Interference of Light
Target Audience: B.Sc. Non-Medical / Physics Students Topic: Wave Optics (Interference)
Question 1
Which of the following is a necessary condition for sustained interference of light?
A) Two sources must be narrow B) Two sources must be broad C) Two sources must be coherent D) Two sources must have different amplitudes
Answer: C) Two sources must be coherent
Detailed Solution: For sustained interference (a stable pattern of bright and dark fringes that does not change with time), the phase difference between the two interfering waves must remain constant over time. Sources that maintain a constant phase difference are called coherent sources. If the sources are incoherent, the phase difference changes randomly and rapidly, averaging out the intensity to a uniform illumination, and no interference pattern is observed.
Question 2
In Young’s Double Slit Experiment (YDSE), if the separation between the slits ($d$) is halved and the distance between the slits and the screen ($D$) is doubled, the fringe width ($\beta$) will:
A) Remain same B) Be halved C) Be doubled D) Become four times
Answer: D) Become four times
Detailed Solution: The formula for fringe width ($\beta$) in YDSE is given by:$$\beta = \frac{\lambda D}{d}$$
Where:
- $\lambda$ = wavelength of light
- $D$ = distance from slits to screen
- $d$ = separation between slits
New conditions: $d’ = \frac{d}{2}$ and $D’ = 2D$. Substituting these into the formula:$$\beta’ = \frac{\lambda (2D)}{(d/2)} = \frac{2\lambda D}{0.5d} = 4 \left( \frac{\lambda D}{d} \right) = 4\beta$$
Thus, the new fringe width is four times the original.
Question 3
When white light is used in Young’s Double Slit Experiment, the central fringe is:
A) Dark B) White C) Red D) Blue
Answer: B) White
Detailed Solution: At the central position on the screen, the path difference between light waves from the two slits is zero for all wavelengths. Since constructive interference occurs for zero path difference regardless of wavelength, all colors of the white light spectrum interfere constructively at the center. The superposition of all colors produces white light. (Note: The fringes surrounding the central white fringe will be colored, with red being on the outer side and violet on the inner side).
Question 4
Two coherent sources of intensity ratio 81:1 produce interference fringes. The ratio of maximum to minimum intensity in the fringe system is:
A) 10:1 B) 9:1 C) 25:16 D) 100:64
Answer: C) 25:16
Detailed Solution: Let $I_1$ and $I_2$ be the intensities. Given: $\frac{I_1}{I_2} = \frac{81}{1}$. Amplitude is proportional to the square root of intensity ($A \propto \sqrt{I}$).$$\frac{A_1}{A_2} = \sqrt{\frac{81}{1}} = \frac{9}{1}$$
Maximum Intensity ($I_{max}$) occurs when amplitudes add up: $(A_1 + A_2)^2$ Minimum Intensity ($I_{min}$) occurs when amplitudes subtract: $(A_1 – A_2)^2$$$\frac{I_{max}}{I_{min}} = \frac{(A_1 + A_2)^2}{(A_1 – A_2)^2} = \frac{(9 + 1)^2}{(9 – 1)^2} = \frac{(10)^2}{(8)^2} = \frac{100}{64} = \frac{25}{16}$$
Question 5
The colors seen in a soap bubble or a thin oil film on water are due to:
A) Dispersion B) Diffraction C) Interference D) Polarization
Answer: C) Interference
Detailed Solution: The brilliant colors observed in thin films (like soap bubbles or oil slicks) arise from the interference of light waves reflected from the top surface and the bottom surface of the film. Depending on the film’s thickness and the angle of viewing, certain wavelengths interfere constructively (appearing bright/colored) while others interfere destructively (disappearing), creating the spectrum of colors.
Question 6
In a thin film of thickness $t$ and refractive index $\mu$, the optical path difference for a ray reflected at a near-normal incidence is:
A) $2\mu t$ B) $2\mu t + \lambda/2$ C) $\mu t$ D) $\mu t + \lambda/2$
Answer: B) $2\mu t + \lambda/2$
Detailed Solution: The geometric path traveled by the light inside the film (down and up) is $2t$. The optical path length is Refractive Index $\times$ Geometric Path = $2\mu t$. However, light reflecting from the top surface (denser medium boundary, air-to-film) undergoes a phase change of $\pi$ (or path difference of $\lambda/2$) due to the Stokes’ relation for reflection from a denser medium. The reflection at the bottom surface (film-to-air) does not undergo this change. Therefore, the total effective path difference is $2\mu t \pm \lambda/2$. (Using $+$ or $-$ yields equivalent physical results).
Question 7
When viewing Newton’s Rings by reflected light, the central spot is usually dark. This is because:
A) The lens touches the glass plate, making thickness zero, and a phase change of $\pi$ occurs. B) The thickness of the air film is large at the center. C) Light is scattered away from the center. D) The lens has a defect at the center.
Answer: A) The lens touches the glass plate, making thickness zero, and a phase change of $\pi$ occurs.
Detailed Solution: At the point of contact between the plano-convex lens and the glass plate, the thickness of the air film ($t$) is effectively zero. The path difference $2\mu t$ becomes zero. However, one of the interfering rays reflects from the bottom glass plate (denser medium), introducing an additional phase shift of $\pi$ (or path difference $\lambda/2$). Total Path Difference $\Delta = 2\mu(0) + \lambda/2 = \lambda/2$. This condition ($\Delta = \lambda/2$) corresponds to destructive interference, resulting in a dark spot.
Question 8
In Newton’s Rings experiment, the diameter of the $n^{th}$ dark ring is proportional to:
A) $n$ B) $n^2$ C) $\sqrt{n}$ D) $1/\sqrt{n}$
Answer: C) $\sqrt{n}$
Detailed Solution: For dark rings in reflected light, the condition is $2t = n\lambda$. From the geometry of the lens (where $R$ is the radius of curvature and $r_n$ is the radius of the ring), we know that $2t \approx \frac{r_n^2}{R}$. Equating the two:$$\frac{r_n^2}{R} = n\lambda \implies r_n^2 = n\lambda R \implies r_n = \sqrt{n\lambda R}$$
Since Diameter $D_n = 2r_n$, we have $D_n \propto \sqrt{n}$. Thus, the diameter of dark rings is proportional to the square root of the natural numbers.
Question 9
If the air film in a Newton’s Rings setup is replaced by a liquid of refractive index $\mu > 1$, the diameter of the rings will:
A) Increase B) Decrease C) Remain unchanged D) Disappear
Answer: B) Decrease
Detailed Solution: The diameter of the $n^{th}$ dark ring in air is $D_{air} = \sqrt{4n\lambda R}$. When a liquid of refractive index $\mu$ is introduced, the optical path changes, and the wavelength effectively becomes $\lambda’ = \lambda / \mu$. The new diameter is $D_{liq} = \sqrt{4n \frac{\lambda}{\mu} R}$. Comparing the two:$$D_{liq} = \frac{D_{air}}{\sqrt{\mu}}$$
Since $\mu > 1$, the denominator is greater than 1, so $D_{liq} < D_{air}$. The rings contract (shrink) and the diameter decreases.
Question 10
Which experiment confirmed the wave nature of light?
A) Photoelectric effect B) Compton effect C) Young’s Double Slit Experiment D) Blackbody radiation
Answer: C) Young’s Double Slit Experiment
Detailed Solution: The Photoelectric effect, Compton effect, and Blackbody radiation are phenomena that demonstrate the particle (quantum) nature of light. Young’s Double Slit Experiment (1801) demonstrated interference, a property unique to waves, thereby establishing the wave nature of light.
Question 11
In a Michelson Interferometer, the purpose of the compensating plate is:
A) To make the two arms equal in length physically. B) To compensate for the path difference introduced by the glass of the beam splitter. C) To change the color of the light. D) To focus the light.
Answer: B) To compensate for the path difference introduced by the glass of the beam splitter.
Detailed Solution: In a Michelson Interferometer, the beam splitter ($G_1$) is a glass plate of finite thickness. One beam (reflected) passes through $G_1$ once, but the other beam (transmitted) passes through $G_1$ three times in total (once entering, twice going to the mirror and back). This introduces an extra optical path in one arm. A compensating plate ($G_2$), made of the same material and thickness as $G_1$, is placed in the path of the first beam to ensure both beams travel through the same thickness of glass. This makes the path difference depend only on the air gap, which is crucial for using white light fringes.
Question 12
Interference of light is a consequence of:
A) Conservation of momentum B) Conservation of energy C) Conservation of charge D) Conservation of mass
Answer: B) Conservation of energy
Detailed Solution: In an interference pattern, energy is not created or destroyed. At bright fringes (constructive interference), the intensity is maximum ($4I_0$), and at dark fringes (destructive interference), it is zero. The average intensity over the pattern remains equal to the sum of the individual intensities ($I_0 + I_0 = 2I_0$). Energy is simply redistributed from the dark regions to the bright regions, consistent with the Law of Conservation of Energy.
Question 13
For constructive interference to take place between two monochromatic waves of wavelength $\lambda$, the path difference must be:
A) $(2n+1) \lambda/2$ B) $(n + 1/2) \lambda$ C) $n\lambda$ D) $(2n-1) \lambda$
Answer: C) $n\lambda$
Detailed Solution: Constructive interference (maximum intensity) occurs when the crest of one wave falls on the crest of another. This happens when the phase difference is an even multiple of $\pi$ ($0, 2\pi, 4\pi…$) or the path difference ($\Delta$) is an integral multiple of the wavelength.$$\Delta = n\lambda$$
where $n = 0, 1, 2, 3…$ Options A and B describe the condition for destructive interference.
Question 14
In Michelson Interferometer, as you move the movable mirror by a distance of $\lambda/4$, the path difference changes by:
A) $\lambda$ B) $\lambda/2$ C) $\lambda/4$ D) $2\lambda$
Answer: B) $\lambda/2$
Detailed Solution: In a Michelson interferometer, the light travels to the movable mirror and reflects back. If the mirror moves by a distance $x$, the additional distance covered by the light is $x$ (forward) + $x$ (return) = $2x$. If the mirror moves by $x = \lambda/4$, the total path difference introduced is:$$\Delta = 2 \times (\lambda/4) = \lambda/2$$
A path difference of $\lambda/2$ shifts the pattern from a maximum to a minimum (or vice versa).
Question 15
The shape of interference fringes in Young’s Double Slit Experiment is: .
A) Circular B) Elliptical C) Hyperbolic D) Parabolic
Answer: C) Hyperbolic
Detailed Solution: The locus of points having a constant path difference from two fixed points (the slits $S_1$ and $S_2$) is a hyperbola. $S_2P – S_1P = \text{constant}$. On a screen placed perpendicular to the plane of the slits, the fringes appear as nearly straight lines near the center because the screen intersects the hyperboloids of revolution at a distance large compared to the slit separation. However, strictly speaking, the geometric locus is hyperbolic.
Question 16
If the monochromatic source in YDSE is replaced by a source of white light, what happens to the fringe width?
A) All fringes become white. B) Fringe width for red color will be greater than that for blue color. C) Fringe width for blue color will be greater than that for red color. D) No fringes will be observed.
Answer: B) Fringe width for red color will be greater than that for blue color.
Detailed Solution: Fringe width is given by $\beta = \frac{\lambda D}{d}$. Since $\beta \propto \lambda$, the color with the larger wavelength will have a larger fringe width. In the visible spectrum, Red ($\lambda \approx 700$ nm) has a longer wavelength than Blue/Violet ($\lambda \approx 400$ nm). Therefore, $\beta_{red} > \beta_{blue}$. This causes the colored fringes to separate, with the red fringes being broader and further apart.
Question 17
Lloyd’s Mirror experiment is an example of interference produced by:
A) Division of amplitude B) Division of wavefront C) Polarization D) Dispersion
Answer: B) Division of wavefront
Detailed Solution: Interference methods are broadly classified into two types:
- Division of Wavefront: Different parts of the same wavefront are separated to form coherent sources (e.g., Young’s Double Slit, Fresnel Biprism, Lloyd’s Mirror).
- Division of Amplitude: A single beam is split into two by partial reflection and refraction (e.g., Newton’s Rings, Michelson Interferometer, Thin Films).
Question 18
In a wedge-shaped film, the interference fringes are:
A) Circular and concentric B) Straight and parallel to the thin edge C) Parabolic D) Irregular
Answer: B) Straight and parallel to the thin edge
Detailed Solution: In a wedge-shaped film (a film of varying thickness), the locus of constant thickness forms straight lines parallel to the apex (the thin edge) of the wedge. Since the condition for constructive/destructive interference depends on the thickness ($t$), the fringes of constant path difference follow the lines of constant thickness. Thus, they appear as straight lines parallel to the edge of the wedge.
Question 19
Which of the following does NOT affect the fringe width in Young’s Double Slit Experiment?
A) Wavelength of light used B) Distance between slits C) Width of the individual slits D) Distance between slits and screen
Answer: C) Width of the individual slits
Detailed Solution: Formula: $\beta = \frac{\lambda D}{d}$.
- $\lambda$ (Wavelength) affects $\beta$.
- $d$ (Separation between slits) affects $\beta$.
- $D$ (Distance to screen) affects $\beta$.
- The width of the individual slits ($a$) affects the diffraction envelope and the brightness of the fringes, but it does NOT appear in the formula for the width ($\beta$) of the interference fringes.
Question 20
Two waves are represented by $y_1 = a \sin(\omega t)$ and $y_2 = a \cos(\omega t)$. The phase difference between them is:
A) $0$ B) $\pi/4$ C) $\pi/2$ D) $\pi$
Answer: C) $\pi/2$
Detailed Solution: To compare phases, both wave equations should be in the same trigonometric form (both sine or both cosine). We know that $\cos(\omega t) = \sin(\omega t + \pi/2)$. So, $y_2$ can be rewritten as $y_2 = a \sin(\omega t + \pi/2)$. Comparing $y_1 = a \sin(\omega t)$ and $y_2 = a \sin(\omega t + \pi/2)$, the phase difference is:$$\phi = (\omega t + \pi/2) – \omega t = \pi/2$$