Question

In a biprism experiment, the fringe width obtained on the screen is $6mm$ from the slits which are $1.5m$away from each other. Keeping the setting unchanged if the eye-piece is moved $20cm$ towards the biprism, find the change in fringe width.
A. $0.90mm$
B. $0.86mm$
C. $0.80mm$
D. $0.53mm$

Answer 1

Use the formula of fringe width to calculate the ratio of wavelength to $d$ once before moving the eye-piece and once after moving the eye-piece. Equate these two values and calculate the new fringe width.

Formula used
$\beta = \dfrac{{\lambda D}}{d}$
Where $\beta$ is the width of the bright fringe, $\lambda$ is the wavelength of light used, $D$is the distance between the slit and the screen and $d$ is the distance between the slits.

Complete step by step answer
A biprism demonstrates the interference phenomenon. It is essentially two prisms each of a very small refracting angle placed base to base. It is necessary to use narrow sources for a biprism experiment because a broad source of light is equivalent to a large number of narrow sources placed side by side. If the slit is broad, the two coherent sources will also be broad. Now each pair of conjugate points on the virtual sources will give rise to an interference pattern. These patterns are slightly displaced from one another which overlap and result in general illumination.
The width of the fringe produced in the biprism interference pattern is given by,
$\beta = \dfrac{{\lambda D}}{d}$
Where $\beta$ is the width of the bright fringe, $\lambda$is the wavelength of light used, $D$is the distance between the slit and the screen and $d$is the distance between the slits.
We are given $D = 1.5m$and $\beta = 6mm$
So we can write,
$\begin{gathered} \dfrac{\lambda }{d} = \dfrac{\beta }{D} \\\ \Rightarrow \dfrac{\lambda }{d} = \dfrac{6}{{1.5}} \\\ \end{gathered}$
Now, if the eye-piece is moved by $20cm = 0.2m$
$D$changes to $\left( {1.5 - 0.2} \right)m = 1.3m$
$\dfrac{\lambda }{d} = \dfrac{\beta }{{1.3}}$
Equating these two values we get,
$\begin{gathered} \dfrac{\beta }{{1.3}} = \dfrac{6}{{1.5}} \\\ \Rightarrow \beta = \dfrac{{6 \times 1.3}}{{1.5}} \\\ \Rightarrow \beta = 5.2mm \\\ \end{gathered}$
So, the change in fringe width is $\left( {6 - 5.2} \right)mm = 0.8mm$

Therefore, the correct option is C.

Note: In biprism, the interference pattern is observed due to the division of wavefront. Since limited portions of the wavefront are used in these devices, diffraction effects are also present along with the interference effects.