Waves and Optics Practice Questions — JAMB UTME — Page 6

Question 101Question →

A progressive wave traveling through a primary medium is represented by the displacement equation $y = 0.04 \sin\left(100\pi t - 2.5\pi x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. If the wave propagates into a secondary medium where its speed increases by $20\%$ , what is the wavelength of the wave in the secondary medium?

0.96\text{ m}

0.67\text{ m}

1.50\text{ m}

0.16\text{ m}

Show answer & explanation

Answer: $0.96\text{ m}$

Answer

The wavelength of the wave in the secondary medium is

0.96\text{ m}

.

Comparing

y = 0.04 \sin(100\pi t - 2.5\pi x)

to the standard form

y = A \sin(\omega t - k x)

, the wave number is

k = 2.5\pi\text{ rad/m}

. The initial wavelength is

\lambda_1 = \frac{2\pi}{2.5\pi} = 0.8\text{ m}

. When a wave moves to a new medium, its frequency stays constant, making wavelength directly proportional to wave speed (

v \propto \lambda

). An increase of

20\%

in wave speed means the new wavelength is

\lambda_2 = 0.8 \times 1.20 = 0.96\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency

\omega

and wave number

k

from the wave equation.

From

y = A \sin(\omega t - k x)

, we identify

\omega = 100\pi\text{ rad/s}

and

k = 2.5\pi\text{ rad/m}

.

Standard wave equation parameters directly define the wave's spatial and temporal frequencies.

2

Calculate the wavelength

\lambda_1

in the initial medium.

$\lambda_1 = \frac{2\pi}{k} = \frac{2\pi}{2.5\pi} = 0.8\text{ m}$.

Wavelength is inversely related to the wave number

k

by

\lambda = \frac{2\pi}{k}

.

3

Apply the boundary conditions of wave refraction across media.

Frequency

f

remains constant across boundary; speed

v

and wavelength

\lambda

scale proportionally.

The frequency of a wave is determined solely by the source and does not change upon entering a new medium.

4

Determine the new wavelength

\lambda_2

in the secondary medium.

$\lambda_2 = \lambda_1 \times (1 + 0.20) = 0.8\text{ m} \times 1.20 = 0.96\text{ m}$.

Since

v = f \lambda

and

f

is constant,

\frac{v_2}{v_1} = \frac{\lambda_2}{\lambda_1} = 1.20

.

Key Concept

Invariance of Wave Frequency across Media and Wave Equation Parameter Extraction

Estimated Time:2m 0s

Question 102Question →

A sound transmitter and a projectile launcher are co-located at a distance of $210\text{ m}$ directly in front of a tall, flat vertical cliff. At time $t = 0\text{ s}$ , a projectile is launched directly away from the cliff at a constant speed of $60\text{ m s}^{-1}$ , while a sound pulse is emitted simultaneously towards the cliff. The sound wave reflects off the cliff face and travels back to overtake the moving projectile. Assuming the speed of sound in air is $340\text{ m s}^{-1}$ , calculate the distance from the cliff face, in meters, to the position where the reflected sound wave intercepts the projectile.

Show answer & explanation

Answer: 300

Answer

The distance of the projectile from the cliff face at the instant of interception is

300\text{ m}

.

The correct calculation accounts for both the two-part path of the sound wave (forward to cliff + back to projectile) and the displacement of the projectile moving away from the cliff over the same time interval, yielding an interception distance of

300\text{ m}

from the cliff.

Step-by-Step Solution

1

Set up the total distance expression for the sound wave from the cliff face

Total sound path =

210\text{ m} + x

The sound pulse must travel

210\text{ m}

forward to hit the cliff face, plus an additional distance

x

away from the cliff face after reflection to reach the projectile.

2

Set up the distance expression for the projectile from its starting point

Projectile path =

x - 210\text{ m}

The projectile starts

210\text{ m}

away from the cliff and moves farther away to position

x

.

3

Equate the time elapsed for both sound propagation and projectile movement

\frac{210 + x}{340} = \frac{x - 210}{60}

Both events happen simultaneously over the exact same time interval

t

.

4

Solve the linear equation for

x

x = 300\text{ m}

Cross-multiplying yields

60(210 + x) = 340(x - 210)

, which simplifies to

28x = 8400

, giving

x = 300\text{ m}

.

Key Concept

Echo reflection path combined with relative linear kinematics

Question 103Question →

A progressive wave traveling along a stretched string is represented by the equation $y = 0.02 \sin(120\pi t - 3\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the velocity of the wave in $\text{m s}^{-1}$ ?

Show answer & explanation

Answer: 40

Answer

The velocity of the wave is

40 \text{ m s}^{-1}

.

The standard progressive wave equation is

y = A \sin(\omega t - kx)

, where

\omega

is the angular frequency and

k

is the wave number (wave vector). Comparing the given equation

y = 0.02 \sin(120\pi t - 3\pi x)

with the standard form yields

\omega = 120\pi \text{ rad s}^{-1}

and

k = 3\pi \text{ m}^{-1}

. The velocity of propagation of the wave is given by

v = \frac{\omega}{k} = \frac{120\pi}{3\pi} = 40 \text{ m s}^{-1}

.

Step-by-Step Solution

1

Compare given equation with the standard progressive wave equation

Matching

y = 0.02 \sin(120\pi t - 3\pi x)

to

y = A \sin(\omega t - kx)

gives

\omega = 120\pi \text{ rad s}^{-1}

and

k = 3\pi \text{ rad m}^{-1}

.

Direct parameter identification from the wave function gives the angular frequency and wave number.

2

Calculate wave velocity

Wave velocity

v = \frac{\omega}{k} = \frac{120\pi}{3\pi} = 40 \text{ m s}^{-1}

.

The ratio of angular frequency to wave number equals the phase velocity of the wave.

Key Concept

Extracting wave parameters (angular frequency and wave number) from the mathematical wave equation to determine wave velocity.

Question 104Question →

An unpolarized light beam with an initial intensity of $80\text{ W/m}^2$ passes through an ideal linear polarizer. What is the intensity of the transmitted light in $\text{W/m}^2$ ?

Show answer & explanation

Answer: 40

Answer

The intensity of the transmitted light is

40\text{ W/m}^2

.

When unpolarized light of initial intensity

I_0

encounters an ideal linear polarizing filter, the transmitted intensity

I

is always equal to half of the incident intensity (

I = \frac{1}{2}I_0

). Substituting

80\text{ W/m}^2

gives

I = 40\text{ W/m}^2

.

Step-by-Step Solution

1

Determine the fraction of unpolarized light intensity transmitted by a polarizer.

Transmitted intensity formula

I = \frac{1}{2} I_0

.

Unpolarized light consists of randomly oriented electric field vectors, resulting in an average transmission factor of one-half.

2

Substitute the incident intensity value into the formula.

I = \frac{80}{2} = 40\text{ W/m}^2

.

Direct mathematical calculation.

Key Concept

Polarization and intensity reduction of unpolarized light upon passing through a linear polarizer.

Question 105Question →

A transverse progressive wave traveling along a stretched string is represented by the displacement equation $y = 0.05 \sin\left(20\pi t - \frac{\pi}{4} x\right)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the phase difference between two points on the string separated by a distance of $2.0\text{ m}$ ?

\frac{\pi}{4}\text{ rad}

\frac{\pi}{2}\text{ rad}

\pi\text{ rad}

2\pi\text{ rad}

Show answer & explanation

Answer: $\frac{\pi}{2}\text{ rad}$

Answer

The phase difference between the two points is

\frac{\pi}{2}\text{ rad}

.

Comparing the given equation

y = 0.05 \sin\left(20\pi t - \frac{\pi}{4} x\right)

to the standard wave equation

y = A \sin(\omega t - kx)

, the wavenumber is

k = \frac{\pi}{4}\text{ rad m}^{-1}

. The phase difference between two points separated by

\Delta x = 2.0\text{ m}

is

\Delta \phi = k \Delta x = \frac{\pi}{4} \times 2.0 = \frac{\pi}{2}\text{ rad}

. Thus, the option stating

\frac{\pi}{2}\text{ rad}

is correct.

Step-by-Step Solution

1

Identify the wavenumber

k

from the standard wave equation

Comparing

y = A \sin(\omega t - kx)

with

y = 0.05 \sin\left(20\pi t - \frac{\pi}{4} x\right)

gives

k = \frac{\pi}{4}\text{ rad m}^{-1}

.

The coefficient of

x

in the wave equation represents the wavenumber

k = \frac{2\pi}{\lambda}

.

2

Calculate the phase difference using

\Delta \phi = k \Delta x

\Delta \phi = \left(\frac{\pi}{4}\text{ rad m}^{-1}\right) \times 2.0\text{ m} = \frac{\pi}{2}\text{ rad}.

Phase difference is directly proportional to the spatial separation between two points along the path of propagation.

Key Concept

Phase difference in a progressive wave

Estimated Time:1m 0s

Question 106Question →

A beam of light traveling in air is incident on a transparent liquid at the polarizing angle (Brewster's angle) of $53.1^\circ$ , where $\tan 53.1^\circ = 1.33$ . What is the critical angle for total internal reflection when light travels from this liquid into air?

\sin^{-1}(0.75)

\sin^{-1}(0.60)

\sin^{-1}(1.33)

53.1^\circ

Show answer & explanation

Answer: $\sin^{-1}(0.75)$

Answer

The critical angle for total internal reflection at the liquid-air boundary is

\sin^{-1}(0.75)

.

According to Brewster's law, the refractive index of the liquid is given by

n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}

. When light travels from the denser liquid medium to the rarer air medium, the critical angle

\theta_c

for total internal reflection satisfies

\sin\theta_c = \frac{1}{n}

. Substituting

n = \frac{4}{3}

gives

\sin\theta_c = \frac{3}{4} = 0.75

, so

\theta_c = \sin^{-1}(0.75)

.

Step-by-Step Solution

1

Determine the refractive index of the liquid using Brewster's law.

n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}

.

Brewster's law states that when light in air (

n_1 = 1

) is incident at the polarizing angle

\theta_B

on a medium of index

n

,

\tan\theta_B = n

.

2

Apply the total internal reflection condition for light passing from liquid to air.

\sin\theta_c = \frac{1}{n} = \frac{1}{4/3} = \frac{3}{4} = 0.75

.

Total internal reflection occurs at an interface when light travels from a denser medium (

n

) to a less dense medium (

1

) at an angle greater than

\theta_c

, where

\sin\theta_c = \frac{1}{n}

.

3

Solve for the critical angle

\theta_c

.

\theta_c = \sin^{-1}(0.75)

.

Taking the inverse sine of

0.75

yields the critical angle.

Key Concept

Synthesizing Brewster's law of polarization with total internal reflection critical angle

Question 107Question →

A survey ship emits an ultrasonic sound pulse vertically downwards into the ocean. The sound wave travels through seawater at a speed of $1500\text{ m/s}$ , and its reflected echo from the seabed is detected by the ship's hydrophone $1.2\text{ s}$ after emission. What is the depth of the ocean floor at that location?

900\text{ m}

1800\text{ m}

450\text{ m}

2700\text{ m}

Show answer & explanation

Answer: $900\text{ m}$

Answer

900\text{ m}

Sound emitted by the ship travels down to the seabed and reflects back to the hydrophone, taking

1.2\text{ s}

for the complete round trip. Using the relation

\text{Depth} = \frac{v \times t}{2}

, the depth is

\frac{1500 \times 1.2}{2} = 900\text{ m}

.

Step-by-Step Solution

1

Calculate total distance traveled by the sound wave

Total distance

s = v \times t = 1500\text{ m/s} \times 1.2\text{ s} = 1800\text{ m}

Distance is the product of speed and total time elapsed.

2

Determine the depth of the seabed

Depth

d = \frac{s}{2} = \frac{1800\text{ m}}{2} = 900\text{ m}

An echo involves the sound traveling down to the ocean floor and back up, so the depth is half the total distance.

Key Concept

Echo Reflection and Distance Calculation

Estimated Time:45s

Question 108Question →

A research vessel emits a high-frequency acoustic pulse vertically downward toward the seabed. The signal reflects off the ocean floor and is detected by the vessel's receiver $1.6\text{ s}$ after transmission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the depth of the ocean floor at this point?

2400\text{ m}

1200\text{ m}

600\text{ m}

937.5\text{ m}

Show answer & explanation

Answer: $1200\text{ m}$

Answer

The depth of the ocean floor is

1200\text{ m}

.

The correct answer is

1200\text{ m}

. Since an echo involves a two-way journey (from the vessel down to the ocean floor and back up), the sound wave takes half of the total time (

0.8\text{ s}

) to reach the bottom. Multiplying the speed of sound in seawater (

1500\text{ m/s}

) by

0.8\text{ s}

yields the correct depth of

1200\text{ m}

.

Step-by-Step Solution

1

Determine the time taken for the sound wave to travel one way to the ocean floor.

t_{\text{one-way}} = \frac{t_{\text{total}}}{2} = \frac{1.6\text{ s}}{2} = 0.8\text{ s}

An echo involves sound traveling from the transmitter to the reflecting surface and back to the receiver.

2

Calculate the depth using the speed of sound in seawater.

\text{Depth } d = v \times t_{\text{one-way}} = 1500\text{ m/s} \times 0.8\text{ s} = 1200\text{ m}

The distance traveled in one direction equals the speed of sound multiplied by the one-way travel time.

Key Concept

Echo location and depth sounding using two-way wave propagation

Estimated Time:1m 0s

Question 109Question →

A research submarine moving underwater at a constant speed of $12.0\text{ m/s}$ directly toward a vertical underwater cliff face emits an ultrasonic acoustic pulse. The echo reflected from the cliff face is detected by the submarine's receiver $2.50\text{ s}$ after emission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the distance between the submarine and the cliff face at the exact moment the echo is detected?

Show answer & explanation

Answer: 1860

Answer

The distance between the submarine and the cliff face at the exact moment the echo is detected is

1860\text{ m}

.

During the

2.50\text{ s}

transit time of the acoustic signal, the sound covers a total path of

3750\text{ m}

(

1500\text{ m/s} \times 2.50\text{ s}

) while the submarine moves

30\text{ m}

closer to the cliff face (

12.0\text{ m/s} \times 2.50\text{ s}

). The total path of the sound consists of the outward journey to the cliff (

d + 30\text{ m}

) and the return journey to the submarine (

d

). Setting

(d + 30) + d = 3750

gives

2d + 30 = 3750

, leading to

d = 1860\text{ m}

.

Step-by-Step Solution

1

Calculate total sound travel distance and submarine displacement during the 2.50 s window.

Sound distance

d_{\text{sound}} = 1500\text{ m/s} \times 2.50\text{ s} = 3750\text{ m}

; Submarine displacement

d_{\text{sub}} = 12.0\text{ m/s} \times 2.50\text{ s} = 30.0\text{ m}

.

Both the acoustic wave and the submarine move continuously throughout the total elapsed transit time.

2

Establish the geometric equation for the sound path relative to the final distance d.

d_{\text{sound}} = 2d + d_{\text{sub}}

, where

d

is the remaining distance to the cliff face at detection time.

The sound pulse travels forward across the initial separation

(d + d_{\text{sub}})

and reflects back across the remaining separation

d

.

3

Solve the linear equation for the final separation distance d.

3750 = 2d + 30 \implies 2d = 3720 \implies d = 1860\text{ m}

.

Subtracting the submarine's forward displacement from the total sound path gives twice the distance to the obstacle at the instant of signal reception.

Key Concept

Echo distance calculations with moving receiver and source

Question 110Question →

A ray of light travels from air into a liquid with a refractive index of $1.33$ . If the sine of the angle of incidence in air is $0.80$ , what is the sine of the angle of refraction in the liquid?

Show answer & explanation

Answer: 0.6

Answer

The sine of the angle of refraction in the liquid is 0.60.

According to Snell's law for light passing from air into a medium, the refractive index

n

is given by

n = \frac{\sin i}{\sin r}

. Rearranging this equation to solve for the sine of the angle of refraction yields

\sin r = \frac{\sin i}{n}

. Substituting

\sin i = 0.80

and

n = 1.33

(or

\frac{4}{3}

) gives

\sin r = \frac{0.80}{4/3} = 0.60

.

Step-by-Step Solution

1

Identify the given physical parameters and state Snell's law

Refractive index

n = 1.33

(or

\frac{4}{3}

),

\sin i = 0.80

. Snell's law:

n = \frac{\sin i}{\sin r}

Snell's law relates the ratio of the sines of the angles of incidence and refraction to the refractive index of the medium.

2

Rearrange the equation to express the sine of the angle of refraction

\sin r = \frac{\sin i}{n}

Algebraically isolating

\sin r

allows direct substitution of the known quantities.

3

Substitute the values and compute the result

\sin r = \frac{0.80}{4/3} = 0.60

Dividing

0.80

by

\frac{4}{3}

gives

0.60

.

Key Concept

Snell's Law of Refraction

Alternative Method

Convert decimal numbers into simple fractions:

n = \frac{4}{3}

and

\sin i = \frac{4}{5}

. Evaluating

\sin r = \frac{4/5}{4/3}

simplifies directly to

\frac{3}{5} = 0.60

.

Estimated Time:45s

Question 111Question →

A ray of light traveling within a dense glass prism of refractive index $1.60$ strikes the boundary with a surrounding transparent liquid. If total internal reflection just occurs at an angle of incidence of $45.0^\circ$ in the glass, what is the refractive index of the liquid? (Take $\sin 45.0^\circ = 0.707$ )

1.13

2.26

0.44

1.60

Show answer & explanation

Answer: $1.13$

Answer

The refractive index of the liquid is

1.13

.

For light traveling from a denser medium (

n_{\text{glass}}

) to a rarer medium (

n_{\text{liquid}}

), the critical angle

\theta_c

is defined by

\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}

. Substituting

n_{\text{glass}} = 1.60

and

\sin 45.0^\circ = 0.707

gives

n_{\text{liquid}} = 1.60 \times 0.707 = 1.13

.

Step-by-Step Solution

1

Identify the given parameters and formula for total internal reflection

Refractive index of denser medium

n_{\text{glass}} = 1.60

, critical angle

\theta_c = 45.0^\circ

, and formula

\sin \theta_c = \frac{n_{\text{rarer}}}{n_{\text{denser}}}

.

Total internal reflection occurs at the critical angle when light travels from an optically denser medium to a less dense (rarer) medium.

2

Rearrange the equation to solve for the refractive index of the liquid (

n_{\text{liquid}}

)

n_{\text{liquid}} = n_{\text{glass}} \times \sin \theta_c

.

Multiplying both sides of the critical angle equation by

n_{\text{glass}}

isolates the target variable.

3

Substitute the values and calculate

n_{\text{liquid}}

n_{\text{liquid}} = 1.60 \times 0.707 = 1.1312 \approx 1.13

.

Carrying out the arithmetic yields the refractive index of the liquid.

Key Concept

Total Internal Reflection and Critical Angle

Question 112Question →

A ray of light strikes a plane mirror at an angle of incidence of $35^\circ$ . What is the angle of deviation of the reflected ray?

110^\circ

70^\circ

55^\circ

145^\circ

Show answer & explanation

Answer: $110^\circ$

Answer

The angle of deviation of the reflected ray is

110^\circ

.

The angle of deviation

d

represents the angle through which a ray of light is turned from its original path. For a plane mirror,

d = 180^\circ - 2i

. Substituting

i = 35^\circ

yields

d = 180^\circ - 70^\circ = 110^\circ

.

Step-by-Step Solution

1

Identify the given angle of incidence

i = 35^\circ

The angle of incidence is measured relative to the normal line.

2

Apply the law of reflection

Angle of reflection

r = i = 35^\circ

The angle of reflection equals the angle of incidence.

3

Calculate the angle of deviation

d = 180^\circ - (i + r) = 180^\circ - 2(35^\circ) = 110^\circ

The angle of deviation measures how much the light ray is turned from its original initial straight path.

Key Concept

Angle of deviation for reflection at a plane surface

Estimated Time:45s

Question 113Question →

A concave mirror forms a real image that is twice the size of an object. When the object is shifted $10\text{ cm}$ closer to the mirror, a virtual image of the same magnification is produced. What is the focal length of the mirror?

10\text{ cm}

15\text{ cm}

20\text{ cm}

30\text{ cm}

Show answer & explanation

Answer: $10\text{ cm}$

Answer

The focal length of the concave mirror is

10\text{ cm}

.

For a concave mirror forming a real image of magnification

2

,

v_1 = 2u_1

, yielding

u_1 = 1.5f

. When the object is moved

10\text{ cm}

closer, a virtual image of magnification

2

is formed, so

v_2 = -2u_2

, yielding

u_2 = 0.5f

. Subtracting the two object positions (

1.5f - 0.5f = 10\text{ cm}

) directly gives

f = 10\text{ cm}

.

Step-by-Step Solution

1

Set up the mirror equation for the first case (real image).

u_1 = \frac{3}{2}f

For a real inverted image with magnification

m = 2

,

v_1 = +2u_1

. Substituting into

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{v_1}

gives

\frac{1}{f} = \frac{1}{u_1} + \frac{1}{2u_1} = \frac{3}{2u_1}

.

2

Set up the mirror equation for the second case (virtual image).

u_2 = \frac{1}{2}f

For a virtual erect image with magnification

m = 2

, sign convention dictates

v_2 = -2u_2

. Substituting into

\frac{1}{f} = \frac{1}{u_2} + \frac{1}{v_2}

gives

\frac{1}{f} = \frac{1}{u_2} - \frac{1}{2u_2} = \frac{1}{2u_2}

.

3

Use the given displacement between the two object positions to solve for

f

.

f = 10\text{ cm}

The object is moved

10\text{ cm}

closer, so

u_1 - u_2 = 10\text{ cm}

. Substituting the expressions yields

\frac{3}{2}f - \frac{1}{2}f = 10\text{ cm} \implies f = 10\text{ cm}

.

Key Concept

Mirror Formula and Sign Convention for Spherical Mirrors

Estimated Time:2m 0s

Question 114Question →

Match each vibrating acoustic system setup on the left with the correct mathematical expression for its resonant frequency ( $f$ ) on the right, where $v$ is the speed of sound in air, $L$ is the physical length of the pipe or string, $e$ is the end correction per open end, $T$ is tension, and $\mu$ is linear mass density.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The fundamental mode of a pipe closed at one end with end correction matches

f = \frac{v}{4(L + e)}

; the fundamental mode of a stretched string matches

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

; the fundamental mode of a pipe open at both ends with end correction at both ends matches

f = \frac{v}{2(L + 2e)}

; and the first overtone of a closed pipe without end correction matches

f = \frac{3v}{4L}

.

Each setup corresponds directly to its derived wave equation: closed pipes produce fundamental frequency

f = \frac{v}{4(L+e)}

for one open end, open pipes produce

f = \frac{v}{2(L+2e)}

for two open ends, stretched strings depend on tension and mass per unit length as

f = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

, and the first overtone of a closed pipe is its third harmonic

f = \frac{3v}{4L}

.

Step-by-Step Solution

1

Analyze boundary conditions and effective acoustic length for closed and open pipes.

A closed pipe has one displacement antinode at the open end and one node at the closed end, adding an effective end correction

e

to its physical length

L

(

L_{\text{eff}} = L + e

). An open pipe has two open ends, giving an effective length

L_{\text{eff}} = L + 2e

.

Air displacement antinodes occur slightly outside open pipe boundaries by a distance

e

per open end.

2

Derive the frequency formula for the fundamental mode of a closed pipe with end correction.

For the fundamental mode,

L + e = \frac{\lambda}{4} \implies \lambda = 4(L + e)

. Frequency

f = \frac{v}{\lambda} = \frac{v}{4(L + e)}

.

The distance between a node and an adjacent antinode is one-quarter of a wavelength.

3

Derive the fundamental frequency for a stretched string fixed at both ends.

L = \frac{\lambda}{2} \implies \lambda = 2L

. Using wave velocity

v = \sqrt{\frac{T}{\mu}}

,

f = \frac{v}{2L} = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

.

Nodes exist at both fixed ends in a vibrating string, making the fundamental wavelength twice the length.

4

Derive the fundamental frequency of an open pipe considering both end corrections.

L + 2e = \frac{\lambda}{2} \implies \lambda = 2(L + 2e)

, so

f = \frac{v}{2(L + 2e)}

.

Antinodes occur at both open ends, placing half a wavelength within the effective acoustic length.

5

Determine the first overtone frequency for a closed pipe without end correction.

The first overtone is the third harmonic (

n = 3

), so

L = \frac{3\lambda}{4} \implies \lambda = \frac{4L}{3}

, which gives

f = \frac{3v}{4L}

.

Closed pipes support only odd integer multiples of the fundamental frequency.

Key Concept

Standing Waves and Resonance in Air Columns and Strings

Question 115Question →

A water wave traveling in deep water has a wavelength of $0.80\text{ m}$ and a speed of $2.4\text{ m/s}$ . Upon entering a shallow region, its speed drops to $1.8\text{ m/s}$ . What are the frequency and wavelength of the wave in the shallow region?

Frequency =

3.0\text{ Hz}

, Wavelength =

0.60\text{ m}

Frequency =

2.25\text{ Hz}

, Wavelength =

0.80\text{ m}

Frequency =

2.25\text{ Hz}

, Wavelength =

0.60\text{ m}

Frequency =

0.33\text{ Hz}

, Wavelength =

0.60\text{ m}

Show answer & explanation

Answer: Frequency = $3.0\text{ Hz}$ , Wavelength = $0.60\text{ m}$

Answer

Frequency =

3.0\text{ Hz}

, Wavelength =

0.60\text{ m}

When a wave passes from deep to shallow water (refraction), its frequency remains unchanged because frequency is fixed by the source. Using

v = f\lambda

, the initial frequency is

f = \frac{2.4}{0.80} = 3.0\text{ Hz}

. In shallow water, the new wavelength is

\lambda' = \frac{1.8}{3.0} = 0.60\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in deep water using the wave equation

v = f \lambda

.

f = \frac{v}{\lambda} = \frac{2.4\text{ m/s}}{0.80\text{ m}} = 3.0\text{ Hz}

.

The frequency depends on the wave source and can be determined from the given initial speed and wavelength.

2

Apply the boundary condition for wave refraction.

The frequency in shallow water remains

f = 3.0\text{ Hz}

.

When a wave travels from one medium to another, its frequency remains constant.

3

Calculate the new wavelength in shallow water using

\lambda' = \frac{v'}{f}

.

\lambda' = \frac{1.8\text{ m/s}}{3.0\text{ Hz}} = 0.60\text{ m}

.

The wavelength changes proportionally with speed when frequency is constant.

Key Concept

Constancy of wave frequency during refraction across medium boundaries

Question 116Question →

A person standing at a stationary position between two tall parallel vertical walls fires a starter pistol. The person hears the first echo reflected from the nearer wall after $1.2\text{ s}$ and the second echo from the farther wall after $1.8\text{ s}$ . Given that the speed of sound in air is $340\text{ m/s}$ , what is the total distance between the two walls in meters?

Show answer & explanation

Answer: 510

Answer

The total distance between the two walls is

510\text{ m}

.

Because sound travels to each wall and reflects back to the observer, the distance to each wall is given by

d = \frac{v \cdot t}{2}

. The distance to the nearer wall is

d_1 = \frac{340 \times 1.2}{2} = 204\text{ m}

, and the distance to the farther wall is

d_2 = \frac{340 \times 1.8}{2} = 306\text{ m}

. Since the observer is between the two walls, the total separation between the walls is

204\text{ m} + 306\text{ m} = 510\text{ m}

.

Step-by-Step Solution

1

Calculate the distance from the observer to the nearer wall.

d_1 = 204\text{ m}

The sound travels to the nearer wall and back in

1.2\text{ s}

, covering twice the distance to that wall.

2

Calculate the distance from the observer to the farther wall.

d_2 = 306\text{ m}

The sound travels to the farther wall and back in

1.8\text{ s}

, covering twice the distance to that wall.

3

Add the two individual distances to find the total distance between the walls.

D = d_1 + d_2 = 510\text{ m}

The observer is positioned between the two walls, so the separation distance is the sum of both distances.

Key Concept

Echo and Speed of Sound Propagation between Parallel Boundaries

Question 117Question →

A rectangular glass block of thickness $6.0\text{ cm}$ has a refractive index of $1.50$ . Calculate the apparent depth, in centimeters, of a mark placed at the bottom of the block when viewed normally from above.

Show answer & explanation

Answer: 4

Answer

The apparent depth of the mark is

4.0\text{ cm}

.

The refractive index of a medium relative to air is given by the ratio of real depth to apparent depth (

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

). Substituting the given values gives

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth for normal view

n = \frac{\text{Real Depth}}{\text{Apparent Depth}}

By definition of optical refraction when looking normally from an optically less dense medium (air) into a denser medium (glass).

2

Substitute the known values (

n = 1.50

,

\text{Real Depth} = 6.0\text{ cm}

) and solve for the apparent depth

\text{Apparent Depth} = \frac{6.0\text{ cm}}{1.50} = 4.0\text{ cm}

Dividing the real thickness by the refractive index yields the perceived (apparent) depth.

Key Concept

Real and Apparent Depth in Refraction

Question 118Question →

A hiker standing at a distance from a tall vertical cliff shouts and hears an echo $0.60\text{ s}$ later. If the speed of sound in air is $330\text{ m/s}$ , what is the distance between the hiker and the cliff?

198\text{ m}

99\text{ m}

550\text{ m}

49.5\text{ m}

Show answer & explanation

Answer: $99\text{ m}$

Answer

The distance between the hiker and the cliff is

99\text{ m}

.

An echo is a reflected sound wave. The sound travels from the hiker to the cliff and back, covering a total distance of

2d

in time

t

. Therefore,

2d = v \times t

, which gives

d = \frac{330\text{ m/s} \times 0.60\text{ s}}{2} = 99\text{ m}

.

Step-by-Step Solution

1

Identify given parameters and echo relation

Speed of sound

v = 330\text{ m/s}

, time elapsed

t = 0.60\text{ s}

. An echo involves sound traveling to the wall and back (

2d

).

The total distance traveled by the sound wave during time

t

is twice the distance to the reflecting surface.

2

Calculate the one-way distance

d = \frac{v \times t}{2} = \frac{330 \times 0.60}{2} = 99\text{ m}

.

Dividing the total distance by 2 yields the actual distance from the hiker to the cliff.

Key Concept

Echo distance calculation

Question 119Question →

A side-view convex mirror on a bus has a radius of curvature of $40\text{ cm}$ . If a motorcycle is located $30\text{ cm}$ in front of the mirror, what is the location of the image formed relative to the mirror?

12\text{ cm}

behind the mirror

12\text{ cm}

in front of the mirror

60\text{ cm}

behind the mirror

60\text{ cm}

in front of the mirror

Show answer & explanation

Answer: $12\text{ cm}$ behind the mirror

Answer

The image is formed

12\text{ cm}

behind the mirror.

For a convex mirror, the focal length is virtual, so

f = -20\text{ cm}

. With an object distance of

u = +30\text{ cm}

, applying the mirror equation

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

gives

\frac{1}{v} = -\frac{1}{20} - \frac{1}{30} = -\frac{1}{12}

, leading to

v = -12\text{ cm}

. The negative sign specifies that the virtual image is located

12\text{ cm}

behind the mirror.

Step-by-Step Solution

1

Determine the focal length of the mirror from its radius of curvature

f = -\frac{R}{2} = -\frac{40\text{ cm}}{2} = -20\text{ cm}

For spherical mirrors, focal length is half the radius of curvature. Convex mirrors have a negative focal length by sign convention.

2

Set up the mirror formula using the given object distance

u = +30\text{ cm}

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies -\frac{1}{20} = \frac{1}{30} + \frac{1}{v}

The mirror equation relates focal length, object distance, and image distance.

3

Solve for the image distance

v

\frac{1}{v} = -\frac{1}{20} - \frac{1}{30} = -\frac{3 + 2}{60} = -\frac{5}{60} = -\frac{1}{12} \implies v = -12\text{ cm}

Algebraic manipulation yields a negative image distance.

4

Interpret the physical meaning of the calculated value

The negative sign indicates a virtual image located

12\text{ cm}

behind the mirror.

Under standard optical sign conventions, negative image distances correspond to virtual images formed behind the mirror.

Key Concept

Mirror equation and sign conventions for convex spherical mirrors

Question 120Question →

Which of the following conditions must be satisfied for light to undergo total internal reflection at the boundary between two transparent media?

The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must exceed the critical angle.

The light ray must travel from an optically less dense medium into an optically denser medium, and the angle of incidence must exceed the critical angle.

The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must be less than the critical angle.

The light ray must travel from an optically less dense medium into an optically denser medium at an angle of incidence equal to

0^\circ

.

Show answer & explanation

Answer: The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must exceed the critical angle.

Answer

The light ray must travel from an optically denser medium into an optically less dense medium, and the angle of incidence must exceed the critical angle.

Total internal reflection occurs only when light passes from a medium of higher optical density into a medium of lower optical density, and the angle of incidence exceeds the critical angle for that boundary.

Step-by-Step Solution

1

Identify the boundary requirement for wave propagation direction in total internal reflection.

Light must travel from an optically denser medium (higher refractive index

n_1

) toward an optically rarer medium (lower refractive index

n_2

).

This direction allows the ray to bend away from the normal, increasing the angle of refraction relative to the angle of incidence.

2

Determine the required angle of incidence at the boundary.

The angle of incidence

i

must be strictly greater than the critical angle

\theta_c

(where

\sin \theta_c = n_2 / n_1

).

When

i > \theta_c

, no refraction can take place because

\sin r > 1

, forcing all light energy to reflect back into the initial denser medium.

Key Concept

Conditions for Total Internal Reflection

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