Waves and Optics Practice Questions — JAMB UTME — Page 2

Question 21Question →

An ultrasonic rangefinder mounted on a drone sends a sound pulse vertically downward to measure its altitude above flat ground. If the echo is detected by the sensor $0.08\text{ s}$ after emission and the speed of sound in air is $340\text{ m s}^{-1}$ , what is the altitude of the drone in meters?

Show answer & explanation

Answer: 13.6

Answer

13.6 meters

The sound pulse emitted by the drone travels down to the ground and reflects back to the sensor. The relationship between speed

v

, total round-trip time

t

, and altitude

d

is given by

2d = v \times t

. Substituting

v = 340\text{ m s}^{-1}

and

t = 0.08\text{ s}

yields

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Step-by-Step Solution

1

Identify the total time taken by the sound pulse for the round trip.

Total round-trip time

t = 0.08\text{ s}

and speed of sound

v = 340\text{ m s}^{-1}

.

Echo detection measures the time for sound to travel to a barrier and return.

2

Apply the echo distance relationship

2d = v \times t

to solve for altitude

d

.

d = \frac{340 \times 0.08}{2} = 13.6\text{ m}

.

Dividing the total path distance by 2 yields the one-way distance to the ground.

Key Concept

Calculation of distance using echoes and two-way sound wave propagation

Alternative Method

Determine the one-way travel time first:

t_{\text{one-way}} = \frac{0.08}{2} = 0.04\text{ s}

. Then calculate altitude directly using distance = speed × one-way time:

d = 340 \times 0.04 = 13.6\text{ m}

.

Estimated Time:1m 0s

Question 22Question →

A motorist travelling at a constant speed of $20\text{ m s}^{-1}$ directly towards a tall vertical cliff sounds a horn. If the motorist hears the echo of the horn $2.0\text{ s}$ later and the speed of sound in air is $340\text{ m s}^{-1}$ , what was the distance of the car from the cliff at the moment the horn was sounded?

Show answer & explanation

Answer: 360

Answer

The distance of the car from the cliff at the instant the horn was sounded was

360\text{ m}

.

When the motorist sounds the horn at an initial distance

D

from the cliff, the sound wave travels toward the cliff. In the

2.0\text{ s}

it takes for the echo to return, the car advances

40\text{ m}

toward the cliff (

20\text{ m s}^{-1} \times 2.0\text{ s}

). The returning echo meets the motorist at a distance of

(D - 40)\text{ m}

from the cliff. Consequently, the sound covers a total distance of

D + (D - 40) = 2D - 40\text{ m}

. Because the sound wave travels at

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

for

2.0\text{ s}

, the actual distance covered by sound is

340 \times 2.0 = 680\text{ m}

. Setting

2D - 40 = 680

gives

2D = 720\text{ m}

, which yields

D = 360\text{ m}

.

Step-by-Step Solution

1

Calculate the distance covered by the car while moving toward the cliff during the echo time interval.

d_{\text{car}} = 20\text{ m s}^{-1} \times 2.0\text{ s} = 40\text{ m}

.

The car continues to move closer to the cliff for the entire

2.0\text{ s}

period.

2

Set up an expression for the total distance covered by the sound wave.

d_{\text{sound}} = D + (D - 40) = 2D - 40\text{ m}

.

The sound travels forward a distance

D

to the cliff and reflects back to the car's updated location, which is

(D - 40)\text{ m}

from the cliff.

3

Calculate the distance travelled by sound using the given speed of sound.

d_{\text{sound}} = 340\text{ m s}^{-1} \times 2.0\text{ s} = 680\text{ m}

.

Sound propagates through air at

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

.

4

Equate the geometric path expression to the physical sound distance and solve for

D

.

2D - 40 = 680 \implies 2D = 720 \implies D = 360\text{ m}

.

Solving the equation yields the initial position of the car relative to the cliff.

Key Concept

Echo distance calculation with a moving observer

Question 23Question →

Match each acoustic or vibrating system operating under boundary conditions on the left with its corresponding fundamental or harmonic frequency relationship on the right (where $v$ is sound speed in air, $T$ is string tension, $\mu$ is linear mass density, $L$ is length, and $r$ is internal pipe radius).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Pipe closed at one end matches

f = \frac{v}{4L}

; Pipe open at both ends matches

f = \frac{v}{2L}

; Stretched string in second harmonic matches

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

; Pipe closed at one end with end-correction matches

f = \frac{v}{4(L + 0.6r)}

.

Each system is correctly matched based on wave mechanics boundary conditions: closed pipes produce quarter-wave fundamental modes (

\lambda = 4L

), open pipes produce half-wave fundamental modes (

\lambda = 2L

), the second harmonic of a string doubles the fundamental frequency

f_1 = \frac{1}{2L}\sqrt{T/\mu}

to yield

f_2 = \frac{1}{L}\sqrt{T/\mu}

, and end-correction increases the effective length of a closed pipe to

L + 0.6r

.

Step-by-Step Solution

1

Analyze boundary conditions for an ideal closed pipe

Displacement node at closed end, antinode at open end. Length

L = \frac{\lambda}{4} \Rightarrow \lambda = 4L

. Frequency

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

.

Determines the fundamental mode frequency formula for a closed pipe without end correction.

2

Analyze boundary conditions for an ideal open pipe

Displacement antinodes at both open ends. Length

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

. Frequency

f = \frac{v}{2L}

.

Determines the fundamental mode frequency formula for an open pipe.

3

Calculate the second harmonic frequency of a stretched string

For wave speed

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

, fundamental

f_1 = \frac{c}{2L}

. Second harmonic is

f_2 = 2f_1 = 2\left(\frac{1}{2L}\sqrt{\frac{T}{\mu}}\right) = \frac{1}{L}\sqrt{\frac{T}{\mu}}

.

Determines the frequency of the first overtone / second harmonic for a vibrating string fixed at both ends.

4

Apply end correction to a closed pipe

End correction

e = 0.6r

adds to physical length

L

at the open top end, giving

L_{eff} = L + 0.6r

. Fundamental frequency is

f = \frac{v}{4L_{eff}} = \frac{v}{4(L + 0.6r)}

.

Accounts for the antinode extending slightly beyond the open end of a real tube.

Key Concept

Boundary conditions, standing waves, harmonics in strings and air columns, and end-correction in resonance pipes

Question 24Question →

A stretched string of length $0.5\text{ m}$ fixed at both ends vibrates in its fundamental mode. If the speed of transverse waves along the string is $200\text{ m/s}$ , calculate the fundamental frequency of the string in hertz.

Show answer & explanation

Answer: 200

Answer

The fundamental frequency of the vibrating string is

200\text{ Hz}

.

For a string fixed at both ends, the fundamental mode corresponds to a standing wave with half a wavelength spanning the length of the string (

L = \frac{\lambda}{2}

, or

\lambda = 2L

). Applying the wave relation

v = f\lambda

, the fundamental frequency is

f = \frac{v}{2L}

. Substituting

v = 200\text{ m/s}

and

L = 0.5\text{ m}

gives

f = \frac{200}{2(0.5)} = 200\text{ Hz}

.

Step-by-Step Solution

1

Identify the relationship between frequency, wave speed, and string length for the fundamental mode.

For a string fixed at both ends, the wavelength of the fundamental harmonic is

\lambda = 2L

, giving the frequency formula

f = \frac{v}{2L}

.

The fundamental standing wave pattern contains nodes at both fixed ends and a single antinode at the center.

2

Substitute the given numerical values into the formula.

f = \frac{200\text{ m/s}}{2 \times 0.5\text{ m}} = \frac{200}{1} = 200\text{ Hz}

.

Dividing the wave speed by twice the length of the string yields the frequency in hertz.

Key Concept

Fundamental frequency of a vibrating string fixed at both ends

Question 25Question →

A train moving at a constant speed of $15\text{ m/s}$ towards a tall vertical cliff emits a whistle signal. The driver hears the echo of the whistle $2.0\text{ s}$ after it was sounded. If the speed of sound in air is $340\text{ m/s}$ , what was the distance between the train and the cliff at the instant the whistle was sounded?

355\text{ m}

340\text{ m}

710\text{ m}

325\text{ m}

Show answer & explanation

Answer: $355\text{ m}$

Answer

The initial distance between the train and the cliff when the whistle was sounded was

355\text{ m}

.

In

2.0\text{ s}

, the sound wave travels

680\text{ m}

while the train advances

30\text{ m}

towards the cliff. The total path of the sound consists of the initial distance

D

to the cliff plus the return distance

(D - 30)\text{ m}

to the moving train. Equating

D + (D - 30) = 680\text{ m}

gives

2D = 710\text{ m}

, which yields an initial distance of

355\text{ m}

.

Step-by-Step Solution

1

Calculate the total distance traveled by the sound wave in the given time interval.

s_{\text{sound}} = v_{\text{sound}} \times t = 340\text{ m/s} \times 2.0\text{ s} = 680\text{ m}

.

Sound travels at a constant speed in air over the total elapsed time of

2.0\text{ s}

.

2

Calculate the distance covered by the moving train during the same time interval.

s_{\text{train}} = v_{\text{train}} \times t = 15\text{ m/s} \times 2.0\text{ s} = 30\text{ m}

.

The train continues moving towards the cliff while the sound wave travels to the cliff and reflects back.

3

Formulate the geometric path equation for the sound wave.

Let

D

be the initial distance to the cliff. Sound travels

D

to the cliff and reflects back a distance of

(D - 30)\text{ m}

to reach the train. Therefore,

D + (D - 30) = 680\text{ m}

.

The echo is received by the driver at a position

30\text{ m}

closer to the cliff than where the sound was emitted.

4

Solve the linear equation for

D

.

2D - 30 = 680 \implies 2D = 710 \implies D = 355\text{ m}

.

Solving for

D

gives the exact distance at the moment the whistle was sounded.

Key Concept

Echo path geometry with a moving sound source

Question 26Question →

An organ pipe open at both ends has a length of $0.60\text{ m}$ . The fundamental frequency of resonance for this pipe is observed to be equal to the frequency of the first overtone of a stretched wire of length $0.40\text{ m}$ fixed at both ends. If the linear mass density of the wire is $4.0 \times 10^{-4}\text{ kg/m}$ and the speed of sound in air is $330\text{ m/s}$ , what is the tension in the wire?

1.21\text{ N}

4.84\text{ N}

9.68\text{ N}

19.36\text{ N}

Show answer & explanation

Answer: $4.84\text{ N}$

Answer

The tension in the wire is

4.84\text{ N}

.

The fundamental frequency of the open pipe is

f = \frac{v}{2L} = \frac{330}{2(0.60)} = 275\text{ Hz}

. Since the wire vibrates in its first overtone (second harmonic,

n=2

), its frequency is

f = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{0.40} = 275\text{ Hz}

. This gives

v_{\text{wire}} = 110\text{ m/s}

. Using

v_{\text{wire}} = \sqrt{T/\mu}

, the tension is

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4})(110)^2 = 4.84\text{ N}

.

Step-by-Step Solution

1

Calculate the fundamental frequency of the pipe open at both ends.

f_{\text{pipe}} = \frac{v_{\text{air}}}{2 L_{\text{pipe}}} = \frac{330}{2 \times 0.60} = 275\text{ Hz}

.

An open pipe supports a fundamental wavelength of

\lambda = 2L_{\text{pipe}}

.

2

Express the frequency of the first overtone of the stretched wire and equate it to the pipe frequency.

f_{\text{wire, 2}} = \frac{2 v_{\text{wire}}}{2 L_{\text{wire}}} = \frac{v_{\text{wire}}}{L_{\text{wire}}} = 275\text{ Hz}

.

For a wire fixed at both ends, the first overtone corresponds to the second harmonic (

n = 2

).

3

Solve for the speed of the transverse wave on the wire.

v_{\text{wire}} = 275 \times 0.40 = 110\text{ m/s}

.

Rearranging

f = v_{\text{wire}} / L_{\text{wire}}

gives

v_{\text{wire}} = f \cdot L_{\text{wire}}

.

4

Calculate the tension in the wire using the wave speed formula.

T = \mu v_{\text{wire}}^2 = (4.0 \times 10^{-4}) \times 110^2 = 4.84\text{ N}

.

The speed of a transverse wave on a stretched string is given by

v = \sqrt{T / \mu}

.

Key Concept

Resonance between air columns and vibrating strings under distinct boundary conditions

Estimated Time:2m 0s

Question 27Question →

A concave mirror has a radius of curvature of $40\text{ cm}$ . What is the focal length of the mirror?

20\text{ cm}

-20\text{ cm}

40\text{ cm}

80\text{ cm}

Show answer & explanation

Answer: $20\text{ cm}$

Answer

The focal length of the concave mirror is

20\text{ cm}

.

For any spherical mirror, the focal length is half of its radius of curvature (

f = \frac{r}{2}

). Since the mirror is concave, its focal length is positive, yielding

\frac{40\text{ cm}}{2} = 20\text{ cm}

.

Step-by-Step Solution

1

Identify the given physical quantity

Radius of curvature

r = 40\text{ cm}

for a concave mirror.

The radius of curvature is the distance from the pole to the center of curvature.

2

Apply the relationship between focal length and radius of curvature

f = \frac{r}{2} = \frac{40\text{ cm}}{2} = 20\text{ cm}

.

For spherical mirrors, the principal focus lies halfway between the pole and the center of curvature.

Key Concept

Focal length of spherical mirrors

Question 28Question →

An acoustic pulse generator located at a fixed point between two parallel rigid reflective barriers emits a sound wave with a frequency of $680\text{ Hz}$ . The first echo from the closer barrier is detected after $0.8\text{ s}$ , and the first echo from the farther barrier is detected after $1.4\text{ s}$ . Assuming the speed of sound in the medium is $340\text{ m s}^{-1}$ , what is the total distance between the two barriers, and how many complete wavelengths of this sound wave fit within this total distance?

374\text{ m}

and

748

wavelengths

748\text{ m}

and

1496

wavelengths

102\text{ m}

and

204

wavelengths

374\text{ m}

and

187

wavelengths

Show answer & explanation

Answer: $374\text{ m}$ and $748$ wavelengths

Answer

The total distance between the barriers is

374\text{ m}

and

748

complete wavelengths fit within this distance.

The distance to the first barrier is

d_1 = \frac{340 \times 0.8}{2} = 136\text{ m}

, and to the second barrier is

d_2 = \frac{340 \times 1.4}{2} = 238\text{ m}

. Adding both distances gives a total separation of

374\text{ m}

. With a wavelength

\lambda = \frac{v}{f} = \frac{340}{680} = 0.5\text{ m}

, the number of full wavelengths fitting in

374\text{ m}

is

\frac{374}{0.5} = 748

.

Step-by-Step Solution

1

Calculate the one-way distance from the generator to each reflective barrier using the echo relationship

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

.

For barrier 1:

d_1 = \frac{340 \text{ m s}^{-1} \times 0.8 \text{ s}}{2} = 136 \text{ m}

. For barrier 2:

d_2 = \frac{340 \text{ m s}^{-1} \times 1.4 \text{ s}}{2} = 238 \text{ m}

.

An echo represents a two-way journey (to the surface and back), so the time taken to travel the one-way distance is half of the echo reception time.

2

Determine the total distance between the two parallel barriers.

D = d_1 + d_2 = 136 \text{ m} + 238 \text{ m} = 374 \text{ m}

.

Since the generator is situated between the two barriers, the total separation distance equals the sum of the individual distances to each barrier.

3

Calculate the wavelength

\lambda

of the sound wave using the wave equation

v = f \lambda

.

\lambda = \frac{v}{f} = \frac{340 \text{ m s}^{-1}}{680 \text{ Hz}} = 0.5 \text{ m}$.

Wavelength is the ratio of wave speed to frequency.

4

Compute the number of complete wavelengths

N

contained within the total separation distance.

N = \frac{D}{\lambda} = \frac{374 \text{ m}}{0.5 \text{ m}} = 748.

Dividing total distance by single wavelength yields the number of full wave cycles occupying that span.

Key Concept

Echo distance calculations and wave speed-frequency-wavelength relationships

Estimated Time:2m 0s

Question 29Question →

A water tank has a real depth of $16.0\text{ cm}$ . If the refractive index of water relative to air is $\frac{4}{3}$ , what is the apparent depth of the tank in centimetres when viewed normally from above?

Show answer & explanation

Answer: 12

Answer

The apparent depth of the water tank is

12.0\text{ cm}

.

When light passes from water to air, refraction causes the apparent depth to be smaller than the real depth by a factor equal to the refractive index

n

. Substituting a real depth of

16.0\text{ cm}

and

n = \frac{4}{3}

into

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

yields

12.0\text{ cm}

.

Step-by-Step Solution

1

Identify the relationship between refractive index, real depth, and apparent depth.

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

Refraction at a plane boundary causes an object immersed in a denser medium to appear closer to the surface when viewed from a rarer medium.

2

Rearrange the equation to isolate Apparent Depth.

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

The unknown quantity requested by the question is the apparent depth.

3

Substitute the known numerical values into the equation.

\text{Apparent Depth} = \frac{16.0}{\frac{4}{3}} = 16.0 \times \frac{3}{4} = 12.0\text{ cm}

Dividing

16.0

by

\frac{4}{3}

gives

12.0

.

Key Concept

Real and Apparent Depth in Light Refraction

Question 30Question →

A worker strikes one end of a long solid aluminum pipeline. A detector at the opposite end records two sound signals—one traveling through the aluminum pipeline and the other through the surrounding air—separated by a time interval of $2.8\text{ s}$ . If the speed of sound in air is $340\text{ m s}^{-1}$ and the speed of sound in aluminum is $5100\text{ m s}^{-1}$ , what is the length of the pipeline in meters?

Show answer & explanation

Answer: 1020

Answer

The length of the pipeline is

1020\text{ m}

.

Sound travels significantly faster through solids like aluminum (

5100\text{ m s}^{-1}

) than through gases like air (

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

). The time taken for sound to travel a distance

L

through air is

t_{\text{air}} = \frac{L}{340}

, while through aluminum it is

t_{\text{metal}} = \frac{L}{5100}

. Setting their difference equal to

2.8\text{ s}

gives

\frac{L}{340} - \frac{L}{5100} = 2.8

, which solves to

L = 1020\text{ m}

.

Step-by-Step Solution

1

Formulate transit time expressions for both media

t_{\text{air}} = \frac{L}{340}

and

t_{\text{metal}} = \frac{L}{5100}

Time taken by a wave to travel distance

L

at constant speed

v

is

t = \frac{L}{v}

.

2

Set up the time difference equation

t_{\text{air}} - t_{\text{metal}} = 2.8\text{ s}

The sound wave travels faster through aluminum than air, so the air pulse arrives later by

2.8\text{ s}

.

3

Solve the algebraic equation for distance

L

L = 1020\text{ m}

Combining terms yields

\frac{14L}{5100} = 2.8

, which simplifies to

L = \frac{2.8 \times 5100}{14} = 1020\text{ m}

.

Key Concept

Propagation speed of sound waves in different physical media

Question 31Question →

A compound microscope in normal adjustment consists of an objective lens with a focal length of $1.5\text{ cm}$ and an eyepiece with a focal length of $5.0\text{ cm}$ . An object is placed at a distance of $1.6\text{ cm}$ in front of the objective lens. Calculate the distance, in centimeters, between the objective lens and the eyepiece.

Show answer & explanation

Answer: 29

Answer

The distance between the objective lens and the eyepiece is 29.0 cm.

Applying the thin lens formula to the objective lens yields

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o}

, giving an image distance

v_o = 24.0\text{ cm}

. Under normal adjustment, the intermediate image falls on the focal point of the eyepiece, making

u_e = f_e = 5.0\text{ cm}

. The total separation between the two lenses is

L = v_o + u_e = 24.0 + 5.0 = 29.0\text{ cm}

.

Step-by-Step Solution

1

Apply the thin lens formula to the objective lens to find the intermediate image position

v_o

.

\frac{1}{1.5} = \frac{1}{1.6} + \frac{1}{v_o} \Rightarrow \frac{1}{v_o} = \frac{2}{3} - \frac{5}{8} = \frac{1}{24}\text{ cm}^{-1} \Rightarrow v_o = 24.0\text{ cm}

The objective lens forms a real, inverted, magnified image at distance

v_o

from the objective.

2

Identify the object distance for the eyepiece

u_e

under normal adjustment.

u_e = f_e = 5.0\text{ cm}

For normal adjustment of a optical instrument, the final image is formed at infinity, requiring the intermediate image to sit exactly at the principal focus of the eyepiece.

3

Sum the intermediate image distance and eyepiece object distance to obtain total lens separation

L

.

L = v_o + u_e = 24.0\text{ cm} + 5.0\text{ cm} = 29.0\text{ cm}

The separation of lenses in a compound microscope is the distance from the objective to the intermediate image plus the distance from the intermediate image to the eyepiece.

Key Concept

Compound microscope optics and lens separation in normal adjustment

Question 32Question →

A convex security mirror installed in a store has a focal length of magnitude $20\text{ cm}$ . If an upright image of a customer is formed with a linear magnification of $0.25$ , at what distance from the mirror is the customer standing?

60\text{ cm}

100\text{ cm}

15\text{ cm}

25\text{ cm}

Show answer & explanation

Answer: $60\text{ cm}$

Answer

The customer is standing at a distance of

60\text{ cm}

from the convex mirror.

By sign convention, a convex mirror has a negative focal length (

f = -20\text{ cm}

). An upright image formed by a mirror has a positive magnification

m = +0.25

. Using

m = -v/u

, we get

v = -u/4

. Substituting these into the mirror equation

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

gives

-\frac{1}{20} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

, which yields

u = 60\text{ cm}

.

Step-by-Step Solution

1

Identify the optical properties and apply sign conventions for a convex mirror

Focal length

f = -20\text{ cm}

(convex mirror focal length is virtual/negative). Magnification

m = +0.25

(upright image).

Convex mirrors always form virtual, upright, diminished images behind the mirror.

2

Relate image distance

v

to object distance

u

using the linear magnification formula

Since

m = -v/u = 1/4

, we obtain

v = -u/4

.

The negative sign in the magnification definition accounts for virtual image distance.

3

Substitute

f = -20\text{ cm}

and

v = -u/4

into the mirror formula

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

\frac{1}{-20} = \frac{1}{u} + \frac{1}{-u/4} \implies -\frac{1}{20} = \frac{1}{u} - \frac{4}{u} = -\frac{3}{u}

.

Combines fractions with a common denominator

u

to solve for the unknown object distance.

4

Solve for the object distance

u

\frac{1}{20} = \frac{3}{u} \implies u = 60\text{ cm}

.

Cross-multiplying yields the real object distance in front of the mirror.

Key Concept

Reflection at Convex Mirrors and Optical Sign Conventions

Question 33Question →

A uniform wire of length $0.60\text{ m}$ and linear mass density $4.0 \times 10^{-3}\text{ kg/m}$ is fixed at both ends under a tension of $360\text{ N}$ . When plucked, the wire vibrates in its second overtone. This frequency is found to be in resonance with the first overtone of an air column in a pipe closed at one end. Taking the speed of sound in air as $340\text{ m/s}$ , what is the length of the pipe?

0.17\text{ m}

0.34\text{ m}

0.51\text{ m}

0.68\text{ m}

Show answer & explanation

Answer: $0.34\text{ m}$

Answer

The length of the pipe is

0.34\text{ m}

.

The correct answer of

0.34\text{ m}

is obtained by finding the wave speed on the string (

300\text{ m/s}

), computing its 3rd harmonic frequency (

750\text{ Hz}

), and equating this to the 3rd harmonic frequency formula for a closed pipe (

f = \frac{3 v_a}{4 L_p}

).

Step-by-Step Solution

1

Calculate the speed of transverse waves on the stretched string.

v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{360\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{90000} = 300\text{ m/s}

Wave speed on a stretched string depends on tension and linear mass density.

2

Determine the fundamental frequency and the 2nd overtone frequency of the string.

Fundamental frequency

f_{1,s} = \frac{v_s}{2 L_s} = \frac{300}{2(0.60)} = 250\text{ Hz}

. Second overtone is the 3rd harmonic (

n=3

), so

f = 3 \times 250 = 750\text{ Hz}

.

For a string fixed at both ends, harmonics are integer multiples of the fundamental, and the 2nd overtone corresponds to

n=3

.

3

Relate the resonance frequency to the length of the closed pipe.

First overtone of a closed pipe is its 3rd harmonic (

m=3

):

f = \frac{3 v_a}{4 L_p} \implies 750 = \frac{3(340)}{4 L_p}

.

Pipes closed at one end only produce odd harmonics (

1, 3, 5, \dots

), so the 1st overtone is

m=3

.

4

Solve for the pipe length

L_p

.

L_p = \frac{3(340)}{4(750)} = \frac{1020}{3000} = 0.34\text{ m}

Algebraic rearrangement yields the required physical length.

Key Concept

Harmonics and overtones in vibrating strings and closed air columns under resonance

Question 34Question →

Match each vibrating system mode on the left with the correct relationship between its standing wavelength ( $\lambda$ ) and length ( $L$ ) on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The mode descriptions match their standing wavelength expressions as follows: Pipe closed at one end in its first overtone matches

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

; Pipe open at both ends in its first overtone matches

\lambda = L

; Stretched string in its second overtone matches

\lambda = \frac{2L}{3}

; Pipe closed at one end in its fundamental mode matches

\lambda = 4L

.

Each pair correctly links the specified boundary condition and mode of vibration to its mathematical relationship between wavelength

\lambda

and physical length

L

.

Step-by-Step Solution

1

Identify boundary conditions and available harmonics for each vibrating system.

Closed pipes support odd harmonics only (

n = 1, 3, 5, \dots

) with

L = \frac{n\lambda}{4}

. Open pipes and fixed strings support all integer harmonics (

n = 1, 2, 3, \dots

) with

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

.

Boundary conditions constrain node and antinode positions, determining allowed harmonic modes.

2

Determine the specific harmonic number

n

corresponding to each specified overtone.

First overtone of closed pipe

\rightarrow n = 3

; First overtone of open pipe

\rightarrow n = 2

; Second overtone of fixed string

\rightarrow n = 3

; Fundamental of closed pipe

\rightarrow n = 1

.

Overtones are higher resonant modes above the fundamental frequency.

3

Solve for wavelength

\lambda

in terms of system length

L

for each item.

For

n = 3

(closed pipe):

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

. For

n = 2

(open pipe):

L = \lambda \Rightarrow \lambda = L

. For

n = 3

(fixed string):

L = \frac{3\lambda}{2} \Rightarrow \lambda = \frac{2L}{3}

. For

n = 1

(closed pipe):

L = \frac{\lambda}{4} \Rightarrow \lambda = 4L

.

Rearranging each expression establishes the correct matching pair.

Key Concept

Boundary conditions and harmonic wavelength relations in pipes and vibrating strings

Question 35Question →

A light ray strikes the first face of a glass prism of refracting angle $30^\circ$ at normal incidence ( $i = 0^\circ$ ). If the refractive index of the glass is $1.50$ , calculate the angle of emergence, in degrees, as the ray leaves the second face into air. (Take $\arcsin(0.75) = 48.6^\circ$ )

Show answer & explanation

Answer: 48.6

Answer

The angle of emergence of the light ray as it exits the prism is

48.6^\circ

.

Because the ray is incident normally at the first surface, it continues undeviated into the glass (

r_1 = 0^\circ

). By prism geometry, the angle of incidence at the second face is equal to the apex angle of the prism (

r_2 = A = 30^\circ

). Applying Snell's Law at the glass-air boundary gives

1.50 \sin(30^\circ) = 1.00 \sin(e)

, leading to

\sin(e) = 0.75

, which evaluates to an emergent angle of

48.6^\circ

.

Step-by-Step Solution

1

Determine the angle of refraction at the first surface

r_1 = 0^\circ

Light entering a surface normally (

i_1 = 0^\circ

) passes straight through without bending.

2

Find the angle of incidence at the second surface inside the prism using prism geometry

r_2 = 30^\circ

For any triangular prism, the refracting angle

A = r_1 + r_2

. Since

r_1 = 0^\circ

,

r_2 = A = 30^\circ

.

3

Apply Snell's Law at the second interface (glass to air)

\sin(e) = 0.75

n_{\text{glass}} \sin(r_2) = n_{\text{air}} \sin(e) \implies 1.50 \times \sin(30^\circ) = 1.00 \times \sin(e)

.

4

Calculate the emergent angle

e

e = 48.6^\circ

Taking the inverse sine of

0.75

yields

e = \arcsin(0.75) = 48.6^\circ

.

Key Concept

Prism Geometry and Snell's Law Refraction

Question 36Question →

A converging lens has a focal length of $25\text{ cm}$ . What is the power of the lens in dioptres?

Show answer & explanation

Answer: 4

Answer

The power of the lens is

4\text{ D}

.

The power

P

of a lens in dioptres is defined as the reciprocal of its focal length

f

expressed in meters (

P = \frac{1}{f}

). Expressing

25\text{ cm}

in meters gives

0.25\text{ m}

. Substituting this value yields

P = \frac{1}{0.25\text{ m}} = 4\text{ D}

.

Step-by-Step Solution

1

Convert the focal length from centimeters to meters

f = 25\text{ cm} = 0.25\text{ m}

The unit of dioptre (

D

) is defined as reciprocal meters (

m^{-1}

), so focal length must be in meters.

2

Apply the lens power formula

P = \frac{1}{f}

and compute the value

P = \frac{1}{0.25} = 4\text{ D}

The power of a converging lens is positive and equal to the reciprocal of its focal length in meters.

Key Concept

Power of a Lens

Question 37Question →

A radio transmitter emits electromagnetic waves with a frequency of $1.0 \times 10^8\text{ Hz}$ . Given that the speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m s}^{-1}$ , what is the wavelength of the emitted wave?

3.0\text{ m}

3.0 \times 10^{16}\text{ m}

0.33\text{ m}

3.0 \times 10^{-1}\text{ m}

Show answer & explanation

Answer: $3.0\text{ m}$

Answer

The wavelength of the emitted radio wave is

3.0\text{ m}

.

According to the wave equation

c = f \lambda

, the wavelength is found by dividing the speed of light by frequency:

\lambda = \frac{3.0 \times 10^8\text{ m s}^{-1}}{1.0 \times 10^8\text{ Hz}} = 3.0\text{ m}

.

Step-by-Step Solution

1

Identify the given parameters and fundamental formula

Speed of light

c = 3.0 \times 10^8\text{ m s}^{-1}

and frequency

f = 1.0 \times 10^8\text{ Hz}

. The electromagnetic wave relation is

c = f \lambda

.

All electromagnetic waves travel at the speed of light in a vacuum.

2

Rearrange the formula to solve for wavelength

\lambda = \frac{c}{f}

Dividing both sides of the wave equation by frequency isolates wavelength.

3

Substitute values into the equation and compute the result

\lambda = \frac{3.0 \times 10^8}{1.0 \times 10^8} = 3.0\text{ m}

Performing the division yields the correct wavelength.

Key Concept

Relationship between wave speed, frequency, and wavelength for electromagnetic radiation in a vacuum.

Question 38Question →

A ray of light traveling inside a glass prism of refractive index $1.60$ strikes the boundary with an adjacent oil medium of refractive index $1.20$ . What is the sine of the critical angle for total internal reflection at this glass-oil interface?

0.75

1.33

0.625

0.80

Show answer & explanation

Answer: $0.75$

Answer

The sine of the critical angle at the glass-oil interface is

0.75

.

For light traveling from a medium of refractive index

n_1

into a medium of refractive index

n_2

(where

n_1 > n_2

), the critical angle

C

satisfies

\sin C = \frac{n_2}{n_1}

. Substituting

n_1 = 1.60

and

n_2 = 1.20

gives

\sin C = \frac{1.20}{1.60} = 0.75

.

Step-by-Step Solution

1

Identify the refractive indices of the two media.

Denser medium (glass):

n_1 = 1.60

; less dense medium (oil):

n_2 = 1.20

.

Total internal reflection occurs when light originates in the optically denser medium and strikes the boundary with a less dense medium.

2

Apply Snell's law at the critical angle

C

.

n_1 \sin C = n_2 \sin 90^\circ \implies \sin C = \frac{n_2}{n_1}

.

At the critical angle of incidence, the angle of refraction in the second medium is

90^\circ

, so

\sin 90^\circ = 1

.

3

Substitute the given numerical values to compute

\sin C

.

\sin C = \frac{1.20}{1.60} = \frac{3}{4} = 0.75

.

Dividing the refractive index of the oil by that of the glass gives the exact sine of the critical angle.

Key Concept

Critical Angle between Two Media

Question 39Question →

A pipe of length $0.80\text{ m}$ is closed at one end and open at the other. If the speed of sound in air is $320\text{ m/s}$ , what is the frequency of its first overtone?

100\text{ Hz}

200\text{ Hz}

300\text{ Hz}

400\text{ Hz}

Show answer & explanation

Answer: $300\text{ Hz}$

Answer

The frequency of the first overtone is

300\text{ Hz}

.

For an air column closed at one end, standing wave resonance occurs only at odd harmonic frequencies given by

f_n = \frac{n v}{4L}

for

n = 1, 3, 5, \dots

. The fundamental frequency (

n = 1

) is

f_1 = \frac{320}{4 \times 0.80} = 100\text{ Hz}

. The first overtone is the very next resonant mode, which corresponds to the third harmonic (

n = 3

), giving

f_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}

.

Step-by-Step Solution

1

Calculate the fundamental frequency of the closed pipe

f_1 = \frac{v}{4L} = \frac{320}{4 \times 0.80} = 100\text{ Hz}

For a pipe closed at one end, the fundamental wavelength is

\lambda_1 = 4L

.

2

Determine the harmonic number for the first overtone

First overtone = 3rd harmonic (

f_3 = 3 f_1

)

A pipe closed at one end produces only odd harmonics (

f_n = n f_1

where

n = 1, 3, 5, \dots

).

3

Compute the first overtone frequency

f_3 = 3 \times 100\text{ Hz} = 300\text{ Hz}

Multiplying the fundamental frequency by 3 yields the first overtone frequency.

Key Concept

Resonance and Harmonics in Closed Air Columns

Question 40Question →

A hypermetropic eye has its unassisted near point situated at $75\text{ cm}$ from the eye. Calculate the focal length, in $\text{cm}$ , of the converging spectacle lens needed to enable the person to read a book comfortably at a distance of $25\text{ cm}$ from the eye.

Show answer & explanation

Answer: 37.5

Answer

The focal length of the required converging lens is

37.5\text{ cm}

.

To correct hypermetropia, the spectacle lens must form a virtual image of an object located at the desired near point (

u = +25\text{ cm}

) at the eye's actual, unassisted near point (

v = -75\text{ cm}

). Substituting

u = +25\text{ cm}

and

v = -75\text{ cm}

into the thin lens formula

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

gives

\frac{1}{f} = \frac{1}{25} - \frac{1}{75} = \frac{2}{75}\text{ cm}^{-1}

. Taking the reciprocal yields

f = 37.5\text{ cm}

.

Step-by-Step Solution

1

Determine the object distance and image distance with sign conventions

u = +25\text{ cm}

and

v = -75\text{ cm}

.

The object is placed at the normal reading distance (

25\text{ cm}

), and the spectacle lens creates a virtual image on the same side of the lens at the person's near point (

75\text{ cm}

).

2

Set up the thin lens formula

\frac{1}{f} = \frac{1}{25} - \frac{1}{75}.

The thin lens formula relates the focal length to object and image distances.

3

Calculate the focal length

f

f = 37.5\text{ cm}.

Simplifying

\frac{3 - 1}{75} = \frac{2}{75}

and taking the reciprocal yields

f = \frac{75}{2} = 37.5\text{ cm}

.

Key Concept

Correction of hypermetropia (farsightedness) using thin lens formula with virtual image sign convention

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