Fresnel's biprism is an optical device used to produce sustained interference fringes from a single monochromatic source by division of wavefront. It consists of two prisms of very small refracting angles placed base-to-base to form a single isosceles prism.
UP TGT ScienceUP PGT PhysicsStudy Notes + Practice
Scope and prerequisites
TGT core
Understanding the construction of the biprism, virtual source separation, fringe width formula, and measurement of wavelength of light.
PGT extension
Derivation of virtual source distance, angular fringe width, effect of inserting a thin transparent film in the path of one of the interfering beams, and white light fringe patterns.
Prerequisites
Huygens' Principle and Wavefronts
Conditions for Sustained Interference
Refraction through a Prism of Small Angle
Construction and Working Principle
A Fresnel biprism is constructed from a single plate of glass ground and polished so that its two adjacent refracting surfaces inclined at a very small angle (typically around 0.5 degrees) meet to form an obtuse angle of nearly 179 degrees.
When a monochromatic slit source is placed symmetrically in front of the obtuse angle, the upper half of the biprism acts as a prism bending light downwards, and the lower half bends light upwards.
The emergent wavefronts appear to diverge from two coherent virtual sources formed behind the biprism, satisfying the requirement for division of wavefront interference.
Formation of Interference Fringes
The two virtual sources S1 and S2 act as coherent sources because they originate from the same primary source S.
In the region where the two emergent beams overlap on a screen placed at a distance from the biprism, alternative dark and bright interference bands are formed.
The central fringe is always bright because the waves reaching the center from both virtual sources travel equal optical paths, resulting in zero phase difference.
Measurement of Wavelength and Source Separation
The fringe width β is given by $\beta = \lambdaD / d$, where λ is the wavelength of light, D is the distance between the source and the screen, and d is the distance between the two virtual sources.
Direct measurement of the separation d between virtual sources is difficult because S1 and S2 are virtual; hence, a convex lens is used to form real images of the virtual sources on a micrometer eyepiece.
Using the displacement method with a convex lens, d is determined using the relation $d = \sqrt(d_{1}d_{2})$, where d1 and d2 are the image separations in two conjugate positions of the lens.
Effect of Thin Film in the Path of Interference
When a transparent thin film of refractive index μ and thickness t is introduced in the path of one of the interfering beams, the entire fringe system gets shifted towards the side of the film.
The lateral shift of the fringes is given by $x = (\mu - 1)t D / d$.
By measuring the fringe shift, either the thickness of the film or its refractive index can be precisely determined.
Derivations
Derivation of Fringe Width β
Conditions: Monochromatic light of wavelength λ, distance between source and screen = D, separation between virtual sources = d.
Let S be the slit source and S1, S2 be the virtual sources separated by distance d.
Let a be the distance from source S to the biprism, and b be the distance from the biprism to the screen. Then total distance $D = a + b$.
Consider a point P on the screen at a distance x from the central maximum O. The path difference Δ between the two interfering waves arriving at P from S1 and S2 is $\Delta = S_{2}P - S_{1}P$.
Using geometric approximation for S2P2 - S1P2, the path difference evaluates to Δ ≈ xd / D.
For constructive interference (bright fringe), $\Delta = n\lambda$, giving xn = nλD / d.
For destructive interference (dark fringe), $\Delta = (n + 1/2)\lambda$, giving xn = (n + 1/2)λD / d.
The fringe width β is the distance between two consecutive bright or dark fringes: $\beta = x_{n+1} - x_{n}$.
Result: \beta = \frac{\lambda D}{d}
Derivation of Separation d Between Virtual Sources
Conditions: Biprism of refractive index μ and small refracting angle α.
A ray from source S at distance a from the biprism strikes the upper face at angle of incidence nearly equal to zero.
The deviation δ produced by a thin prism of small angle α and refractive index μ is given by $\delta = (\mu - 1)\alpha$.
The two virtual sources S1 and S2 are formed at distance a from the source (or a + b from the biprism depending on convention, here taken at distance a from S).
The angle subtended by S1S2 at the biprism is 2δ.
Therefore, the separation between the virtual sources is $d = 2 a \delta = 2 a (\mu - 1)\alpha$.
Result: $d = 2 a (\mu - 1)\alpha$
Formula sheet
Formula / law
Equation
Variables and units
Conditions
Fringe Width
$$\beta = \frac{\lambda D}{d}$$
$\beta = fringe width$, $\lambda = wavelength$, D = distance from source to screen, d = separation between virtual sources m
Valid for small angles and monochromatic light
Virtual Source Separation
$$d = 2 a (\mu - 1)\alpha$$
a = distance from slit to biprism, $\mu = refractive index of glass$, $\alpha = refracting angle of biprism$ m
Thin prism approximation
Lens Method for Source Separation
$$d = \sqrt{d_1 d_2}$$
d1 = image separation in first conjugate position, d2 = image separation in second conjugate position m
Convex lens placed between biprism and screen
Fringe Shift due to Thin Film
$$x_0 = \frac{(\mu - 1) t D}{d}$$
t = thickness of thin film, $\mu = refractive index of film$ m
Film placed normal to one interfering beam
Wavelength Determination Formula
$$\lambda = \frac{\beta d}{D}$$
$\beta = measured fringe width$ m
Standard experimental setup
Path Difference
$$\Delta = \frac{xd}{D}$$
x = distance from central maximum on screen m
x << D
Angular Fringe Width
$$\theta = \frac{\lambda}{d}$$
$\theta = angular width of a fringe$ rad
Independent of distance D
Refractive Index from Film Shift
$$\mu = 1 + \frac{x_0 d}{t D}$$
$x_0 = shift in fringe pattern$ dimensionless
Measured shift method
Graphs and experiments
Fresnel's Biprism Experimental Arrangement
Draw: An optical bench layout showing sodium lamp, narrow vertical slit S, Fresnel's biprism, convex lens for measuring d, and micrometer eyepiece on the optical bench rail.
Interpret: Illustrates the physical alignment of components required to observe interference fringes by division of wavefront.
Exam use: Tested in viva voce and conceptual matching questions regarding bench alignment.
Intensity Distribution of Interference Fringes
Draw: A graph of Light Intensity (I) on the y-axis versus Position on Screen (x) on the x-axis, showing uniform periodic maxima and minima with equal spacing β.
Interpret: Demonstrates that interference fringes have equal width and equal intensity across the pattern, unlike diffraction patterns.
Exam use: Distinguishing interference patterns from diffraction patterns in multiple-choice questions.
Conjugate Focus Method for d Measurement
Draw: A ray diagram showing two positions of a convex lens forming magnified (d1) and diminished (d2) real images of virtual sources S1 and S2 in an eyepiece.
Interpret: Proves that the source separation can be accurately calculated as the geometric mean of two image sizes using the principle of reversibility.
Exam use: Numerical problem formulation and conceptual understanding of optical magnification.
Solved numericals
TGT core
In a Fresnel's biprism experiment, the distance between the virtual sources is 0.6 mm and the screen is placed 1.0 m away from the sources. If the wavelength of light used is 589 nm, calculate the fringe width.
Identify given values: $d = 0.6 mm = 0.6 \times 10^{-3} m$, D = 1.0 m, $\lambda = 589 nm = 589 \times 10^{-9} m$.
Write the formula for fringe width: $\beta = (\lambdaD) / d$.
Calculate the result: $\beta = 9.816 \times 10^{-4} m = 0.982 mm$.
Answer: 0.982 mm
PGT extension
A Fresnel's biprism of refracting angle 1 degree and refractive index 1.5 is placed at a distance of 20 cm from a slit illuminated by light of wavelength 6000 Å. Find the fringe width on a screen placed 80 cm from the biprism.
Given: $\alpha = 1° = \pi / 180 radians$, $\mu = 1.5$, a = 20 cm = 0.2 m, b = 80 cm = 0.8 m, $\lambda = 6000 Å = 6 \times 10^{-7} m$.
Total distance from source to screen $D = a + b = 0.2 + 0.8 = 1.0 m$.
Compute final value: β ≈ 1.72 × 10-4 m = 0.172 mm.
Answer: 0.172 mm
PGT extension
When a mica sheet of refractive index 1.58 and thickness 6 × 10-6 m is placed in the path of one of the interfering beams in a biprism setup, the fringes shift by a distance equal to 15 fringe widths. Find the wavelength of light used.
The refracting angle of the biprism is assumed to be very small, which validates the thin prism deviation formula $\delta = (\mu - 1)\alpha$.
The source slit must be extremely narrow and parallel to the refracting edge of the biprism to ensure good fringe visibility and contrast.
The distance D from source to screen is much greater than the source separation d (D >> d), validating the approximations used in deriving the path difference.
Common misconceptions
Common error: Fresnel's biprism produces interference by division of amplitude.
Correct: It produces interference by division of wavefront.
A single wavefront from the slit is split into two wavefronts by the two halves of the biprism, acting as two virtual sources.
Common error: The central fringe in a biprism interference pattern is dark.
Correct: The central fringe is always bright.
The waves from virtual sources travel equal optical paths to reach the center of the screen, resulting in zero path difference and constructive interference.
Common error: Fringe width depends on the distance of the biprism from the source (a) and screen (b) individually.
Correct: Fringe width depends only on the total distance $D = a + b between the source$ and the screen.
Mathematically, $\beta = \lambdaD / d$, where D is the total distance from the primary source to the screen.
Quick revision
Fresnel's biprism uses division of wavefront to form two coherent virtual sources.
Fringe width expression is $\beta = \lambdaD / d$.
Virtual source separation is given by $d = 2a(\mu - 1)\alpha$.
Central fringe in biprism interference is always bright.
Introduction of a transparent film shifts the entire fringe pattern towards the film side without altering fringe width.
Must remember
Fresnel's biprism produces interference by the division of wavefront, creating two coherent virtual sources from a single primary slit source.
The central fringe in the interference pattern formed by a Fresnel biprism is always bright due to zero optical path difference.
The fringe width is given by $\beta = \frac{\lambda D}{d}$, where $D$ is the total distance from the source to the screen and $d$ is the separation between virtual sources.
Direct measurement of virtual source separation $d$ is impossible; it is determined using a convex lens in two conjugate positions via $d = \sqrt{d_1 d_2}$.
When a transparent thin film of refractive index $\mu$ and thickness $t$ is introduced in the path of one beam, the fringe pattern shifts by $x_0 = \frac{(\mu - 1)t D}{d} .
Glossary
Fresnel's BiprismA single glass prism with a very obtuse angle (nearly $179^\circ$) used to produce interference by division of wavefront.
Fringe Width ($\beta$)The distance between two consecutive bright or dark interference fringes, expressed as $\beta = \frac{\lambda D}{d}$.
Virtual Source Separation ($d$)The distance between the two coherent virtual sources $S_1$ and $S_2$ formed behind the biprism.
Conjugate Foci MethodAn experimental technique using a convex lens to determine source separation $d$ using $d = \sqrt{d_1 d_2}$.
Fringe Shift ($x_0$)The lateral displacement of the entire interference pattern caused by introducing a transparent thin film in one of the paths.
Division of Wavefront vs. Division of Amplitude
Feature
Division of Wavefront
Division of Amplitude
Mechanism
A single wavefront is split into parts by reflection, refraction, or diffraction.
A light beam is split into partial reflection and transmission by a thin film.
Newton's Rings, Michelson Interferometer, Thin Film Interference
Source Requirement
Requires a narrow point or line source
Can use an extended source under proper conditions
Practice Quiz
Choose an option to reveal the explanation.
1. Which of the following optical phenomena is responsible for the formation of interference fringes in Fresnel's biprism?
Explanation: Fresnel's biprism splits a single incident wavefront from a narrow slit into two parts using refraction through two adjacent small-angled prisms, achieving interference by the division of wavefront.
2. What is the approximate value of the obtuse angle of a Fresnel biprism?
Explanation: A Fresnel biprism is constructed from a glass plate such that its two refracting surfaces are inclined at a very small angle, making the obtuse angle nearly equal to $179^\circ$.
3. What is the nature of the central interference fringe in a Fresnel's biprism experiment?
Explanation: The central fringe is always bright because the waves reaching the center from both virtual sources travel equal optical paths, resulting in zero phase difference and constructive interference.
4. What is the correct formula for the fringe width $\beta$ in a Fresnel biprism setup?
Explanation: The fringe width is given by $\beta = \frac{\lambda D}{d}$, where $\lambda$ is the wavelength, $D$ is the total distance from source to screen, and $d$ is the virtual source separation.
5. Why is a convex lens used in the Fresnel biprism experiment?
Explanation: Because $S_1$ and $S_2$ are virtual sources and cannot be measured directly, a convex lens is used to form real images of these sources on a micrometer eyepiece to calculate $d$.
6. In the lens method for finding virtual source separation, if $d_1$ and $d_2$ are the image separations in two conjugate positions, then $d$ is given by:
Explanation: Using the displacement method with a convex lens, the actual separation between virtual sources is given by the geometric mean, $d = \sqrt{d_1 d_2}$.
7. What happens to the entire interference fringe pattern when a transparent thin film of refractive index $\mu$ and thickness $t$ is introduced in the path of one interfering beam?
Explanation: Introducing a transparent thin film introduces an additional optical path, causing the entire fringe pattern to shift laterally towards the side of the film.
8. What is the expression for the lateral shift $x_0$ of the fringe pattern when a thin film of thickness $t$ and refractive index $\mu$ is introduced?
Explanation: The lateral shift of the fringes is given by $x_0 = \frac{(\mu - 1)t D}{d}$.
9. How does the fringe width $\beta$ change if white light is used instead of monochromatic light in a Fresnel biprism?
Explanation: With white light, the central fringe is white (since path difference is zero for all wavelengths), while surrounding fringes are colored because fringe width depends on wavelength.
10. What is the angular fringe width $\theta$ for a Fresnel biprism setup?
Explanation: The angular fringe width is given by $\theta = \frac{\beta}{D} = \frac{\lambda}{d}$, which is independent of the distance $D$ from source to screen.
11. If the distance between the source and the biprism is $a$ and between the biprism and the screen is $b$, what is the total distance $D$ appearing in the fringe width formula?
Explanation: The total distance $D$ from the primary source to the screen is the sum of the distance from source to biprism ($a$) and biprism to screen ($b$), so $D = a + b$.
12. Which of the following parameters does NOT affect the fringe width $\beta$ directly?
Explanation: The formula $\beta = \frac{\lambda D}{d}$ shows dependence on $\lambda$, $D$, and $d$. While $\mu$ determines $d$ via $d = 2a(\mu - 1)\alpha$, it does not appear as an independent primary variable in the fringe width equation.
13. In a Fresnel biprism experiment, if the wavelength of light is doubled, what happens to the fringe width?
Explanation: Since fringe width $\beta$ is directly proportional to the wavelength $\lambda$ ($\beta \propto \lambda$), doubling the wavelength doubles the fringe width.
14. What happens to the fringe pattern if the slit width of the primary monochromatic source is made excessively large?
Explanation: An excessively wide source acts as multiple independent point sources, producing overlapping fringe systems that wash out and destroy the visible interference pattern.
15. The virtual source separation $d$ in a Fresnel biprism of refracting angle $\alpha$ and refractive index $\mu$, placed at a distance $a$ from the slit, is given by:
Explanation: Using the thin prism approximation, each half acts as a prism of angle $\alpha$, giving a deviation $\delta = (\mu - 1)\alpha$. The distance between the two virtual sources is $d = 2a(\mu - 1)\alpha$.
PYQ Patterns
These are pattern-based practice questions. Authentic year labels are only used when verified from a source.
1. In a Fresnel's biprism experiment, the distance between the source and screen is $1.5\text{ m}$ and the wavelength of light is $5893\text{ \AA}$. If the fringe width is $0.0925\text{ mm}$, what is the separation between the two virtual sources?
2. A convex lens is introduced between a Fresnel biprism and a screen to measure the virtual source separation $d$. In two conjugate positions of the lens, the measured image separations are $d_1 = 4.0\text{ mm}$ and $d_2 = 0.9\text{ mm}$. The actual separation $d$ is:
Explanation: Using $d = \sqrt{d_1 d_2} = \sqrt{4.0 \times 0.9} = \sqrt{3.6} \approx 1.897\text{ mm}$ (closest to $1.9\text{ mm}$ or exact calculation if numbers adjusted; let's check: $\sqrt{4 \times 0.9} = \sqrt{3.6} = 1.897$. Wait, let's use $d_1=4.0$ and $d_2=1.0$, then $\sqrt{4}=2.0$. Let's ensure clean numbers: if $d_1=4\text{ mm}$ and $d_2=1\text{ mm}$, $d=2\text{ mm}$. Let's select options accordingly. Correction: Option $1.9\text{ mm}$ or let's pick values $d_1=3.6\text{ mm}, d_2=0.9\text{ mm}$ giving $\sqrt{3.24} = 1.8\text{ mm}$. Let's use $d_1=4\text{ mm}, d_2=1\text{ mm}$, $d=2\text{ mm}$.)
3. A thin transparent sheet of refractive index $1.5$ and thickness $6 \times 10^{-6}\text{ m}$ is placed in the path of one interfering beam in a biprism experiment. If $\lambda = 6000\text{ \AA}$ and $d = 1\text{ mm}$, $D = 1\text{ m}$, what is the fringe shift?
4. In a biprism experiment, fringes of width $0.4\text{ mm}$ are obtained on a screen. When the screen is moved away from the biprism by $0.2\text{ m}$, the fringe width becomes $0.44\text{ mm}$. If the initial distance between source and screen was $D$, what is $D$?
Explanation: We have $\beta_1 = \frac{\lambda D}{d}$ and $\beta_2 = \frac{\lambda (D + 0.2)}{d}$. Taking ratio: $\frac{0.4}{0.44} = \frac{D}{D + 0.2} \implies 10D = 9(D + 0.2) \implies D = 1.8\text{ m}$ (Let's adjust numbers to make $D=2.0\text{ m}$: if $\beta_1 = 0.4$, $\beta_2 = 0.44$, ratio is $10/11$). Let's use standard values.
5. Why does Fresnel's biprism form interference fringes whereas an ordinary glass prism placed in front of a slit does not form visible interference fringes on a screen?
Explanation: An ordinary prism has a single refracting surface acting in one direction, whereas a biprism has two adjacent small-angled opposing refracting surfaces that split the wavefront into two overlapping virtual sources.
6. If the refracting angle of a Fresnel biprism is $1^\circ$ and its refractive index is $1.5$, and the slit is placed $20\text{ cm}$ from the biprism, what is the separation $d$ between the virtual sources? (Use $\alpha$ in radians: $1^\circ = \frac{\pi}{180}\text{ rad}$)
3. What is the primary reason why direct measurement of virtual source separation $d$ is impossible?
Explanation: The virtual sources $S_1$ and $S_2$ are imaginary points behind the biprism, necessitating an optical method like a convex lens to image them.
4. In the lens method for finding $d$, if the two conjugate positions of the lens yield image separations $d_1$ and $d_2$, the formula $d = \sqrt{d_1 d_2}$ is a consequence of:
Explanation: The product of linear magnifications in conjugate positions of a lens equals $1$, leading to $d = \sqrt{d_1 d_2}$.
5. If a mica sheet of refractive index $1.6$ and thickness $t$ is introduced in one path, and the fringe shift is equal to 3 fringe widths, what is the thickness $t$?
6. What happens to the fringe pattern if the monochromatic source is replaced by a white light source?
Explanation: White light produces a central white fringe (zero path difference for all wavelengths) flanked by colored fringes.
7. If the distance $D$ between source and screen is doubled while keeping $d$ and $\lambda$ constant, the fringe width becomes:
Explanation: Since $\beta \propto D$, doubling $D$ doubles the fringe width.
8. Which of the following describes the shape of interference fringes formed by a Fresnel biprism with a linear slit source?
Explanation: A narrow linear slit source produces straight, parallel, equidistant interference fringes on a screen placed parallel to the biprism.
9. What is the condition for obtaining clear and well-defined interference fringes in a Fresnel biprism experiment?
Explanation: Interference requires coherent sources that are narrow and monochromatic to maintain a stable phase relationship and visibility.
10. In a Fresnel biprism experiment, the fringe width $\beta$ is measured as $0.3\text{ mm}$. If the entire apparatus is immersed in water (refractive index $4/3$), what will be the new fringe width?
Explanation: When immersed in water, wavelength becomes $\lambda' = \lambda / \mu_w$. Therefore, the new fringe width $\beta' = \beta / \mu_w = 0.3 / (4/3) = 0.3 \times 3 / 4 = 0.225\text{ mm}$.