Question

In a biprism experiment, interference bands are observed at a distance of one metre from the slit. A convex lens put between the slit and the eye piece gibes two images of slit 0.7 cm apart, the lens being 30 cm from the slit. What is the fringe width if a light of wavelength 6000°A is used?
1. 0.4 mm
2. 0.2 mm
3. 0.6 mm
4. 0.9 mm

Answer 1

Correct Answer - Option 2 : 0.2 mm

Concept:

In a biprism experiment, the fringe width is given by

β = λD/d;

When convex lenses are put between the slits and the eyepiece and nearer to the slits, the magnified images will be formed on the eyepiece.

When convex lenses are put nearer to the eyepiece, the diminished images will be formed on the eyepiece.

From the property of conjugate foci of the convex lens,

\(\frac{{Size\;of\;object}}{{Size\;of\;image}} = \frac{{Object\;Distance}}{{Image\;Distance}}\)

\( \Rightarrow \frac{d}{{{d_1}}} = \frac{x}{y}\)

Where d – the actual distance between S₁ and S₂; d₁ – Size of the magnified image;

x – the distance between lens and slits, y – the distance between the lens and eyepiece;

Calculation:

Given d₁ = 0.7 cm, x = 30 cm, λ = 6000°A, D = 1 m;

As x + y = D ⇒ y = 70 cm;

From the property of conjugate foci of the convex lens,

\(\Rightarrow \frac{d}{{0.7}} = \frac{{30}}{{70}}\)

⇒ d = 0.3 cm;

Now the fringe width will be 

\(\beta = \frac{{\lambda D}}{d} = \frac{{6 \times {{10}^{ - 7}} \times 100}}{{0.3}} = 0.2\;mm\)