Question

A narrow slit is illuminated by a parallel beam of monochromatic light of wavelength λ equals to 6000 Å and the angular width of the central maxima in the resulting diffraction pattern is measured. When the slit is next illuminated by light of wavelength λ’, the angular width decreases by 30%. Calculate the value of the wavelength λ’.

Answer 1

The wavelength of incident radiation 6000 Å

The decrease in the wavelength is of 30%

We know that,

d sin θ = λ (First diffraction)

where, d is the slit width,

θ is the angle of incidence,

λ is the wavelength of the incident light.

In case of the small angle, sinθ ≈ θ,

So, d θ = λ

Then we can say half angular width will be,

\(\theta = \frac \lambda d\)

Then, Full- angular width will be,

\(w = 2\theta = \frac{2\lambda} d\)    .....(1)

For the second wavelength,

\(w' = \frac{2\lambda'}d\)    ......(2)

So, by dividing the two equations wae get,

\(\frac{\lambda '}{\lambda} = \frac{w'}{w}\)

\({\lambda '} = \frac{w'}{w}\times {\lambda}\)

Substituting the values, we get,

λ' = 0.7 × 6000

= 4200 Å

The wavelength of light will be 4200 Å.

Answer 2

Angular width 2ϴ = 2λ/d

Given λ = 6000 Å

In Case of new λ (assumed λ’ here), angular width decreases by 30% 

2 λ’/d = 0.70 X (2 λ/d)

∴ λ’= 4200 Å