Understand Wave Phenomena - IB Physics

In a Young's double-slit experiment, the separation between adjacent bright fringes is $s$ . The entire apparatus is then submerged in a transparent liquid with a refractive index $n > 1$ . What is the new fringe separation?
Answer: $s/n$ The fringe separation is given by $s = \frac{\lambda D}{d}$ . When the apparatus is submerged in a liquid of refractive index $n$ , the wavelength of the light changes to $\lambda' = \frac{\lambda}{n}$ . The distance to the screen $D$ and the slit separation $d$ remain unchanged. The new fringe separation $s'$ will be $s' = \frac{\lambda' D}{d} = \frac{(\lambda/n) D}{d} = \frac{1}{n} \left(\frac{\lambda D}{d}\right) = \frac{s}{n}$ .
A wave traveling in medium 1 with speed $v_1 = 300$ m/s strikes a boundary with medium 2 at an angle of 30° to the normal. The refracted wave travels at 45° to the normal in medium 2. If a wave pulse in medium 1 has a frequency of 150 Hz, what will be the wavelength of the transmitted pulse in medium 2?
Answer: 2.83 m Using Snell's law: $\frac{\sin 30°}{v_1} = \frac{\sin 45°}{v_2}$ , so $\frac{0.5}{300} = \frac{0.707}{v_2}$ . Solving: $v_2 = \frac{0.707 \times 300}{0.5} = 424$ m/s. The frequency remains constant at 150 Hz during transmission. The wavelength in medium 2 is $\lambda_2 = \frac{v_2}{f} = \frac{424}{150} = 2.83$ m. Choice A uses the wrong speed calculation. Choice B assumes equal speeds. Choice D incorrectly multiplies instead of dividing.
Which of the following phenomena provides the strongest evidence for the wave nature of light rather than its particle nature?
Answer: Diffraction through a narrow slit Diffraction is the spreading of waves as they pass through an aperture or around an obstacle. This phenomenon is characteristic of waves and cannot be explained by a simple particle model of light (where particles would travel in straight lines). Reflection and refraction can be explained by both wave and particle models (Newton's corpuscular theory explained them). The photoelectric effect provides strong evidence for the particle nature of light (photons).
Two coherent point sources S1 and S2 emit waves of the same wavelength $\lambda$ . A point P is located such that the distance S1P is 12.5 $\lambda$ and the distance S2P is 10.0 $\lambda$ . What is the nature of the interference at point P?
Answer: Destructive, producing a minimum. Interference depends on the path difference between the two waves. The path difference is $\Delta x = |S1P - S2P| = |12.5\lambda - 10.0\lambda| = 2.5\lambda$ . Destructive interference occurs when the path difference is an odd integer multiple of half a wavelength, i.e., $(m + \frac{1}{2})\lambda$ . Here, $2.5\lambda = (2 + \frac{1}{2})\lambda$ , which fits the condition for destructive interference with $m=2$ .
In a double-slit experiment, red light ( $\lambda_R = 700$ nm) produces a first-order bright fringe at a certain position on a screen. If the red light is replaced by blue light ( $\lambda_B = 400$ nm), which bright fringe of the blue light will be formed closest to the original position of the red light's first-order fringe?
Answer: Second-order The position of the $n$ -th order bright fringe is given by $x = \frac{n\lambda D}{d}$ . The position of the first-order red fringe is $x_R = \frac{1 \cdot \lambda_R D}{d} = \frac{700 D}{d}$ (in arbitrary units). We want to find an integer $n_B$ such that the position of the $n_B$ -th blue fringe, $x_B = \frac{n_B \lambda_B D}{d} = \frac{n_B \cdot 400 D}{d}$ , is closest to $x_R$ . We set $x_B \approx x_R$ , so $n_B \cdot 400 \approx 700$ . This gives $n_B \approx 700/400 = 1.75$ . The closest integer to 1.75 is 2. Therefore, the second-order blue fringe will be closest.
A ray of light travels from medium X to medium Y. Total internal reflection is observed to be possible. Which statement provides a sufficient condition for this observation?
Answer: The refractive index of medium X is greater than that of medium Y. Total internal reflection (TIR) can only occur when light travels from a medium with a higher refractive index to a medium with a lower refractive index. This is the definition of moving from a more optically dense to a less optically dense medium. Therefore, $n_X > n_Y$ is the necessary condition for TIR to be possible. The angle of incidence must also be greater than the critical angle, but the possibility of TIR rests on the refractive indices.
Significant diffraction occurs when a wave passes through an aperture. By what factor must the frequency of a wave be changed for it to diffract to the same extent through an aperture that is twice as wide? Assume the wave speed is constant.
Answer: It must be halved. The extent of diffraction depends on the ratio of the wavelength to the aperture size, $\lambda/b$ . For the diffraction to be the same, this ratio must remain constant. If the aperture width $b$ is doubled ( $b' = 2b$ ), the wavelength $\lambda$ must also be doubled ( $\lambda' = 2\lambda$ ) to keep the ratio $\lambda'/b'$ the same. Since wave speed $v$ is constant and $v = f\lambda$ , frequency is inversely proportional to wavelength ( $f = v/\lambda$ ). To double the wavelength, the frequency must be halved.
For two light sources to produce a stable interference pattern, they must be coherent. Which statement gives the essential conditions for coherence?
Answer: The sources have the same frequency and a constant phase difference. Coherence has two requirements: 1) the waves must have the same frequency (be monochromatic), and 2) the phase difference between the waves at any given point must be constant over time. The phase difference does not need to be zero. Amplitude and intensity affect the visibility of the fringes, and polarization affects whether interference can occur, but they are not the defining conditions for coherence itself.
When unpolarized light is incident on the surface of water at the Brewster angle, the reflected light is completely plane-polarized. This phenomenon is an application of
Answer: reflection and refraction. Brewster's angle is a specific angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. This effect occurs at the boundary between two media and involves both reflection and refraction. At this angle, the reflected ray and the refracted ray are perpendicular to each other. It is fundamentally a consequence of the laws of reflection and refraction.
A light ray travels in a material with refractive index $n_1 = 2.0$ . It is incident on a boundary with a second material with refractive index $n_2 = 1.4$ . What is the approximate critical angle for total internal reflection?
Answer: 44° Total internal reflection can occur because the light is traveling from a higher refractive index medium to a lower one ( $n_1 > n_2$ ). The critical angle $\theta_c$ is defined by the equation $n_1 \sin\theta_c = n_2 \sin(90°)$ . This simplifies to $\sin\theta_c = n_2 / n_1$ . Plugging in the values: $\sin\theta_c = 1.4 / 2.0 = 0.70$ . To find the angle, we take the inverse sine: $\theta_c = \arcsin(0.70) \approx 44.4°$ . The closest answer is 44°.
Coherent light of wavelength $\lambda$ is incident on two slits separated by a distance $d$ . A screen is placed a distance $D$ from the slits. The path difference between the waves arriving at the second-order dark fringe from the central maximum is
Answer: $1.5 \lambda$ Destructive interference, which creates a dark fringe, occurs when the path difference is an odd integer multiple of half a wavelength, i.e., path difference = $(m + \frac{1}{2})\lambda$ , where $m = 0, 1, 2, ...$ . The first-order dark fringe corresponds to $m=0$ (path difference $0.5\lambda$ ). The second-order dark fringe corresponds to $m=1$ , so the path difference is $(1 + \frac{1}{2})\lambda = 1.5\lambda$ .
Two wave pulses travel towards each other on a string. One pulse has a displacement that varies from 0 to +5 cm. The other pulse has a displacement that varies from 0 to -3 cm. What is the maximum possible magnitude of the displacement of the string during the interaction?
Answer: 5 cm The principle of superposition states that the resultant displacement is the algebraic sum of the individual displacements. The question asks for the maximum possible magnitude. The interaction will produce various displacements as the pulses pass through each other. The maximum positive displacement occurs when the peak of the +5 cm pulse aligns with a zero-displacement part of the -3 cm pulse, giving +5 cm. The maximum negative displacement occurs when the trough of the -3 cm pulse aligns with a zero-displacement part of the +5 cm pulse, giving -3 cm. The magnitude of the displacement is maximized at 5 cm.
Light with a wavelength of 600 nm in a vacuum (n=1.00) enters a block of glass with a refractive index of 1.50. It then immediately enters a second material with a refractive index of 1.20. What is the wavelength of the light in the second material?
Answer: 500 nm The frequency of light remains constant when it changes medium. The wavelength in a medium with refractive index $n$ is given by $\lambda_{medium} = \frac{\lambda_{vacuum}}{n}$ . The properties of the first block of glass are irrelevant to the final wavelength in the second material, as long as the reference vacuum wavelength is known. Therefore, the wavelength in the second material is $\lambda_2 = \frac{600 \text{ nm}}{1.20} = 500 \text{ nm}$ .
A sound wave travels from warm air into colder air. The speed of sound is higher in warm air than in cold air. As the wave crosses the boundary, what happens to its frequency and wavelength?
Answer: Frequency is constant; Wavelength decreases The frequency of a wave is determined by its source and remains constant as it propagates through different media. The wave relationship is $v = f\lambda$ . Since the sound wave travels from warm air (higher speed $v$ ) to cold air (lower speed $v$ ), its speed decreases. Because $f$ is constant, if $v$ decreases, the wavelength $\lambda$ must also decrease to maintain the equality.
A circular water wave spreads outward from a point source. At a distance of 2.0 m from the source, the wave amplitude is 5.0 cm. Assuming no energy loss, what will be the amplitude when the wave reaches a distance of 8.0 m from the source?
Answer: 1.25 cm For a circular wave spreading in two dimensions, energy is conserved but distributed over an increasing circumference. The intensity (energy per unit area) decreases as $\frac{1}{r^2}$ , and since intensity is proportional to amplitude squared, amplitude decreases as $\frac{1}{r}$ . Therefore: $\frac{A_2}{A_1} = \frac{r_1}{r_2} = \frac{2.0}{8.0} = 0.25$ . So $A_2 = 5.0 \times 0.25 = 1.25$ cm. Choice B uses $\frac{1}{\sqrt{r}}$ relationship. Choice C assumes no amplitude change. Choice D incorrectly uses inverse relationship.
A standing wave is formed on a string fixed at both ends. The string has length 1.2 m and the third harmonic has a frequency of 450 Hz. If the string tension is doubled while keeping the linear mass density constant, what will be the frequency of the second harmonic?
Answer: 424 Hz For a string fixed at both ends, $f_n = \frac{n}{2L}\sqrt{\frac{T}{\mu}}$ . Initially, $f_3 = 450$ Hz, so the fundamental frequency is $f_1 = \frac{450}{3} = 150$ Hz. When tension doubles, the wave speed increases by $\sqrt{2}$ , so all frequencies increase by $\sqrt{2}$ . The new fundamental frequency is $150\sqrt{2} \approx 212$ Hz. The second harmonic is $2 \times 212 = 424$ Hz. Choice B assumes linear relationship with tension. Choice C uses $\sqrt{2} \times 450$ . Choice D doubles the original third harmonic frequency.
A diffraction grating has N slits per unit length. For a fixed wavelength, what is the effect of increasing N on the angular separation of the principal maxima and on their sharpness?
Answer: Angular separation increases; Sharpness increases The slit separation $d$ is the reciprocal of the number of slits per unit length, $d = 1/N$ . The grating equation is $d \sin\theta = n\lambda$ , or $\sin\theta = n\lambda N$ . If N increases, $\sin\theta$ increases for a given order $n$ , meaning the angular separation between orders increases. The sharpness of the principal maxima is proportional to the total number of slits illuminated. Increasing N means more slits are illuminated over the same beam width, which causes more effective destructive interference between the maxima, making them sharper (narrower).
In a single-slit diffraction experiment, the angle subtended by the first minimum is $\theta$ . If the experiment is repeated with light of double the wavelength and a slit of four times the width, what is the new angle $\theta'$ subtended by the first minimum?
Answer: $\theta/2$ For a single slit, the condition for the first minimum is $b \sin\theta = \lambda$ . For small angles, this is approximated as $b\theta \approx \lambda$ , so $\theta \approx \lambda/b$ . The initial angle is $\theta \approx \lambda/b$ . The new parameters are $\lambda' = 2\lambda$ and $b' = 4b$ . The new angle is $\theta' \approx \lambda'/b' = (2\lambda)/(4b) = \frac{1}{2}(\lambda/b) = \theta/2$ .
Monochromatic light is incident on a diffraction grating with slit separation $d$ . The third-order maximum is observed at an angle of 30° to the central maximum. What is the ratio of the wavelength to the slit separation, $\lambda/d$ ?
Answer: 1/6 The condition for a maximum in a diffraction grating is given by the equation $d \sin\theta = n\lambda$ , where $n$ is the order of the maximum. We are given $n=3$ and $\theta = 30°$ . Substituting these values gives $d \sin(30°) = 3\lambda$ . Since $\sin(30°) = 0.5 = 1/2$ , the equation becomes $d (1/2) = 3\lambda$ . We are asked for the ratio $\lambda/d$ . Rearranging the equation gives $\frac{\lambda}{d} = \frac{1/2}{3} = \frac{1}{6}$ .
A wave pulse travels along a string and encounters a boundary where the string density changes. The incident pulse has amplitude A and the reflected pulse has amplitude 0.6A. What fraction of the incident wave's energy is transmitted through the boundary?
Answer: 0.64 Wave energy is proportional to the square of amplitude. The reflected amplitude is 0.6A, so the reflected energy fraction is $(0.6)^2 = 0.36$ . By conservation of energy, the transmitted energy fraction is $1 - 0.36 = 0.64$ . Choice A incorrectly uses only the reflected energy. Choice C uses the amplitude ratio directly. Choice D incorrectly assumes amplitude and energy have the same relationship.

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