Question

In a Young 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 cm. Determine the wavelength of light used in the experiment.

Answer 1

Given,
distance between screen and slits , D = 1.4m
distance between slits , d = 0.28mm = 0.28 × 10^-3 m

A/C to question,
the distance between central fringe and the fourth fringe is measured to be 1.2cm.
so, fringe width = 1.2/4 = 0.3cm = 3 × 10^-3m

we know,
$\omega=\frac{\lambda D}{d}$

here $\Omega$ is fringe width.

so, $\lambda=\frac{\omega d}{D}$

so, $\lambda$ = 3 × 10^-3 × 0.28 × 10^-3/1.4
= 0.06 × 10^-5 m
= 6 × 10^-7 m
= 600 × 10^-9 m
= 600nm

Answer 2

Given :

Separation between slits =d=0.28mm=0.28 x 10⁻³ m

Distance between screen and slit D=1.4m

Distance between central bright and fourth bright fringes:

x=1.2cm =1.2 x 10⁻²m

number of fringes =n=4

For constructive interference x=n Dλ/d

1.2 x 10⁻²= 4x1.4 xλ /0.28 x10⁻³

∴Wavelength =λ=1.2x10⁻²x0.28 x10⁻³/4x1.4

λ=6x10⁻⁷m

λ=600x 10⁻⁹

λ=600nm [since 1nm=10⁻⁹m]

Thus the wavelength of light is 6x10⁻⁷m