Question

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Answer 1

QUESTION :-

In the Young Double's Slit Experiment the slits are 0.5 mm apart and interference is observed on a screen placed at a distance of 100 cm from the slits . it is found that the 9th bright fringe is at a distance of 7.5 mm from the second dark fringe from the centre of the fringe pattern. what is the wavelength of light used ?

ANSWER :

Given :

Distance between the slits (d) = 0.5 mm .

Distance of the interference (D) = 100 cm .

d = 0.5 mm ⇒ 0.5 × 10⁻¹ cm ⇒ 5 × 10⁻⁴ m ⇒ 5/10000 m

⇒ 0.0005 m

D = 100 cm ⇒ 100 × 10⁻² m ⇒ 1 m .

We know that the distance of the 9 th ring = 9 λ D / d .

Distance of the 2 nd ring = ( 2×2 - 1 ) λ D / 2 d

⇒ ( 4 - 1 ) λ D / 2 d

⇒ 3 λ D / 2 d

Distance between 9 th and 2 nd ring :

⇒ 9 λ D / d - 3 λ D / 2 d

⇒ ( 18 λ D - 3 λ D ) / 2 d

⇒ 15 λ D / 2 d

⇒ 7.5 λ D / d

Given :

7.5 λ D / d = 7.5 mm

⇒ λ D / d = 1 mm

⇒ λ D / d = ( 1/1000 ) m

⇒ λ D / d = 0.001 m

⇒ λ = 0.001 m × d / D

⇒ λ = 0.001 × 0.0005 / 1  

⇒ λ = 10⁻³ × 5 × 10⁻⁴ m

⇒ λ = 10⁻⁷ × 5 m

⇒ λ = 5 × 10⁻⁷ × 10¹⁰ A°

⇒ λ = 5 × 10³ A°

⇒ λ = 5000 A°

Correct answer is OPTION C .

NOTE :

1 A° = 10¹⁰ m .

So first convert everything to metres and then later on we can reconvert the units to angstrom .