In a biprism experiment, the slit is illuminated by the wavelength of 6000 A°. The distance between the slit and the eyepiece is 0.6 m. The two virtual images of the slit are formed 0.3 mm apart. If eyepiece is displaced 20 cm away from the slit, the change in fringe width will be

Question

Question

Answer 1

Concept:

F2 J.K 19.5.20 Pallavi D8

If Fresnel’s biprism experiment, the fringe width is given by

\(\beta= \frac{{λ D}}{d}\)

Calculation:

Given λ = 6000 A° = 6 × 10^-4 mm; D = 0.6 m = 600 mm; d = 0.3 mm;

Now the fringe width will be

\(\beta= \frac{{6 \times 10^{-4} \times 600}}{0.3} = 1.2 \; mm\)

Now eyepiece is displaced 20 cm away from the slit,

D = 600 + 200 = 800 mm;

Fringe width in this case will be

\(\beta= \frac{{6 \times 10^{-4} \times 800}}{0.3} = 1.6 \; mm\)

The change in fringe width is 1.6 - 1.2 = 0.4 mm