In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed I m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water ? (μ water=4/3)

Question

In a double slit experiment, when light of wavelength 400 nm was used, the angular width of the first minima formed on a screen placed I m away, was found to be 0.2°. What will be the angular width of the first minima, if the entire experimental apparatus is immersed in water ? (μ water=4/3) (A) 0.05° (B) 0.1° (C) 0.266° (D) 0.15°

Tardigrade Admin · Accepted Answer

In air angular fringe width θ0 = (β/D) Angular fringe width in water θw = (β/μ D) = (θ0/μ) = (0.2°/((4)3)) = 0.15°

0.05°

$0. 1^{\circ}$

$0.26 6^{\circ}$

$0.1 5^{\circ}$

In air angular fringe width $θ_{0} = \frac{β}{D}$
Angular fringe width in water
$θ_{w} = \frac{β}{μ D} = \frac{θ _{0}}{μ}$
$= \frac{0. 2 ^{\circ}}{( \frac{4}{3} )} = 0.1 5^{\circ}$

0.05°

0.1∘

0.266∘

0.15∘

In air angular fringe width θ0​=Dβ​ Angular fringe width in water θw​=μDβ​=μθ0​​ =(34​)0.2∘​=0.15∘

$0. 1^{\circ}$

$0.26 6^{\circ}$

$0.1 5^{\circ}$

In air angular fringe width $θ_{0} = \frac{β}{D}$
Angular fringe width in water
$θ_{w} = \frac{β}{μ D} = \frac{θ _{0}}{μ}$
$= \frac{0. 2 ^{\circ}}{( \frac{4}{3} )} = 0.1 5^{\circ}$