Waves and Oscillations

We need to find the ratio of extra waves introduced in the glass to the number of waves in the same length of air.

Given:
- Wavelength in air: $\lambda$
- Refractive index: $n = \frac{c_{air}}{c_{glass}}$
- Glass thickness: $t$
Step 1: Number of waves in air

In air, over length $t$, the number of wavelengths is: $$N_{air} = \frac{t}{\lambda}$$

Step 2: Wavelength in glass

When light enters glass, its speed decreases: $c_{glass} = \frac{c_{air}}{n}$

Since frequency $f$ remains constant, the wavelength in glass is: $$\lambda_{glass} = \frac{c_{glass}}{f} = \frac{c_{air}}{nf} = \frac{\lambda}{n}$$

Step 3: Number of waves in glass

In glass, over the same length $t$, the number of wavelengths is: $$N_{glass} = \frac{t}{\lambda_{glass}} = \frac{t}{\lambda/n} = \frac{nt}{\lambda} = n \cdot N_{air}$$

Step 4: Extra waves introduced

The extra number of waves in glass compared to air is: $$\Delta N = N_{glass} - N_{air} = nN_{air} - N_{air} = (n-1)N_{air}$$

Step 5: Required ratio

The ratio of extra waves to the number of waves in air is: $$\text{Ratio} = \frac{\Delta N}{N_{air}} = \frac{(n-1)N_{air}}{N_{air}} = n-1$$

Therefore, the answer is A ($(n-1)$).

This problem involves single-slit diffraction. The angular spread of the diffraction pattern is related to the wavelength and slit width.

Diffraction formula:

For single-slit diffraction, the first minimum occurs at angle $\theta$ where: $$\sin \theta = \frac{\lambda}{a}$$

where $\lambda$ is the wavelength and $a$ is the slit width.

The angular spread of the diffraction pattern is proportional to $\frac{\lambda}{a}$ - specifically, the spread increases with wavelength and decreases with slit width.

Effect of water:

When light enters water from air, its speed decreases due to the refractive index of water ($n_{water} \approx 1.33$). The frequency remains constant, but the wavelength changes:

$$\lambda_{water} = \frac{\lambda_{air}}{n_{water}}$$

Since $n_{water} > 1$, we have $\lambda_{water} < \lambda_{air}$.

Angular spread in water:

With the smaller wavelength in water: $$\sin \theta_{water} = \frac{\lambda_{water}}{a} = \frac{\lambda_{air}}{n_{water} \cdot a} = \frac{\sin \theta_{air}}{n_{water}}$$

Since $\sin \theta_{water} < \sin \theta_{air}$, the diffraction angle is smaller in water.

Therefore, the diffracted beam is less spread out in water than in air, which corresponds to answer C.

This problem involves diffraction through a circular aperture and how intensity changes with hole size.

Given: - Initial hole area: A
- Initial intensity at center: I0
- When diameter halves, angular width doubles (given relationship)
Key principle: For diffraction through a circular aperture (Airy disk), the angular width of the central maximum is inversely proportional to the diameter: $$\theta \propto \frac{1}{d}$$

When diameter is halved ($d \rightarrow d/2$): $$\theta \rightarrow 2\theta$$ This confirms the given relationship.

Effect on power through the hole:

The power passing through the hole is proportional to its area. Since area scales with diameter squared: $$A \propto d^2$$

When diameter halves: $d \rightarrow d/2$ $$A' \propto \left(\frac{d}{2}\right)^2 = \frac{d^2}{4} = \frac{A}{4}$$

Effect on intensity at center:

Intensity = Power/Area. The power is reduced by factor of 4, but the beam is now spread over a larger area due to diffraction. The central spot area scales as $\theta^2$, so: $$\text{Central spot area} \propto \theta^2$$

When $\theta \rightarrow 2\theta$: $$\text{Central spot area} \rightarrow 4 \times \text{original area}$$

Combined effect: $$I' = \frac{\text{Power}/4}{\text{Area} \times 4} = \frac{I_0}{16}$$

Therefore, the answer is D ($I_0/16$).

This problem involves scaling relationships for reverberation time in acoustically similar rooms.

Given: $$T = \frac{kV}{\alpha A}$$

where:
- $T$ = reverberation time
- $V$ = room volume
- $A$ = surface area
- $k$ = constant
- $\alpha$ = mean absorption coefficient
Scaling for identical shapes:

Rooms of identical shape have the same proportions, meaning all linear dimensions scale by the same factor.

For a characteristic length $L$: $$V \propto L^3 \quad \text{and} \quad A \propto L^2$$

Scaling by factor of 10:

When the room is "10 times longer" (all linear dimensions scale by 10): $$L' = 10L$$ $$V' = (10L)^3 = 1000L^3 = 1000V$$ $$A' = (10L)^2 = 100L^2 = 100A$$

New reverberation time:

$$T' = \frac{kV'}{\alpha A'} = \frac{k(1000V)}{\alpha(100A)} = \frac{1000}{100} \cdot \frac{kV}{\alpha A} = 10T$$

The reverberation time increases by a factor of 10, which corresponds to answer C.

This problem involves total internal reflection and the critical angle for viewing through water.

Given: - Tank diameter: 30 cm, so radius $r = 15$ cm
- Critical angle for water: $\theta_c = 48^\circ$
- Fish at center of circular tank, at depth $d$
- Fish can see entire hemisphere ($180^{\circ}$ view) above water
Physics principle:

For the fish to see the full hemisphere above water, light from the entire horizon must reach the fish's eye. The limiting case is light from the edge of the tank's rim. If light from beyond the rim cannot reach the fish (due to total internal reflection at the water-air interface), the fish's view is limited.

Light can only reach the fish from directions within the critical angle of the vertical. For a $180^{\circ}$ (hemispherical) view, the edge of the tank must be exactly at the critical angle.

Geometric analysis:

From the fish's position, the angle to the tank rim is: $$\tan\theta = \frac{\text{horizontal distance}}{\text{vertical depth}} = \frac{r}{d} = \frac{15}{d}$$

For full hemispherical view: $\theta = \theta_c = 48^\circ$

$$\tan(48^\circ) = \frac{15}{d}$$

$$d = \frac{15}{\tan(48^\circ)}$$

Using $\tan(48^\circ) \approx 1.11$: $$d \approx \frac{15}{1.11} \approx 13.5 \text{ cm}$$

Therefore, the fish must be at depth 13.5 cm, which corresponds to answer B.

This problem involves wave energy and amplitude as a circular wave spreads.

Given:
- Wave spreads in circle of radius r
- Energy per unit length: $\frac{dE}{d\ell} \propto \frac{1}{r}$
- Energy of wave: $E \propto A^2$ (proportional to amplitude squared)
- Find new amplitude when r increases by factor of 4
Energy analysis:

For a circular wave, the total energy E is distributed around the circumference $C = 2\pi r$.

The energy per unit length is: $$\frac{E}{C} = \frac{E}{2\pi r} \propto \frac{1}{r}$$

Amplitude relationship:

The problem states that the energy of the wave is proportional to the square of its amplitude: $$E \propto A^2$$

However, we need to consider what "energy of the wave" means. For the energy per unit length: $$\frac{E}{C} \propto A^2$$

But the energy per unit length decreases as 1/r: $$\frac{E}{C} \propto \frac{1}{r}$$

Therefore: $A^2 \propto \frac{1}{r}$

$$A \propto \frac{1}{\sqrt{r}}$$

When radius increases by factor of 4:

$$r' = 4r$$

$$A' = A \cdot \frac{1}{\sqrt{4}} = A \cdot \frac{1}{2} = \frac{A}{2}$$

The new amplitude is A/2.

Answer: A (A/2)

This problem tests knowledge of optical phenomena.

Definitions of each option: A. Dispersion: The separation of white light into its component colors (spectrum) due to wavelength-dependent refraction. This is what happens in a prism. B. Diffraction: The bending of waves around obstacles or through apertures, causing interference patterns. C. Reflection: The bouncing of light off a surface at an angle equal to the incident angle. D. Refraction: The bending of light as it passes from one medium to another due to speed change. Analysis:

When white light enters a prism:
- Different wavelengths (colors) travel at different speeds in glass
- This causes different amounts of bending (refraction) for each color
- The light spreads out into a rainbow spectrum
This separation of colors due to wavelength-dependent refraction is called dispersion.

While refraction is involved in the process, the specific phenomenon of splitting into colors is dispersion.

Answer: A (Dispersion)

This problem involves multiple reflections in angled plane mirrors.

Given:
- Two plane mirrors at angle $\theta = 15^{\circ}$
- Object O placed at equal distance from both mirrors
- Find total number of images visible
Mirror image formula:

For two mirrors at angle $\theta$, the number of images is: $$N = \frac{360^{\circ}}{\theta} - 1$$

Derivation:

Each reflection creates a new image. The images are arranged in a circle around the intersection point of the mirrors. The angular separation between consecutive images equals the mirror angle.

Calculation:

$$N = \frac{360^{\circ}}{15^{\circ}} - 1$$

$$N = 24 - 1 = 23$$

Including the original object:

Total visible = Number of images + Original object

$$\text{Total} = 23 + 1 = 24$$

Verification:

With mirrors at $15^{\circ}$:
- $360^{\circ} / 15^{\circ} = 24$ positions around the circle
- One position is the actual object
- 23 positions contain images
- Total: 24 visible (object + 23 images)
Note: If $360^{\circ}/\theta$ is not an integer, the formula is more complex, but $3$rac6015$ = 24$ exactly, so this simple formula applies.

Answer: B (24)

This problem involves cesium atomic clock radiation wavelength.

Given:
- 9,192,631,770 periods = 1 second
- Cesium-133 transition
Frequency:

$$f = 9,192,631,770 \text{ Hz} \approx 9.19 \times 10^9 \text{ Hz}$$

Wavelength:

$$\lambda = \frac{c}{f} = \frac{3.0 \times 10^8}{9.19 \times 10^9}$$

$$\lambda = 0.033 \text{ m} = 3.3 \text{ cm}$$

Answer: A (3.3 cm)

This is in the microwave region, which is why atomic clocks use microwave cavities!

This problem involves wave physics.

The question mentions:
- Waves cancelling (destructive interference)
- Waves adding together (constructive interference)
- Stationary waves (standing waves)
The fundamental principle behind all these phenomena is superposition - the principle that when two or more waves overlap, the resultant displacement is the sum of the individual displacements.

While the question mentions interference and standing waves, the broader principle that explains all these effects is superposition.

Answer: E (superposition)

This problem involves wave speed and wavelength calculation.

Given:
- Frequency: $f = 60$ kHz $= 60 \times 10^{3}$ Hz
- Speed of sound in air: $v = 330$ m/s
Wave equation: $$v = f \lambda$$

where $\lambda$ is the wavelength.

Solving for wavelength: $$\lambda = \frac{v}{f} = \frac{330}{60 \times 10^{3}}$$

$$\lambda = 5.5 \times 10^{-3} \text{ m}$$

$$\lambda = 5.5 \text{ mm}$$

Physical insight:

High frequency sound has short wavelengths. 60 kHz is ultrasonic (above human hearing range of $\sim$20 kHz).

Answer: D (5.5 mm)

This problem involves wave phase and interference.

Given:
- Frequency: $f = 165$ Hz
- Distance: $d = 157$ m
- Speed of sound: $v = 330$ m/s
Wavelength calculation: $$\lambda = \frac{v}{f} = \frac{330}{165} = 2 \text{ m}$$ Phase difference:

The phase difference depends on how many wavelengths fit into the distance: $$N = \frac{d}{\lambda} = \frac{157}{2} = 78.5$$

This means:
- 78 complete wavelengths ($360^{\circ}$ each)
- Plus 0.5 wavelength = $180^{\circ}$
Phase calculation: $$\phi = 0.5 \times 360^{\circ} = 180^{\circ}$$

Physical interpretation:

The received signal is completely out of phase ($180^{\circ}$) with the source. If this were a single frequency, destructive interference would occur.

Answer: E ($180^{\circ}$)