Question

A double-slit interference pattern is created by two narrow slits spaced 0.25 mm apart. Thedistance between the first and the fifth minimum on a screen 60 cm behind the slits is 5.5mm. What is the wavelength (in nm) of the light used in this experiment

Answer 1

Step-by-step explanation:

Slit distance d = .25 x 10⁻³ m .

Distance of screen D = 60 x 10⁻² m

Distance between first and fifth minimum = 5.5 x 10⁻³ m .

There will be four bright fringes between first and 5 th minimum so

distance between them = 4 x λ D / d where λ is the wavelength of light .

Substituting the values

4 x λ x 60 x 10⁻² m / .25 x 10⁻³ m = 5.5 x 10⁻³ m

4 x λ x 60 x 10⁻² m = .25 x 10⁻³ m x 5.5 x 10⁻³ m = 1.375 x 10⁻⁶ m

λ = .005729 x 10⁻⁴

= 572.9 x 10⁻⁹ m

= 572.9 nm .