In young's double slit experiment, the width of the fringes obtained with light of wavelength 6000 mathringA is 2 mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33 ?

Question

In young's double slit experiment, the width of the fringes obtained with light of wavelength 6000 mathringA is 2 mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33 ? (A) 5 mm (B) 3 mm (C) 1.5 mm (D) 1 mm

Tardigrade Admin · Accepted Answer

β=(λ D / d ), β prime=(λ prime D / d ) ∴ (β prime/β)=(λ prime D/d) × (d/λ D)=(λ prime/λ)=(1/μ) β prime=(β/μ)=(2/1.33)=1.5 mm

Q. In young's double slit experiment, the width of the fringes obtained with light of wavelength $6000 \overset{˚}{A}$ is $2 mm$ . What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index $1.33 ?$

$5 mm$

$3 mm$

$1.5 mm$

$1 mm$

$β = \frac{λ D}{d}, β^{'} = \frac{λ ^{'} D}{d}$
$∴ \frac{β ^{'}}{β} = \frac{λ ^{'} D}{d} \times \frac{d}{λ D} = \frac{λ ^{'}}{λ} = \frac{1}{μ}$
$β^{'} = \frac{β}{μ} = \frac{2}{1.33} = 1.5 mm$

Q. In young's double slit experiment, the width of the fringes obtained with light of wavelength 6000A˚ is 2mm. What will be the fringe width, if the entire apparatus is immersed in a liquid of refractive index 1.33?

5mm

3mm

1.5mm

1mm