =2.3 × 10-5 cm =2.3 × 10-7 m

Q. If in a Young's double slit experiment, the slit width is $3 c m$ , the separation between slits and screen is $7 c m$ and wavelength of light is $1000 \overset{˚}{A}$ , then fringe width will be $(α = 1^{\circ}, μ = 1.5)$ :

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

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

$0.2 \times 1 0^{- 5} m$

none of these

Fringe width
$β = \frac{λ D}{d} ... (1)$
Given : $λ = 1000 \overset{˚}{A} = 1 0^{- 5} c m,$
$d = 3 c m$ ,
$D = 7 c m$
Putting given value in eq (i)
$∴ β = \frac{1 0 ^{- 5} \times 7}{3}$
$= 2.3 \times 1 0^{- 5} c m$
$= 2.3 \times 1 0^{- 7} m$

Q. If in a Young's double slit experiment, the slit width is 3cm, the separation between slits and screen is 7cm and wavelength of light is 1000A˚, then fringe width will be (α=1∘,μ=1.5) :

2×10−5m

2×10−3m

0.2×10−5m