Question

In Fresnel’s bi-prism experiment, a mica sheet of refractive index$\;1.5$ and thickness $6 \times 10^{-6}\;m$ is placed in the path of one of interfering beams as a result of which the central fringe gets shifted through five fringe widths. The wavelength of light used is:
A). $6000\;A^{\circ}$
B). $8000\;A^{\circ}$
C). $4000\;A^{\circ}$
D). $2000\;A^{\circ}$

Answer 1

Hint: We know that introducing a mica sheet produces a path difference in the beam. In such a case, arrive at an expression for the path difference in terms of the material introduced and another expression for the same in terms of the geometric configuration of the beam in the experimental setup. Equate the two expressions and evaluate them to arrive at an expression for the wavelength of the light used, following which you can plug in the given values to arrive at the appropriate result.
Formula Used:
Fringe width: $\beta = \dfrac{\lambda D}{d}$
Path difference: $\Delta x = \dfrac{xd}{D}$ or $\Delta x = (\mu-1)t$

Complete step-by-step solution:
We know that a Fresnel’s biprism is essentially an optical device that is used to observe interference of light and can be used to determine the wavelength of light incident on it. The biprism is made by joining the bases of two thin prisms together. When a monochromatic light source$\;S$ of wavelength $\lambda$ is incident on the biprism, two virtual coherent light sources $S_1$ and $S_2$ are produced, separated by a distance$\;d$. The light coming from these sources get refracted by the biprism and form an interference pattern at a distance$\;D$ from the virtual sources, with alternating bright and dark fringes.

The fringes formed are of equal width, and the fringe width $\beta$ can be given as:
$\beta = \dfrac{\lambda D}{d}$
We are given that a mica sheet of refractive index $\mu = 1.5$ and thickness $t = 6 \times 10^{-6}\;m$ is introduced into the path of one of the interfering beams such that it creates a difference in the beam path. The path difference $\Delta x$ in the beam as a result on introducing the mica sheet can be given as:
$\Delta x = (\mu-1)t$
But we know that if$\;x$ is the original path or position of the beam, the path difference can also be given as
$\Delta x = \dfrac{xd}{D}$
Equating the above two expressions:
$\dfrac{xd}{D} = (\mu-1)t $
$\Rightarrow x = (\mu-1)t \times \dfrac{D}{d}$
Multiplying and dividing the RHS of the above equation with $\lambda$, we get:
$x = \dfrac{(\mu-1)t}{\lambda} \times \dfrac{\lambda D}{d}$
But we know that $\dfrac{\lambda D}{d} = \beta$. So the above equation becomes:
$x = \dfrac{(\mu-1)t\beta}{\lambda}$
Now, given that the central fringe gets shifted by 5 fringe widths, this means that the position of the beam also gets shifted through this width, i.e.,
$x = 5\beta$
Plugging this in to our previous equation we get:
 $5\beta = \dfrac{(\mu-1)t\beta}{\lambda}$
$\Rightarrow \lambda = \dfrac{(\mu-1)t\beta}{5\beta} = \dfrac{(\mu-1)t}{5}$
Substituting the given values into the above equation, we can get the wavelength of the light used as:
$\lambda = \dfrac{(1.5 - 1) \times 6 \times 10^{-6}}{5} = \dfrac{0.5 \times 6 \times 10^{-6}}{5}$
$\Rightarrow \lambda = 0.6 \times 10^{-6} = 6000 \times 10^{-10}\;m = 6000\;A^{\circ}$
Therefore, the correct choice is A. $600\;A^{\circ}$

Note: Let us understand the role of the biprism in the above experiment. The Fresnel biprism is used to essentially divide the wavefront of a monochromatic beam of light in such a way that it produces images that serve as two coherent virtual sources of the same light to consequently obtain a well resolved interference pattern. This is why experiments involving finding the wavelength of a source of light preferably involve Fresnel’s biprism instead of the Young's double slit apparatus. The biprism eliminates the difficulty associated with setting up the extended secondary slits to produce coherent beams of light. However, the biprism experiment is entirely based on the principles of Young’s experiment and follows calculations consistent with the latter.