Question

A single slit of width 0.14 mm is illuminated normally by monochromatic light and diffraction bands
are observed on a screen 2 m away. If the centre of the second dark band is 1.6 cm from the middle of
the central bright band, deduce the wavelength of light.
Solution​

Answer 1

Answer:

here is the answer

Explanation:

Explanation: 16×10^-3×0.14×10^-3

2× 2

= 0.56 × 10^-6 m

= 5600 A°

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Answer 2

The wavelength of the light is 5600 A°.

Given,

Width of slit = 0.14 mm

Screen distance = 2 m

Distance between centre of dark and bright band = 1.6 cm

To Find,

Wavelength of light

Solution,

We know that the condition of minima in the case of single slit diffraction is -

a*Sin θ = nλ

where a = Width of slit = 0.14 mm = 0.14 x 10⁻³ m,

n = Number of fringes = 2 (in this case)

and λ = Wavelength of light

We also know that Sin θ can be written as -

Sin θ = y/D

where y = Distance between centre of dark and bright band = 1.6 cm = 1.6 x 10⁻² m

and D = Screen distance = 2 m

So we can write the new expression as -

(a*y)/D = nλ

a*y = nλ*D

(0.14 x 10⁻³) x (1.6 x 10⁻²) = 2 x λ x 2

0.224 x 10⁻⁵ = 4λ

λ = $\frac{0.224}{4}$ x 10⁻⁵

λ = 0.056 x 10⁻⁵

λ = 5600 x 10⁻¹⁰

λ = 5600 A°

Hence, the wavelength of the light is 5600 A°.

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