Waves and Optics Practice Questions — JAMB UTME

Question 1Question →

A progressive transverse mechanical wave travels along a stretched string with a wavelength of $\lambda = 0.80\text{ m}$ . If the maximum speed of an oscillating particle on the string is equal to one-quarter ( $\frac{1}{4}$ ) of the propagation speed of the wave, what is the amplitude of the wave, and how are the particle vibration and energy propagation directions oriented relative to each other?

The amplitude is

\frac{1}{10\pi}\text{ m}

, and particles vibrate perpendicular to the direction of energy propagation.

The amplitude is

\frac{1}{5\pi}\text{ m}

, and particles vibrate perpendicular to the direction of energy propagation.

The amplitude is

0.20\text{ m}

, and particles vibrate parallel to the direction of energy propagation.

The amplitude is

\frac{1}{10\pi}\text{ m}

, and particles vibrate parallel to the direction of energy propagation.

Show answer & explanation

Answer: The amplitude is $\frac{1}{10\pi}\text{ m}$ , and particles vibrate perpendicular to the direction of energy propagation.

Answer

The amplitude of the wave is

\frac{1}{10\pi}\text{ m}

, and particles vibrate perpendicular to the direction of energy propagation.

The maximum speed of a particle in simple harmonic wave motion is

v_{p,\text{max}} = \omega A = 2\pi f A

. The wave propagation speed is

v = f \lambda

. Given

v_{p,\text{max}} = \frac{1}{4}v

, we set

2\pi f A = \frac{1}{4} f \lambda

, which yields

A = \frac{\lambda}{8\pi} = \frac{0.80}{8\pi} = \frac{1}{10\pi}\text{ m}

. Because the wave is transverse, particle vibrations occur perpendicular to the direction of wave energy propagation.

Step-by-Step Solution

1

Relate maximum particle speed to wave parameters.

Maximum transverse particle speed

v_{p,\text{max}} = \omega A = 2\pi f A

.

Particles in simple harmonic wave motion have maximum speed given by the product of angular frequency

\omega

and amplitude

A

.

2

Express wave propagation speed in terms of frequency and wavelength.

Wave speed

v = f \lambda

.

The fundamental wave equation relates wave speed

v

directly to frequency

f

and wavelength

\lambda

.

3

Set up the given proportion and solve for amplitude

A

.

2\pi f A = \frac{1}{4} (f \lambda) \implies 2\pi A = \frac{\lambda}{4} \implies A = \frac{\lambda}{8\pi} = \frac{0.80}{8\pi} = \frac{1}{10\pi}\text{ m}

.

Canceling frequency

f

from both sides allows direct evaluation of

A

.

4

Classify the direction of particle motion relative to energy propagation for a transverse wave.

Particles vibrate perpendicular to the direction of wave travel.

By definition, transverse mechanical waves involve oscillations perpendicular to the direction of wave energy propagation.

Key Concept

Relationship between particle velocity and wave velocity in transverse mechanical waves

Question 2Question →

A tube closed at one end has a length of $0.85\text{ m}$ . If the speed of sound in air is $340\text{ m/s}$ , what is the fundamental frequency of the sound wave produced in the tube?

100\text{ Hz}

200\text{ Hz}

400\text{ Hz}

289\text{ Hz}

Show answer & explanation

Answer: $100\text{ Hz}$

Answer

100\text{ Hz}

For a pipe closed at one end, the fundamental frequency is given by

f = \frac{v}{4L}

. Substituting

v = 340\text{ m/s}

and

L = 0.85\text{ m}

yields

f = \frac{340}{4 \times 0.85} = \frac{340}{3.4} = 100\text{ Hz}

.

Step-by-Step Solution

1

Determine the relationship between pipe length and wavelength for the fundamental mode of a closed pipe.

For a pipe closed at one end,

\lambda = 4L = 4 \times 0.85\text{ m} = 3.4\text{ m}

.

A closed tube forms a node at the closed end and an antinode at the open end, corresponding to one quarter of a full wavelength.

2

Calculate fundamental frequency using the wave equation

v = f\lambda

.

f = \frac{v}{\lambda} = \frac{340\text{ m/s}}{3.4\text{ m}} = 100\text{ Hz}

.

Frequency equals wave velocity divided by fundamental wavelength.

Key Concept

Fundamental frequency of a pipe closed at one end

Question 3Question →

In a resonance tube experiment using a tuning fork of constant frequency, the first two consecutive resonant lengths of the air column above the water level are measured to be $23.5\text{ cm}$ and $73.5\text{ cm}$ respectively. What is the end correction of the tube in centimeters?

Show answer & explanation

Answer: 1.5

Answer

The end correction of the tube is

1.5\text{ cm}

.

In a resonance tube closed at one end by water, consecutive resonances occur when the air column length increases by half a wavelength. Subtracting the first resonant length from the second gives

\frac{\lambda}{2} = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}

, which yields

\lambda = 100.0\text{ cm}

and

\frac{\lambda}{4} = 25.0\text{ cm}

. The first resonance condition accounts for end correction through

L_1 + e = \frac{\lambda}{4}

. Substituting

L_1 = 23.5\text{ cm}

gives

e = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}

.

Step-by-Step Solution

1

Determine the wavelength using consecutive resonant positions

$\frac{\lambda}{2} = L_2 - L_1 = 73.5\text{ cm} - 23.5\text{ cm} = 50.0\text{ cm}$, so $\lambda = 100.0\text{ cm}$

For a column closed at one end, consecutive resonances occur at intervals of half a wavelength.

2

Calculate the quarter-wavelength value

$\frac{\lambda}{4} = \frac{100.0\text{ cm}}{4} = 25.0\text{ cm}$

The fundamental mode position of the displacement antinode corresponds to a distance of one quarter-wavelength from the closed end.

3

Calculate the end correction

$e = \frac{\lambda}{4} - L_1 = 25.0\text{ cm} - 23.5\text{ cm} = 1.5\text{ cm}$

The effective length for the first resonance includes the physical length plus the end correction.

Key Concept

End Correction in Resonance Air Columns

Question 4Question →

Which of the following conditions is necessary for total internal reflection to occur when light travels between two media?

Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be greater than the critical angle.

Light must travel from an optically less dense medium to an optically denser medium, and the angle of incidence must be greater than the critical angle.

Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be less than the critical angle.

Light must travel from an optically less dense medium to an optically denser medium, and the angle of incidence must equal zero degrees.

Show answer & explanation

Answer: Light must travel from an optically denser medium to an optically less dense medium, and the angle of incidence must be greater than the critical angle.

Answer

Total internal reflection requires light to travel from an optically denser medium to an optically less dense medium with an angle of incidence greater than the critical angle.

For total internal reflection to happen, two strict conditions must be satisfied: (1) light must travel from a medium with a higher refractive index to one with a lower refractive index, and (2) the angle of incidence must be strictly greater than the critical angle for that boundary.

Step-by-Step Solution

1

Identify the optical density requirement

Light must attempt to pass from a medium of higher refractive index (denser) into a medium of lower refractive index (less dense) so that the refracted ray bends away from the normal.

Bending away from the normal allows the angle of refraction to reach

90^\circ

at the critical angle.

2

Identify the angle of incidence requirement

The angle of incidence in the denser medium must exceed the critical angle (

i > C

).

When the incidence angle exceeds the critical angle, no refraction is possible and all energy is reflected back into the denser medium.

Key Concept

Conditions for Total Internal Reflection

Question 5Question →

A wave generator in a ripple tank creates periodic surface waves. An observer counts 45 complete wave crests passing a fixed marker in $15\text{ seconds}$ . If the distance measured between the first crest and the seventh crest is $1.2\text{ m}$ , what is the speed of propagation of the wave in $\text{m/s}$ ?

Show answer & explanation

Answer: 0.6

Answer

The speed of propagation of the wave is

0.6\text{ m/s}

.

The frequency is calculated as

f = \frac{45\text{ crests}}{15\text{ s}} = 3\text{ Hz}

. The distance between the 1st and 7th crest spans 6 complete wavelengths (

6\lambda = 1.2\text{ m}

), giving

\lambda = 0.2\text{ m}

. Applying the wave speed formula

v = f\lambda

yields

v = 3 \times 0.2 = 0.6\text{ m/s}

.

Step-by-Step Solution

1

Calculate the frequency of the wave

Frequency

f = 3\text{ Hz}

Frequency is the rate at which complete wave cycles pass a fixed point, calculated as

f = \frac{45}{15\text{ s}} = 3\text{ Hz}

.

2

Calculate the wavelength from crest spacing

Wavelength

\lambda = 0.2\text{ m}

The distance between

n

successive crests contains

(n-1)

full wavelength intervals. Thus, between the 1st and 7th crest there are

7 - 1 = 6

wavelengths, so

6\lambda = 1.2\text{ m} \implies \lambda = 0.2\text{ m}

.

3

Calculate wave speed using the fundamental wave equation

Speed

v = 0.6\text{ m/s}

Wave propagation speed is the product of frequency and wavelength:

v = f\lambda = 3\text{ Hz} \times 0.2\text{ m} = 0.6\text{ m/s}

.

Key Concept

Wave speed calculation using frequency and wavelength derived from crest counting (

v = f\lambda

).

Question 6Question →

A longitudinal mechanical wave travels through Medium X with a propagation speed of $340\text{ m/s}$ . The distance between two consecutive compressions in Medium X is $0.68\text{ m}$ . When the wave propagates across a boundary into Medium Y, the distance between a compression and the immediately adjacent rarefaction is measured as $1.70\text{ m}$ . What is the speed of propagation of the wave in Medium Y?

1700\text{ m/s}

850\text{ m/s}

340\text{ m/s}

680\text{ m/s}

Show answer & explanation

Answer: $1700\text{ m/s}$

Answer

The speed of propagation of the wave in Medium Y is

1700\text{ m/s}

.

In Medium X, the distance between consecutive compressions gives a full wavelength of

0.68\text{ m}

, yielding a wave frequency of

f = \frac{340\text{ m/s}}{0.68\text{ m}} = 500\text{ Hz}

. Because frequency depends solely on the source, it remains

500\text{ Hz}

in Medium Y. In Medium Y, the distance between a compression and the adjacent rarefaction is half a wavelength, making

\lambda_Y = 2 \times 1.70\text{ m} = 3.40\text{ m}

. Multiplying frequency by the new wavelength gives the speed in Medium Y as

v_Y = 500\text{ Hz} \times 3.40\text{ m} = 1700\text{ m/s}

.

Step-by-Step Solution

1

Determine the frequency of the wave using parameters from Medium X.

In longitudinal waves, the distance between consecutive compressions equals one wavelength, so

\lambda_X = 0.68\text{ m}

. Using

v = f \cdot \lambda

, frequency

f = \frac{340\text{ m/s}}{0.68\text{ m}} = 500\text{ Hz}

.

Frequency is determined by the wave source and remains unchanged when a wave transitions between different media.

2

Calculate the wavelength of the wave in Medium Y.

The distance from a compression to the immediately adjacent rarefaction is half of a wavelength (

\frac{\lambda_Y}{2} = 1.70\text{ m}

). Therefore,

\lambda_Y = 2 \times 1.70\text{ m} = 3.40\text{ m}

.

A full wavelength spans from compression to compression or rarefaction to rarefaction.

3

Calculate the wave propagation speed in Medium Y.

v_Y = f \cdot \lambda_Y = 500\text{ Hz} \times 3.40\text{ m} = 1700\text{ m/s}

.

Applying the wave speed formula with constant frequency.

Key Concept

Wave Propagation across Boundaries and Longitudinal Wave Characteristics

Estimated Time:2m 0s

Question 7Question →

A progressive sinusoidal wave traveling in a primary medium is governed by the equation $y = 0.08 \sin(120\pi t - 4\pi x)$ , where $x$ and $y$ are measured in meters and $t$ in seconds. When the wave passes into a second medium, its propagation speed changes to $15\text{ m/s}$ . What is the wavelength of the wave in the second medium?

0.50\text{ m}

0.25\text{ m}

0.125\text{ m}

4.00\text{ m}

Show answer & explanation

Answer: $0.25\text{ m}$

Answer

0.25\text{ m}

Comparing the given equation

y = 0.08 \sin(120\pi t - 4\pi x)

to the standard wave equation

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

gives an angular frequency

\omega = 120\pi\text{ rad/s}

. The wave frequency is therefore

f = \frac{120\pi}{2\pi} = 60\text{ Hz}

. Since the frequency of a wave is determined solely by its source, it remains invariant when transmitting into a new medium. Given the new speed

v_2 = 15\text{ m/s}

, the new wavelength is

\lambda_2 = \frac{v_2}{f} = \frac{15}{60} = 0.25\text{ m}

.

Step-by-Step Solution

1

Extract angular frequency (

\omega

) from the wave equation

Standard form

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

gives

\omega = 120\pi\text{ rad/s}

.

The coefficient of

t

in the argument of the sine function represents angular frequency.

2

Calculate the frequency of the wave in the primary medium

f = \frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60\text{ Hz}

.

Frequency

f

is related to angular frequency by

\omega = 2\pi f

.

3

Apply the principle of frequency invariance across boundary media

Frequency in the second medium

f_2 = 60\text{ Hz}

.

When a wave passes from one medium to another, its frequency depends only on the source and remains constant across boundaries.

4

Compute the wavelength in the second medium using the wave equation

v = f\lambda

\lambda_2 = \frac{v_2}{f_2} = \frac{15\text{ m/s}}{60\text{ Hz}} = 0.25\text{ m}

.

Wavelength is inversely proportional to frequency for a given wave speed in that medium.

Key Concept

Wave Equation Parameter Extraction and Frequency Invariance across Media Boundaries

Question 8Question →

Monochromatic light of wavelength $500\text{ nm}$ is incident normally on a plane diffraction grating. If the second-order principal maximum is observed at an angle of $30^\circ$ to the normal, calculate the number of lines per millimeter ruled on the grating.

Show answer & explanation

Answer: 500

Answer

500 lines/mm

Using the diffraction grating equation

d \sin\theta = n\lambda

with

n = 2

,

\lambda = 5.0 \times 10^{-7}\text{ m}

, and

\sin(30^\circ) = 0.5

gives a slit separation of

d = 2.0 \times 10^{-6}\text{ m}

. Converting to grating ruling density per millimeter yields

\frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}

.

Step-by-Step Solution

1

Identify the given physical parameters and convert wavelength to standard SI meters.

Order

n = 2

, diffraction angle

\theta = 30^\circ

, wavelength

\lambda = 500 \times 10^{-9}\text{ m} = 5.0 \times 10^{-7}\text{ m}

.

Standard SI unit conversion is required to perform wave equation calculations accurately.

2

Apply the diffraction grating equation

d \sin\theta = n\lambda

to compute the grating spacing

d

.

d = \frac{2 \times 5.0 \times 10^{-7}\text{ m}}{\sin 30^\circ} = \frac{1.0 \times 10^{-6}}{0.5} = 2.0 \times 10^{-6}\text{ m}

.

The condition for principal constructive interference maxima is

d \sin\theta = n\lambda

.

3

Calculate the number of lines per millimeter by dividing

1\text{ mm}

(

10^{-3}\text{ m}

) by the grating spacing

d

.

N_{\text{mm}} = \frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}

.

Grating density (lines per unit length) is the reciprocal of the slit separation

d

.

Key Concept

Diffraction Grating Principal Maxima Condition

Estimated Time:2m 0s

Question 9Question →

The speed of light in medium $A$ is $1.5 \times 10^8\text{ m/s}$ and in medium $B$ is $3.0 \times 10^8\text{ m/s}$ . What is the critical angle for a light ray traveling from medium $A$ towards medium $B$ ?

30^\circ

45^\circ

60^\circ

90^\circ

Show answer & explanation

Answer: $30^\circ$

Answer

The critical angle for light traveling from medium

A

to medium

B

is

30^\circ

.

The critical angle

C

is the angle of incidence in the denser medium for which the angle of refraction in the rarer medium is

90^\circ

. Using the wave speed form of Snell's law,

\sin C = \frac{v_A}{v_B} = \frac{1.5 \times 10^8}{3.0 \times 10^8} = 0.5

. Taking the inverse sine yields

C = 30^\circ

.

Step-by-Step Solution

1

Determine the relationship between critical angle and wave speeds in the two media.

\sin C = \frac{n_B}{n_A} = \frac{v_A}{v_B}

By Snell's Law, critical angle occurs when the angle of refraction is

90^\circ

, giving

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

. Since refractive index

n = \frac{c}{v}

, the ratio

\frac{n_B}{n_A}

simplifies to

\frac{v_A}{v_B}

.

2

Substitute the given wave speeds into the equation.

\sin C = \frac{1.5 \times 10^8\text{ m/s}}{3.0 \times 10^8\text{ m/s}} = 0.5

Light is traveling from the optically denser medium

A

(lower speed) to the less dense medium

B

(higher speed).

3

Calculate the inverse sine to find the critical angle

C

.

C = \arcsin(0.5) = 30^\circ

The angle whose sine is

0.5

is

30^\circ

.

Key Concept

Critical Angle and Total Internal Reflection in terms of Wave Speed

Question 10Question →

Match each defect of vision to its correct image formation characteristic and mode of optical correction.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Hypermetropia pairs with light focusing behind the retina (corrected with a converging lens). Myopia pairs with light focusing in front of the retina (corrected with a diverging lens). Presbyopia pairs with age-related loss of accommodation (corrected with bifocal/converging lenses). Astigmatism pairs with unequal corneal curvature in different planes (corrected with a cylindrical lens).

Each vision defect uniquely corresponds to a specific optical fault and corrective lens: Hypermetropia focuses images behind the retina (convex lens), Myopia focuses images in front of the retina (concave lens), Presbyopia involves age-related elasticity loss (bifocal/convex lens), and Astigmatism stems from asymmetric corneal curvature (cylindrical lens).

Step-by-Step Solution

1

Identify the structural defect and focus point for Hypermetropia.

Near point moves farther away, so light rays from standard near point focus behind the retina. A converging (convex) lens is required to bend light rays inward prior to entering the eye.

Hypermetropic eyes lack sufficient refractive power for near vision.

2

Identify the structural defect and focus point for Myopia.

Far point is reduced, so light rays from distant objects focus in front of the retina. A diverging (concave) lens is required to diverge incoming parallel rays.

Myopic eyes possess excessive refractive power or an elongated eyeball.

3

Identify the cause and correction for Presbyopia.

Caused by aging of ciliary muscles and hardening of the eye lens. Corrected with converging or bifocal lenses.

Presbyopia specifically relates to loss of elastic accommodation in older individuals.

4

Identify the cause and correction for Astigmatism.

Caused by non-uniform curvature of the cornea along horizontal vs vertical axes. Corrected using cylindrical lenses.

Cylindrical lenses compensate for asymmetrical focal lengths along specific axes.

Key Concept

Defects of Vision and Corrective Lenses

Question 11Question →

An object is placed $12\text{ cm}$ in front of a diverging lens having a focal length of $18\text{ cm}$ . What is the magnitude of the image distance from the lens?

7.2\text{ cm}

36.0\text{ cm}

30.0\text{ cm}

6.0\text{ cm}

Show answer & explanation

Answer: $7.2\text{ cm}$

Answer

The magnitude of the image distance is

7.2\text{ cm}

.

By the thin lens equation

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

, substituting

f = -18\text{ cm}

(due to the lens being diverging) and

u = +12\text{ cm}

gives

\frac{1}{v} = -\frac{1}{18} - \frac{1}{12} = -\frac{5}{36}\text{ cm}^{-1}

, which yields

v = -7.2\text{ cm}

. Thus, the image is located

7.2\text{ cm}

from the lens.

Step-by-Step Solution

1

Identify the given values and assign the correct signs according to the real-is-positive sign convention.

Object distance

u = +12\text{ cm}

; Focal length of diverging (concave) lens

f = -18\text{ cm}

.

Diverging lenses have virtual focus, so their focal length must be taken as negative.

2

Set up the thin lens formula

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

to solve for image distance

v

.

\frac{1}{-18} = \frac{1}{12} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{18} - \frac{1}{12}

.

Rearranging the equation isolates the reciprocal of the image distance.

3

Find a common denominator and solve for

v

.

\frac{1}{v} = \frac{-2 - 3}{36} = -\frac{5}{36}\text{ cm}^{-1} \implies v = -\frac{36}{5} = -7.2\text{ cm}

.

The negative sign indicates that the image formed is virtual and located on the same side of the lens as the object.

4

Take the magnitude of the image distance.

|v| = 7.2\text{ cm}

.

The question asks for the distance, which is a scalar magnitude.

Key Concept

Thin Lens Formula & Sign Convention for Concave Lenses

Estimated Time:1m 30s

Question 12Question →

An object is placed $10\text{ cm}$ in front of a converging lens of focal length $15\text{ cm}$ . What is the distance of the image from the lens and its nature?

30\text{ cm}

from the lens and virtual

30\text{ cm}

from the lens and real

6\text{ cm}

from the lens and real

6\text{ cm}

from the lens and virtual

Show answer & explanation

Answer: $30\text{ cm}$ from the lens and virtual

Answer

The image is

30\text{ cm}

from the lens and virtual.

Using the thin lens equation

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

with

f = +15\text{ cm}

and

u = +10\text{ cm}

yields

\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = -\frac{1}{30}\text{ cm}^{-1}

. Taking the reciprocal gives

v = -30\text{ cm}

. The magnitude of the image distance is

30\text{ cm}

, and the negative sign confirms that the image formed is virtual.

Step-by-Step Solution

1

Identify the given optical parameters and sign conventions.

Object distance

u = +10\text{ cm}

, Focal length for converging lens

f = +15\text{ cm}

.

By convention for real objects and converging lenses, both

u

and

f

are positive.

2

Substitute the values into the thin lens formula

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

.

\frac{1}{15} = \frac{1}{10} + \frac{1}{v}

The thin lens formula relates focal length, object distance, and image distance.

3

Rearrange to solve for the image distance reciprocal

\frac{1}{v}

.

\frac{1}{v} = \frac{1}{15} - \frac{1}{10} = \frac{2 - 3}{30} = -\frac{1}{30}\text{ cm}^{-1}

Subtracting

\frac{1}{10}

from

\frac{1}{15}

requires finding a common denominator of 30.

4

Take the reciprocal to find

v

and interpret its sign.

v = -30\text{ cm}

The magnitude of the distance is

30\text{ cm}

. The negative sign indicates that the image is formed on the same side as the object, making it virtual.

Key Concept

Thin lens formula and sign convention for image formation

Estimated Time:1m 30s

Question 13Question →

Match each eye defect to its corresponding optical cause and method of vision correction.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Myopia corresponds to image formation in front of the retina (corrected with a concave lens); Hypermetropia corresponds to image formation behind the retina (corrected with a convex lens); Astigmatism corresponds to uneven corneal curvature (corrected with a cylindrical lens); Presbyopia corresponds to age-related loss of accommodation (corrected with bifocal lenses).

Each vision defect is matched precisely to its biological cause and optical correction method: Myopia with image in front of retina and concave lens, Hypermetropia with image behind retina and convex lens, Astigmatism with non-spherical cornea and cylindrical lens, and Presbyopia with age-related loss of accommodation and bifocal lenses.

Step-by-Step Solution

1

Identify the cause and corrective lens for Myopia.

Myopia causes light to focus in front of the retina due to excessive converging power or an elongated eyeball, requiring a concave (diverging) lens.

A concave lens diverges incoming light rays prior to entry into the eye so the focal point moves back to the retina.

2

Identify the cause and corrective lens for Hypermetropia.

Hypermetropia causes light to focus behind the retina due to insufficient converging power or a shortened eyeball, requiring a convex (converging) lens.

A convex lens provides extra converging power to bring the focal point forward onto the retina.

3

Identify the cause and corrective lens for Astigmatism.

Astigmatism arises from irregular corneal curvature, requiring a cylindrical lens.

Cylindrical lenses vary focal power along a specific axis to equalize uneven light refraction.

4

Identify the cause and corrective lens for Presbyopia.

Presbyopia is caused by loss of elasticity in the lens with age, requiring bifocal lenses.

Bifocal lenses provide dual correction zones for distance and near vision.

Key Concept

Vision Defects and Lens Corrections

Question 14Question →

A progressive transverse wave with a frequency of $50\text{ Hz}$ travels through a medium with a wave speed of $300\text{ m/s}$ . If the wave crosses a boundary into a second medium where its speed decreases to $150\text{ m/s}$ , what is the frequency of the wave in the second medium?

50\text{ Hz}

25\text{ Hz}

100\text{ Hz}

75\text{ Hz}

Show answer & explanation

Answer: $50\text{ Hz}$

Answer

The frequency of the wave in the second medium is

50\text{ Hz}

.

The frequency of a wave is determined entirely by the vibrating source that creates it. When a wave passes from one medium to another, its speed and wavelength change, but its frequency remains constant. Therefore, the frequency remains

50\text{ Hz}

.

Step-by-Step Solution

1

Identify the given physical parameters.

Initial frequency

f_1 = 50\text{ Hz}

, initial speed

v_1 = 300\text{ m/s}

, new speed

v_2 = 150\text{ m/s}

.

Recognize which wave properties are given and which property is being asked for.

2

Apply the fundamental principle of wave propagation across different media boundaries.

Frequency

f

depends only on the source of vibration and remains constant during refraction/transmission across media boundaries.

While wave speed

v

and wavelength

\lambda

change when entering a new medium, frequency

f

remains unaffected.

3

Determine the frequency in the second medium.

f_2 = f_1 = 50\text{ Hz}

.

Since frequency is invariant across media boundaries,

f_2

must equal

50\text{ Hz}

.

Key Concept

Invariance of wave frequency across media boundaries

Estimated Time:45s

Question 15Question →

A ship uses a sonar device to determine the depth of the ocean floor. A sound pulse sent vertically downward reflects off the seabed and is detected $0.6\text{ s}$ after transmission. If the speed of sound in seawater is $1500\text{ m/s}$ , what is the depth of the ocean floor at that location?

450\text{ m}

900\text{ m}

1800\text{ m}

2500\text{ m}

Show answer & explanation

Answer: $450\text{ m}$

Answer

450\text{ m}

The correct answer is

450\text{ m}

because sound travels to the ocean floor and back, covering twice the actual depth. Dividing the total distance of

900\text{ m}

by

2

yields the one-way depth of

450\text{ m}

.

Step-by-Step Solution

1

Calculate the total distance traveled by the sound wave

d_{\text{total}} = 1500\text{ m/s} \times 0.6\text{ s} = 900\text{ m}

Total distance traveled by the wave equals speed multiplied by total elapsed time.

2

Determine the one-way depth of the ocean floor

\text{Depth} = \frac{d_{\text{total}}}{2} = \frac{900\text{ m}}{2} = 450\text{ m}

An echo travels down to the seabed and back to the receiver, so the ocean depth is half the total distance covered.

Key Concept

Echo distance calculation

Estimated Time:45s

Question 16Question →

Match each physical wave description on the left with its definitive wave classification on the right based on propagation mechanisms and oscillation directions.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings are: Electromagnetic field oscillations without a medium match with Non-mechanical transverse progressive wave; Periodic compressions/rarefactions parallel to wave motion match with Longitudinal mechanical progressive wave; Superposition of opposing identical waves with zero net energy flow matches with Stationary (standing) mechanical wave; Perpendicular particle vibrations in an elastic medium match with Transverse mechanical progressive wave.

Each wave type is categorized by three primary criteria: medium requirement (mechanical requires a medium, non-mechanical does not), vibration alignment (transverse is perpendicular, longitudinal is parallel), and energy movement (progressive transfers energy continuously, stationary confines energy between nodes). Oscillating fields in vacuum represent non-mechanical transverse waves; compressions/rarefactions represent longitudinal mechanical waves; opposing wave superposition forms stationary waves; and perpendicular elastic vibrations represent transverse mechanical waves.

Step-by-Step Solution

1

Classify the wave mechanism based on medium requirement.

Waves requiring an elastic medium (particle vibrations) are mechanical; waves capable of propagating through a vacuum via field oscillations are non-mechanical.

Determines whether the wave is mechanical or non-mechanical.

2

Classify the direction of particle or field oscillation relative to propagation direction.

Oscillations parallel to wave motion form longitudinal waves; oscillations perpendicular to wave motion form transverse waves.

Distinguishes between longitudinal and transverse wave modes.

3

Determine energy transfer characteristics (progressive vs. stationary).

Continuous outward transfer of energy indicates a progressive wave, whereas trapped energy bounded by fixed nodes/antinodes resulting from wave superposition indicates a standing (stationary) wave.

Differentiates traveling waves from standing wave patterns.

4

Map each wave description to its complete wave classification.

All four wave descriptions are uniquely matched with their corresponding taxonomy classifications.

Completes the matching process.

Key Concept

Classification of waves by medium requirement, oscillation direction, and energy propagation mode

Estimated Time:1m 30s

Question 17Question →

An electromagnetic wave has a wavelength of $1.5 \times 10^{-7}\text{ m}$ in a vacuum. It enters an optical medium where its speed decreases to $2.0 \times 10^8\text{ m/s}$ . Given that the speed of light in vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the wavelength of the wave in the medium, and to which spectral region does its frequency belong?

1.0 \times 10^{-7}\text{ m}

and Ultraviolet region

1.0 \times 10^{-7}\text{ m}

and Visible light region

2.25 \times 10^{-7}\text{ m}

and Ultraviolet region

1.5 \times 10^{-7}\text{ m}

and Infrared region

Show answer & explanation

Answer: $1.0 \times 10^{-7}\text{ m}$ and Ultraviolet region

Answer

The wavelength in the medium is

1.0 \times 10^{-7}\text{ m}

and its frequency belongs to the Ultraviolet region.

When an electromagnetic wave transitions into a medium, its frequency

f

remains constant at

2.0 \times 10^{15}\text{ Hz}

, placing it strictly in the Ultraviolet spectrum. Its wavelength in the medium shrinks to

\lambda_{med} = \frac{v}{f} = \frac{2.0 \times 10^8}{2.0 \times 10^{15}} = 1.0 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in vacuum using the wave equation

c = f \lambda_{vac}

.

f = \frac{3.0 \times 10^8\text{ m/s}}{1.5 \times 10^{-7}\text{ m}} = 2.0 \times 10^{15}\text{ Hz}

.

Frequency is an intrinsic property determined by the source and does not change when entering a new medium.

2

Identify the electromagnetic spectrum region corresponding to the calculated frequency.

A frequency of

2.0 \times 10^{15}\text{ Hz}

falls in the range

7.5 \times 10^{14}\text{ Hz}

to

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

, which corresponds to the Ultraviolet region.

Spectral classifications are uniquely determined by frequency (or vacuum wavelength).

3

Determine the wavelength in the medium using

\lambda_{med} = \frac{v}{f}

.

\lambda_{med} = \frac{2.0 \times 10^8\text{ m/s}}{2.0 \times 10^{15}\text{ Hz}} = 1.0 \times 10^{-7}\text{ m}

.

The wavelength changes proportionally with phase velocity in a medium.

Key Concept

Invariance of electromagnetic wave frequency across media boundaries and wave equation relationship

v = f \lambda

.

Estimated Time:2m 0s

Question 18Question →

A man standing at a distance in front of a tall vertical cliff claps his hands and hears the echo after $1.2\text{ s}$ . If the speed of sound in air is $340\text{ m/s}$ , what is the distance between the man and the cliff in meters?

Show answer & explanation

Answer: 204

Answer

The distance between the man and the cliff is

204\text{ m}

.

An echo is formed when sound travels to an obstacle and reflects back. The time taken for the sound to travel to the cliff and back is

1.2\text{ s}

. The total distance covered by sound is

v \times t = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

. Since this distance covers two equal trips (to the cliff and back), the distance to the cliff is

408\text{ m} / 2 = 204\text{ m}

.

Step-by-Step Solution

1

State the relationship between sound speed, total echo time, and distance.

Total distance traveled by the sound is twice the distance to the cliff:

2d = v \times t

.

An echo involves sound traveling from the source to the reflecting barrier and back.

2

Substitute the given values into the equation.

2d = 340\text{ m/s} \times 1.2\text{ s} = 408\text{ m}

.

To find the total distance traversed by the sound wave.

3

Solve for the distance

d

.

d = \frac{408\text{ m}}{2} = 204\text{ m}

.

The one-way distance to the cliff is half of the total distance traveled by the echo.

Key Concept

Calculation of echo distance using

d = \frac{v t}{2}

Question 19Question →

A student stands between two tall, parallel vertical walls and claps her hands once. She hears the first echo after $1.0\text{ s}$ and the second echo after $1.5\text{ s}$ . Taking the speed of sound in air to be $340\text{ m s}^{-1}$ , what is the distance between the two walls?

425\text{ m}

850\text{ m}

510\text{ m}

255\text{ m}

Show answer & explanation

Answer: $425\text{ m}$

Answer

The distance between the two walls is

425\text{ m}

.

Sound travels from the student to each wall and reflects back. The distance to the first wall is

\frac{340 \times 1.0}{2} = 170\text{ m}

and to the second wall is

\frac{340 \times 1.5}{2} = 255\text{ m}

. Adding both distances gives the total separation between the walls as

425\text{ m}

.

Step-by-Step Solution

1

Calculate the distance from the student to the closer wall (

d_1

)

d_1 = \frac{v \times t_1}{2} = \frac{340 \text{ m s}^{-1} \times 1.0 \text{ s}}{2} = 170\text{ m}

Sound travels to the wall and reflects back, so the time given corresponds to twice the distance.

2

Calculate the distance from the student to the further wall (

d_2

)

d_2 = \frac{v \times t_2}{2} = \frac{340 \text{ m s}^{-1} \times 1.5 \text{ s}}{2} = 255\text{ m}

The sound for the second echo travels to the second wall and back.

3

Determine the total distance between the two parallel walls

D = d_1 + d_2 = 170\text{ m} + 255\text{ m} = 425\text{ m}

Since the student is positioned between the two walls, the separation of the walls is the sum of both individual distances.

Key Concept

Echo distance relation

2d = v t

for sound reflection from barriers.

Estimated Time:1m 30s

Question 20Question →

An observer standing at a position between two tall parallel vertical cliffs fires a signal pistol. The observer hears the first echo reflected from the nearer cliff after $1.2\text{ s}$ and the second echo reflected from the farther cliff after $1.8\text{ s}$ . If the speed of sound in air is $340\text{ m s}^{-1}$ , what is the total distance between the two cliffs?

102\text{ m}

510\text{ m}

816\text{ m}

1020\text{ m}

Show answer & explanation

Answer: $510\text{ m}$

Answer

The total distance between the two cliffs is

510\text{ m}

.

Because an echo involves two-way travel of sound, the distance

d

from an observer to a reflecting barrier is given by

d = \frac{v t}{2}

. For the nearer cliff, the distance is

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

. For the farther cliff, the distance is

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

. The total distance between the parallel cliffs is

d_1 + d_2 = 204\text{ m} + 306\text{ m} = 510\text{ m}

.

Step-by-Step Solution

1

Calculate the distance from the observer to the nearer cliff

d_1 = \frac{v \times t_1}{2} = \frac{340\text{ m s}^{-1} \times 1.2\text{ s}}{2} = 204\text{ m}

An echo involves sound traveling from the observer to the cliff and back, so the one-way distance is half the total path length.

2

Calculate the distance from the observer to the farther cliff

d_2 = \frac{v \times t_2}{2} = \frac{340\text{ m s}^{-1} \times 1.8\text{ s}}{2} = 306\text{ m}

Similarly, the sound travels to the farther cliff and back in

1.8\text{ s}

.

3

Sum the two one-way distances to find the total distance between the two cliffs

D = d_1 + d_2 = 204\text{ m} + 306\text{ m} = 510\text{ m}

Since the observer is positioned between the two parallel cliffs, the total separation distance is the sum of the individual distances to each cliff.

Key Concept

Echo distance calculation involving two-way sound propagation

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