Fresnel's Biprism

Fresnel's Biprism

Fresnel's Biprism is an elegant optical experiment developed by Augustin-Jean Fresnel to demonstrate the wave nature of light. This experiment uses a specially designed biprism, which effectively involves two prisms joined at their bases to split a single light source into two coherent beams. These beams overlap and interfere, creating a pattern of bright and dark fringes on a screen. Fresnel's Biprism experiment not only confirms the wave theory of light but also allows precise measurement of light's wavelength. It is an optical device for producing interference of light Fresnel's biprism is made by joining base to base two thin prism of very small angle. When a monochromatic light source is kept in front of biprism two coherent virtual sources ${\mathrm{s}}_1$ and ${\mathrm{s}}_2$ are produced. Interference fringes are found on the screen placed behind the biprism interference fringes are formed in the limited region which can be observed with the help of an eyepiece.

Fringes are of equal width and their value is $\beta =\frac{\lambda D}{d}$

Let the separation between ${\mathrm{s}}_1$ and ${\mathrm{s}}_2$ be d and the distance of slits and the screen from the biprism be a and b respectively i.e. $D=(a+b)$. If the angle of the prism is $A$ and the refractive index is $\mu$ then $d=2a(\mu -1)A$

$\therefore \lambda =\frac{\beta [2a(\mu -1)A]}{(a+b)}\Rightarrow \beta =\frac{(a+b)\lambda }{2a(\mu -1)A}$

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Solved Examples Based on Fresnel's Biprism

Example 1: In a Fresnel's biprism experiment, the distance between the source and eyepiece is given 100 m and two slits are kept at a separation of 1.5 m. The observed fringe width is 0.5 mm. What will be the wavelength of light?

1) $1.5\times {10}^{-5}\mathrm{m}$
2) $1.5\times {10}^{-6}\mathrm{m}$
3) $7.5\times {10}^{-6}\mathrm{m}$
4) $7.5\times {10}^{-5}\mathrm{m}$

Solution:

Fresnel's Biprism

Fringes width has value as $\beta =\frac{\lambda D}{d}$

Given,

D= 100 m

d= 1.5 m

$\beta$= 0.5 mm

Using the formula of fringe width: $\beta =\frac{\lambda D}{d}$

Therefore, $\lambda =\frac{\beta d}{D}$

Putting the given values into the formula

$\begin{array}{rl}& \lambda =\frac{0.5\times {10}^{-3}\times 1.5}{100}\\ & \lambda =7.5\times {10}^{-6}\mathrm{m}.\end{array}$

Hence, the answer is the option (4).

Example 2: In a Fresnel's Biprism experiment Interference Fringes are observed with a biprism of refracting angle of 2° and refractive index of 1.5 on a screen which is 100m away from the source. If the distance between the source and the biprism is 20m and the fringe width is 0.10 mm what is the wavelength of light?

1)69.7 nm

2)6.97 mm

3)697 nm

4)None of these

Solution:

First of all, find deviation, $\delta =(\mu -1)A$

we know,
where a is the distance between source and biprism. e.g. a = 20m
so, $d=2\times \frac{\pi }{180}\times 20=\frac{2\pi }{9}m$

now, use the formula of fringe width,

here,
$\mathrm{D}=100\mathrm{m},\mathrm{d}=\frac{2\pi }{9}m$

$\begin{array}{rl}& \text{so,}0.1\times {10}^{-3}=\lambda \times \left(\frac{100}{\frac{2\pi }{9}}\right)d\\ & \Rightarrow >{10}^{-4}\times \frac{2\pi }{9}\times {10}^{-2}=\lambda \\ & \Rightarrow \lambda =697\mathrm{nm}\end{array}$

Hence, the answer is the option (3).

Example 3: If the prism angle $\alpha =1^{\circ },\mu =1.54$, the distance between screen and prism (b) $=0.7\mathrm{m}$, and the distance between prism and source, $\mathrm{a}=0.3\mathrm{m},\lambda =180\pi \mathrm{nm}$ then in Fresnel biprism find the value of $\beta$ (fringe width) :

1) ${10}^{-4}\mathrm{m}$
2) ${10}^{-3}\mathrm{mm}$
3) ${10}^{-4}\times \pi \mathrm{m}$
4) ${10}^{-4}\times \pi \mathrm{m}$

Solution:

Fringe width $\beta$ is given by

$\begin{array}{rl}& \Rightarrow \beta =\frac{(a+b)\lambda }{2a(\mu -1)\alpha }\\ & =\frac{(0.3+0.7)\times 180\pi \times {10}^{-9}}{2\times 0.3(1.54-1)\times (1\times \frac{\pi }{180})}={10}^{-4}\mathrm{m}\end{array}$

Hence, the answer is the option (1).

Example 4: How large can be the aperture opening to work with laws of ray optics using monochromatic light of wavelength 450 nm to a distance of around 20 m?

1)6 mm

2)3 mm

3)2 mm

4)8 mm

Solution:

Now, here we are given Fresnel distance equal to $20\mathrm{m}\lambda =450\mathrm{nm}$, we have to estimate the value of a.
Putting ${\mathrm{Z}}_{\mathrm{F}}=\frac{{\mathrm{a}}^2}{\lambda },\mathrm{a}=\sqrt{{\mathrm{Z}}_{\mathrm{F}}\lambda }=3\times {10}^{-3}\mathrm{m}$ or 3 mm

Hence, the answer is the option (2).

Example 5: How large can be aperture opening to work with laws of ray optics using a monochromatic light of wavelength 800 nm to a distance of around 20 m?

1) 6 mm

2) 4 mm

3) 2 mm

4) 8 mm

Solution:

Now, here we are given Fresnel distance equal to $20\mathrm{m}\lambda =800\mathrm{nm}$, we have to estimate the value of a.
Putting ${\mathrm{Z}}_{\mathrm{F}}=\frac{{\mathrm{a}}^2}{\lambda },\mathrm{a}=\sqrt{{\mathrm{Z}}_{\mathrm{F}}\lambda }=4\times {10}^{-3}\mathrm{m}$ or 4 mm

Hence, the answer is the option (2).

Summary

Fresnel's Biprism experiment effectively demonstrates the wave nature of light by producing interference patterns from a single light source split into two coherent beams. The experiment involves calculating fringe width and understanding the relationships between the wavelength of light, the distances involved, and the properties of the biprism. It has practical applications in fields such as optics and photonics, influencing technologies like holography and advanced microscopy. The solved examples illustrate the use of formulas to determine fringe width and wavelength, showcasing the precision and importance of this optical phenomenon.