Waves and Optics Practice Questions — JAMB UTME — Page 3

Question 41Question →

A stretched string of length $0.50\text{ m}$ fixed at both ends vibrates in its third harmonic mode at a frequency of $450\text{ Hz}$ . What is the speed of the transverse wave along the string in $\text{m/s}$ ?

Show answer & explanation

Answer: 150

Answer

The speed of the transverse wave along the string is

150\text{ m/s}

.

For a string fixed at both ends, standing wave modes produce harmonics given by

f_n = \frac{n v}{2L}

. Given

n = 3

,

L = 0.50\text{ m}

, and

f_3 = 450\text{ Hz}

, substituting these into the equation yields

450 = \frac{3v}{2(0.50)} = 3v

, leading to

v = 150\text{ m/s}

.

Step-by-Step Solution

1

Identify the standing wave frequency equation for a string fixed at both ends.

The frequency of the

n

-th harmonic is

f_n = \frac{n v}{2L}

, where

n

is the harmonic number,

v

is the wave speed, and

L

is the string length.

Fixed ends require nodes at both boundaries, producing standing wave modes with wavelengths

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

.

2

Substitute the given physical quantities into the harmonic equation.

450 = \frac{3 \times v}{2 \times 0.50}

.

The question specifies

L = 0.50\text{ m}

, third harmonic mode (

n = 3

), and frequency

f_3 = 450\text{ Hz}

.

3

Solve for the wave speed

v

.

v = 150\text{ m/s}

.

Simplifying

2 \times 0.50 = 1.0

gives

3v = 450

, so

v = \frac{450}{3} = 150\text{ m/s}

.

Key Concept

Standing Waves and Harmonics in Vibrating Strings

Question 42Question →

A transverse wave propagating through a stretched string has a frequency of $250\text{ Hz}$ and a wavelength of $1.4\text{ m}$ . What is the speed of propagation of the wave in $\text{m/s}$ ?

Show answer & explanation

Answer: 350

Answer

The speed of propagation of the wave is

350\text{ m/s}

.

The speed of propagation of a progressive wave is determined using the wave equation

v = f \lambda

. Substituting the given values

f = 250\text{ Hz}

and

\lambda = 1.4\text{ m}

gives

v = 250 \times 1.4 = 350\text{ m/s}

.

Step-by-Step Solution

1

Identify the given physical quantities

Frequency

f = 250\text{ Hz}

, Wavelength

\lambda = 1.4\text{ m}

.

These parameters are required to calculate the speed of propagation.

2

Apply the fundamental wave equation

v = f \lambda

The speed of a progressive wave equals the product of its frequency and its wavelength.

3

Perform the numerical calculation

v = 250 \times 1.4 = 350\text{ m/s}

Multiplying

250\text{ Hz}

by

1.4\text{ m}

yields the wave velocity in meters per second.

Key Concept

Wave Propagation Speed Equation (

v = f \lambda

)

Question 43Question →

An infrared sensor detects electromagnetic radiation with a frequency of $4.0 \times 10^{14}\text{ Hz}$ . A second sensor detects ultraviolet radiation with a wavelength of $300\text{ nm}$ . Given that the speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the ratio of the wavelength of the infrared radiation to the wavelength of the ultraviolet radiation?

2.5

0.4

25.0

1.33

Show answer & explanation

Answer: $2.5$

Answer

The ratio of the wavelength of the infrared radiation to the wavelength of the ultraviolet radiation is 2.5.

Using

c = f \lambda

, the infrared wavelength is

\lambda = \frac{3.0 \times 10^8}{4.0 \times 10^{14}} = 7.5 \times 10^{-7}\text{ m}

. Comparing this to the ultraviolet wavelength of

300\text{ nm} = 3.0 \times 10^{-7}\text{ m}

gives a ratio of

\frac{7.5 \times 10^{-7}}{3.0 \times 10^{-7}} = 2.5

.

Step-by-Step Solution

1

Calculate the wavelength of the infrared radiation using the wave equation

c = f \lambda

.

\lambda_{\text{IR}} = \frac{3.0 \times 10^8\text{ m/s}}{4.0 \times 10^{14}\text{ Hz}} = 7.5 \times 10^{-7}\text{ m}

The speed of all electromagnetic waves in a vacuum is constant (

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

).

2

Convert the ultraviolet wavelength to meters for consistent units.

\lambda_{\text{UV}} = 300\text{ nm} = 300 \times 10^{-9}\text{ m} = 3.0 \times 10^{-7}\text{ m}

Units must be in standard meters before calculating the dimensionless ratio.

3

Compute the ratio of the infrared wavelength to the ultraviolet wavelength.

\text{Ratio} = \frac{7.5 \times 10^{-7}\text{ m}}{3.0 \times 10^{-7}\text{ m}} = 2.5

Dividing the two wavelengths in the same units yields the desired ratio.

Key Concept

Wave Equation and Electromagnetic Spectrum Properties

Estimated Time:2m 0s

Question 44Question →

An object is placed $12\text{ cm}$ in front of a concave mirror having a focal length of $20\text{ cm}$ . What is the distance of the image from the mirror and its nature?

30\text{ cm}

behind the mirror, virtual

30\text{ cm}

in front of the mirror, real

7.5\text{ cm}

in front of the mirror, real

7.5\text{ cm}

behind the mirror, virtual

Show answer & explanation

Answer: $30\text{ cm}$ behind the mirror, virtual

Answer

The image is formed

30\text{ cm}

behind the mirror and is virtual.

Applying the mirror formula

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

with

f = 20\text{ cm}

and

u = 12\text{ cm}

yields

\frac{1}{v} = \frac{1}{20} - \frac{1}{12} = -\frac{1}{30}\text{ cm}^{-1}

, resulting in

v = -30\text{ cm}

. The negative sign confirms that the image is virtual and located

30\text{ cm}

behind the mirror.

Step-by-Step Solution

1

Identify given quantities and apply standard mirror sign conventions.

Object distance

u = +12\text{ cm}

, focal length

f = +20\text{ cm}

.

For a concave mirror, real objects and real focal points carry positive values.

2

Set up the mirror equation

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

.

\frac{1}{20} = \frac{1}{12} + \frac{1}{v}

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

3

Rearrange the equation to isolate

\frac{1}{v}

and solve.

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

Finding a common denominator of 60 allows exact calculation of the negative reciprocal.

4

Determine image characteristics based on the sign of

v

.

The image distance magnitude is

30\text{ cm}

behind the mirror, and the image is virtual.

A negative value for image distance

v

signifies a virtual image located behind the mirror surface.

Key Concept

Concave Mirror Formula and Virtual Image Formation

Question 45Question →

A uniform string of length $0.50\text{ m}$ and mass $2.0\text{ g}$ is fixed at both ends under a tension of $90\text{ N}$ . When vibrating, the second harmonic of this string resonates with the first overtone of an air column in a pipe closed at one end. Assuming the speed of sound in air is $340\text{ m/s}$ , calculate the length of the pipe in meters.

Show answer & explanation

Answer: 0.85

Answer

The length of the pipe is

0.85\text{ m}

.

The wave speed on the string is computed from tension and linear mass density as $150\text{ m/s}$, yielding a second harmonic frequency of $300\text{ Hz}$. Equating this to the first overtone (third harmonic) frequency formula of a closed air pipe, $f = \frac{3v_{air}}{4L_p}$, yields an exact pipe length of $0.85\text{ m}$.

Step-by-Step Solution

1

Calculate the linear mass density ($\mu$) of the string

$\mu = \frac{m}{L_s} = \frac{0.0020\text{ kg}}{0.50\text{ m}} = 4.0 \times 10^{-3}\text{ kg/m}$

Mass must be converted to kilograms before determining mass per unit length.

2

Determine the wave speed ($v_s$) along the stretched string

$v_s = \sqrt{\frac{T}{\mu}} = \sqrt{\frac{90\text{ N}}{4.0 \times 10^{-3}\text{ kg/m}}} = \sqrt{22500} = 150\text{ m/s}$

The velocity of a transverse wave on a string depends on tension and linear mass density.

3

Calculate the second harmonic frequency of the string ($f_{2,s}$)

$f_{2,s} = \frac{v_s}{L_s} = \frac{150\text{ m/s}}{0.50\text{ m}} = 300\text{ Hz}$

The fundamental frequency is $f_{1,s} = \frac{v_s}{2L_s} = 150\text{ Hz}$, so the second harmonic is twice the fundamental frequency.

4

Set up the resonance equation for the first overtone of a closed pipe

$f_{3,p} = \frac{3 v_{air}}{4 L_p} = 300\text{ Hz}$

A pipe closed at one end produces only odd harmonics, so the first overtone is the 3rd harmonic.

5

Solve for the length of the closed pipe ($L_p$)

$L_p = \frac{3 \times 340\text{ m/s}}{4 \times 300\text{ Hz}} = \frac{1020}{1200} = 0.85\text{ m}$

Rearranging the frequency formula gives the required air column length.

Key Concept

Coupled resonance between standing waves on strings and air columns in closed pipes

Question 46Question →

For any real object positioned in front of a convex spherical mirror, the image formed is always virtual, upright, and diminished, located between the pole and the principal focus.

True

False

Show answer & explanation

Answer: True

Answer

The statement is true. A convex mirror always produces a virtual, erect, and diminished image positioned between the pole and the focus for any real object placed in front of it.

The statement accurately expresses the fundamental optics rule for convex mirrors: reflecting surface geometry causes incident parallel rays to diverge, producing a virtual, erect, and diminished image located between the mirror's pole and principal focus for all real object positions.

Step-by-Step Solution

1

Analyze ray tracing principles for a convex mirror.

Ray 1, traveling parallel to the principal axis, reflects such that it appears to originate from the principal focus

F

behind the mirror. Ray 2, directed toward the center of curvature

C

, reflects back along its original path.

Tracing these rays determines where their virtual extensions intersect.

2

Determine the location and characteristics of the virtual intersection.

The backward extensions of the reflected rays intersect behind the reflecting surface between the pole

P

and the principal focus

F

.

Since the light rays diverge and only their extensions meet, the image formed is virtual, erect, and diminished.

Key Concept

Image characteristics of convex spherical mirrors

Question 47Question →

An open pipe of physical length $0.58\text{ m}$ and a pipe closed at one end emit sound at the same frequency when both are vibrating in their first overtone mode. If the end correction at each open end is $0.01\text{ m}$ , what is the physical length of the closed pipe?

0.29\text{ m}

0.43\text{ m}

0.44\text{ m}

0.45\text{ m}

Show answer & explanation

Answer: $0.44\text{ m}$

Answer

The physical length of the closed pipe is

0.44\text{ m}

.

The effective length of an open pipe with two open ends is

L_{1,\text{eff}} = 0.58 + 2(0.01) = 0.60\text{ m}

. The first overtone for an open pipe is its second harmonic (

n=2

), giving

f = \frac{v}{0.60}

. For a pipe closed at one end, the first overtone is its third harmonic (

m=3

), giving

f = \frac{3v}{4L_{2,\text{eff}}}

. Equating frequencies gives

L_{2,\text{eff}} = 0.45\text{ m}

. Subtracting the single end correction (

e = 0.01\text{ m}

) yields the physical length

0.44\text{ m}

.

Step-by-Step Solution

1

Calculate the effective length and first overtone frequency of the open pipe.

L_{1,\text{eff}} = L_1 + 2e = 0.58\text{ m} + 2(0.01\text{ m}) = 0.60\text{ m}

, so

f_{\text{open,1st overtone}} = \frac{2v}{2L_{1,\text{eff}}} = \frac{v}{0.60}

.

An open pipe has two open ends, so end correction is applied at both ends (

2e

). Its first overtone corresponds to the second harmonic (

n=2

).

2

Express the first overtone frequency of the closed pipe in terms of its effective length.

f_{\text{closed,1st overtone}} = \frac{3v}{4L_{2,\text{eff}}}

.

A pipe closed at one end produces only odd harmonics (

m=1, 3, 5, \dots

). The first overtone corresponds to the third harmonic (

m=3

).

3

Equate the two frequencies and solve for the effective length of the closed pipe.

\frac{v}{0.60} = \frac{3v}{4L_{2,\text{eff}}} \implies 4L_{2,\text{eff}} = 1.80\text{ m} \implies L_{2,\text{eff}} = 0.45\text{ m}

.

Both pipes emit sound at the same frequency in their first overtone modes.

4

Calculate the physical length of the closed pipe.

L_2 = L_{2,\text{eff}} - e = 0.45\text{ m} - 0.01\text{ m} = 0.44\text{ m}

.

A closed pipe has only one open end, so its effective length is

L_2 + e

.

Key Concept

Standing Waves and End Correction in Open and Closed Organ Pipes

Question 48Question →

A progressive wave traveling along a stretched string is represented by the equation $y = 0.02 \sin(50\pi t - 4\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the speed of the wave in meters per second ( $\text{m/s}$ )?

Show answer & explanation

Answer: 12.5

Answer

The speed of the wave is

12.5\text{ m/s}

.

Comparing the wave equation

y = 0.02 \sin(50\pi t - 4\pi x)

with the standard form

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

shows that the angular frequency is

\omega = 50\pi\text{ rad/s}

and the wave number is

k = 4\pi\text{ rad/m}

. Using the relation

v = \frac{\omega}{k}

, the speed of the wave is calculated as

v = \frac{50\pi}{4\pi} = 12.5\text{ m/s}

.

Step-by-Step Solution

1

Extract angular frequency and wave number from the equation

Comparing

y = 0.02 \sin(50\pi t - 4\pi x)

with

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

gives

\omega = 50\pi\text{ rad/s}

and

k = 4\pi\text{ rad/m}

.

Matching corresponding terms yields the wave parameters.

2

Calculate the wave propagation speed

v = \frac{\omega}{k} = \frac{50\pi}{4\pi} = 12.5\text{ m/s}

.

Wave speed equals the ratio of angular frequency to wave number.

Key Concept

Calculating wave speed from a mathematical wave equation

Question 49Question →

A dentist uses a small concave mirror with a focal length of $20\text{ mm}$ to inspect a patient's tooth. If the mirror produces an upright image that is magnified $4$ times, how far from the tooth is the mirror placed?

15\text{ mm}

25\text{ mm}

80\text{ mm}

5\text{ mm}

Show answer & explanation

Answer: $15\text{ mm}$

Answer

The mirror must be placed

15\text{ mm}

from the tooth.

For a concave mirror, an upright image is virtual. Linear magnification

m = 4

implies

v = -4u

. Substituting

f = 20\text{ mm}

into the mirror formula

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

gives

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

, solving to

u = 15\text{ mm}

.

Step-by-Step Solution

1

Identify given values and apply sign conventions.

Focal length of concave mirror

f = +20\text{ mm}

. Magnification

m = +4

because the image is upright (virtual).

Upright images formed by single optical mirrors are always virtual, requiring a negative image distance.

2

Express image distance

v

in terms of object distance

u

.

Linear magnification

m = -\frac{v}{u} \implies +4 = -\frac{v}{u} \implies v = -4u

.

The linear magnification formula relates orientation, object distance, and image distance.

3

Substitute

f

and

v

into the mirror formula.

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{20} = \frac{1}{u} - \frac{1}{4u} = \frac{3}{4u}.

The mirror equation determines object and image locations relative to the focal length.

4

Solve for the object distance

u

.

4u = 60 \implies u = 15\text{ mm}.

Cross-multiplying gives the required distance between the mirror and the tooth.

Key Concept

Reflection and image formation by concave spherical mirrors (virtual magnified image)

Question 50Question →

A light wave of frequency $5.00 \times 10^{14}\text{ Hz}$ travels from air into a glass medium with a refractive index of $1.50$ . What is the frequency of the light wave inside the glass medium?

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

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

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

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

Show answer & explanation

Answer: $5.00 \times 10^{14}\text{ Hz}$

Answer

The frequency of the light wave inside the glass medium remains

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

.

When light passes from one medium into another, its speed and wavelength change, but its frequency remains constant because it depends solely on the source producing the wave.

Step-by-Step Solution

1

Identify the given physical parameters.

Initial frequency

f = 5.00 \times 10^{14}\text{ Hz}

and refractive index

n = 1.50

.

Understanding provided quantities is the starting point for wave refraction analysis.

2

Apply the principle of frequency conservation across optical boundaries.

Frequency in glass medium

f_{\text{glass}} = f_{\text{air}} = 5.00 \times 10^{14}\text{ Hz}

.

When a wave passes into a different medium, its speed and wavelength alter in proportion to the refractive index, but its frequency is determined solely by the periodic source and remains constant.

Key Concept

Invariance of Wave Frequency across Refracting Media

Question 51Question →

A ray of light passes symmetrically through an equilateral glass prism of refractive index $\sqrt{2}$ . What is the angle of minimum deviation of the ray, in degrees?

Show answer & explanation

Answer: 30

Answer

The angle of minimum deviation of the ray is

30^\circ

.

For an equilateral triangular prism, the apex angle

A

is

60^\circ

. At minimum deviation, light travels symmetrically through the prism and obeys the exact relation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

. Substituting

n = \sqrt{2}

and

A = 60^\circ

gives

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{\sqrt{2}}{2} = \sin(45^\circ)

. Equating arguments yields

\frac{60^\circ + D_m}{2} = 45^\circ

, leading directly to

D_m = 30^\circ

.

Step-by-Step Solution

1

Determine the refracting angle of the prism

A = 60^\circ

An equilateral prism has interior angles of

60^\circ

each, so the apex angle

A = 60^\circ

.

2

Set up the prism formula for minimum deviation

n = \frac{\sin\left(\frac{A + D_m}{2}\right)}{\sin\left(\frac{A}{2}\right)}

When light passes symmetrically through a prism, the deviation is at its minimum value

D_m

.

3

Substitute known values into the equation

\sqrt{2} = \frac{\sin\left(\frac{60^\circ + D_m}{2}\right)}{\sin(30^\circ)}

Given

n = \sqrt{2}

and

A = 60^\circ

, with

\sin(30^\circ) = 0.5

.

4

Calculate the sine of the half-angle

\sin\left(\frac{60^\circ + D_m}{2}\right) = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

Multiplying both sides by

\sin(30^\circ) = 0.5

.

5

Solve for the minimum deviation angle

D_m

D_m = 30^\circ

Since

\arcsin\left(\frac{\sqrt{2}}{2}\right) = 45^\circ

, we set

\frac{60^\circ + D_m}{2} = 45^\circ \implies 60^\circ + D_m = 90^\circ \implies D_m = 30^\circ

.

Key Concept

Refraction of light through a prism at the angle of minimum deviation

Question 52Question →

A marine navigational radar system emits electromagnetic pulses with a wavelength of $3.0\text{ cm}$ . Given that the speed of electromagnetic waves in air is $3.0 \times 10^8\text{ m/s}$ , what is the frequency of the radar signal in gigahertz ( $\text{GHz}$ )?

Show answer & explanation

Answer: 10

Answer

The frequency of the radar signal is

10\text{ GHz}

.

Converting

3.0\text{ cm}

to meters gives

\lambda = 0.03\text{ m}

. Using

f = c / \lambda

,

f = (3.0 \times 10^8\text{ m/s}) / 0.03\text{ m} = 1.0 \times 10^{10}\text{ Hz}

. Dividing by

10^9\text{ Hz/GHz}

yields

10\text{ GHz}

.

Step-by-Step Solution

1

Convert the given wavelength into fundamental SI units (meters).

\lambda = 3.0\text{ cm} = 3.0 \times 10^{-2}\text{ m}

The speed of light

c

is given in meters per second, so the wavelength must be expressed in meters to keep units consistent.

2

Rearrange the electromagnetic wave speed formula

c = f \lambda

to solve for frequency

f

.

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{3.0 \times 10^{-2}\text{ m}} = 1.0 \times 10^{10}\text{ Hz}

Frequency is the ratio of wave propagation speed to wavelength.

3

Convert the frequency from hertz to gigahertz.

f = \frac{1.0 \times 10^{10}\text{ Hz}}{10^9\text{ Hz/GHz}} = 10\text{ GHz}

The question explicitly asks for the answer in gigahertz (

\text{GHz}

).

Key Concept

Electromagnetic wave propagation equation (

c = f \lambda

) and unit metric prefixes.

Question 53Question →

A progressive transverse wave traveling through a primary medium is governed by the mathematical wave equation $y = 0.05 \sin\left(100\pi t - 2.5\pi x\right)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. Upon entering a secondary medium, the wave undergoes refraction such that its wavelength decreases by $20\%$ . What is the speed of the wave in the secondary medium in $\text{m/s}$ ?

Show answer & explanation

Answer: 32

Answer

The speed of the wave in the secondary medium is

32\text{ m/s}

.

Comparing

y = 0.05 \sin\left(100\pi t - 2.5\pi x\right)

with

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

yields

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

and

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

. This gives a source frequency

f = \frac{\omega}{2\pi} = 50\text{ Hz}

and initial wavelength

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

. Because wave frequency is invariant across boundaries,

f

remains

50\text{ Hz}

in the second medium. The new wavelength is

\lambda_2 = 0.8\text{ m} \times 0.80 = 0.64\text{ m}

. Consequently, the wave speed in the second medium is

v_2 = 50\text{ Hz} \times 0.64\text{ m} = 32\text{ m/s}

.

Step-by-Step Solution

1

Extract angular frequency and wave number from the wave equation

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

and

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

Matching coefficients in

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

allows determination of temporal and spatial characteristics.

2

Determine wave frequency and original wavelength

f = 50\text{ Hz}

and

\lambda_1 = 0.8\text{ m}

Using fundamental relationships

f = \frac{\omega}{2\pi}

and

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

.

3

Calculate the refracted wavelength under constant frequency

\lambda_2 = 0.64\text{ m}

while

f

remains

50\text{ Hz}

Wave frequency depends solely on the wave source and remains unchanged during medium transitions.

4

Compute wave propagation speed in the new medium

v_2 = 32\text{ m/s}

Applying the wave equation

v = f\lambda

with the updated wavelength.

Key Concept

Wave equation parameter extraction and frequency invariance during refraction

Estimated Time:2m 0s

Question 54Question →

A coin lies at the bottom of a vessel filled with a liquid to a depth of $14.0\text{ cm}$ . If the refractive index of the liquid relative to air is $1.40$ , calculate the apparent upward displacement of the coin in centimeters when viewed vertically from directly above.

Show answer & explanation

Answer: 4

Answer

The apparent upward displacement of the coin is

4.0\text{ cm}

.

Refraction at the liquid-air boundary makes an object at real depth

h = 14.0\text{ cm}

appear at an apparent depth

h' = \frac{h}{n} = \frac{14.0}{1.40} = 10.0\text{ cm}

. The apparent upward displacement is the difference between real depth and apparent depth:

d = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

.

Step-by-Step Solution

1

Identify the given values and formula for refractive index in terms of depth.

Real depth

h = 14.0\text{ cm}

, refractive index

n = 1.40

. Formula:

n = \frac{\text{Real depth}}{\text{Apparent depth}} = \frac{h}{h'}

.

Light rays bending away from the normal upon leaving the denser liquid cause the coin to appear closer to the surface.

2

Calculate the apparent depth (

h'

).

h' = \frac{14.0\text{ cm}}{1.40} = 10.0\text{ cm}

.

Rearranging the refractive index formula gives

h' = \frac{h}{n}

.

3

Calculate the apparent upward displacement (

d

).

d = h - h' = 14.0\text{ cm} - 10.0\text{ cm} = 4.0\text{ cm}

.

The displacement is the distance between the actual position at the bottom and the virtual image position.

Key Concept

Real depth, apparent depth, and apparent displacement

Question 55Question →

A ray of light traveling in air strikes the flat surface of a transparent glass slab at an angle of incidence of $60^\circ$ . If the refractive index of the glass slab relative to air is $\sqrt{3}$ , what is the angle of refraction inside the glass slab in degrees?

Show answer & explanation

Answer: 30

Answer

The angle of refraction inside the glass slab is

30^\circ

.

According to Snell's Law,

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

. Substituting

n = \sqrt{3}

and

i = 60^\circ

gives

\sqrt{3} = \frac{\sin 60^\circ}{\sin r}

. Since

\sin 60^\circ = \frac{\sqrt{3}}{2}

, rearranging gives

\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = 0.5

. Taking the inverse sine of

0.5

gives

r = 30^\circ

.

Step-by-Step Solution

1

Identify Snell's law formula relating the angle of incidence and angle of refraction.

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

Snell's law describes how light bends when crossing the boundary between two optical media.

2

Substitute the given values

i = 60^\circ

and

n = \sqrt{3}

into the equation.

\sqrt{3} = \frac{\sin 60^\circ}{\sin r}

Plugging in the known parameters allows us to isolate the unknown sine of the angle of refraction.

3

Substitute

\sin 60^\circ = \frac{\sqrt{3}}{2}

and rearrange for

\sin r

.

\sin r = \frac{\sqrt{3}/2}{\sqrt{3}} = 0.5

Canceling

\sqrt{3}

from both sides yields a simple numerical value for

\sin r

.

4

Take the inverse sine of

0.5

to find

r

.

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

The angle whose sine is

0.5

is

30^\circ

.

Key Concept

Snell's Law of Refraction

Question 56Question →

An astronomical telescope in normal adjustment consists of an objective lens with a focal length of $60\text{ cm}$ and an eyepiece with a focal length of $5\text{ cm}$ . What is the angular magnification produced by the telescope and the separation distance between the two lenses?

12

and

65\text{ cm}

0.083

and

65\text{ cm}

12

and

55\text{ cm}

300

and

65\text{ cm}

Show answer & explanation

Answer: $12$ and $65\text{ cm}$

Answer

The angular magnification is

12

and the separation distance between the lenses is

65\text{ cm}

.

For an astronomical telescope in normal adjustment, the magnification is given by

M = \frac{f_o}{f_e} = \frac{60}{5} = 12

, and the length of the telescope tube (distance between lenses) is

L = f_o + f_e = 60 + 5 = 65\text{ cm}

.

Step-by-Step Solution

1

Calculate the angular magnification (

M

) of the telescope

M = \frac{f_o}{f_e} = \frac{60}{5} = 12

In an astronomical telescope in normal adjustment, angular magnification is given by the ratio of the focal length of the objective lens to that of the eyepiece.

2

Calculate the separation distance (

L

) between the two lenses

L = f_o + f_e = 60 + 5 = 65\text{ cm}

When in normal adjustment, the final image is formed at infinity, so the distance between the objective lens and eyepiece equals the sum of their focal lengths.

Key Concept

Astronomical telescope in normal adjustment (magnification and lens separation)

Estimated Time:1m 0s

Question 57Question →

A diagnostic medical device emits electromagnetic radiation with a frequency of $6.0 \times 10^{18}\text{ Hz}$ in a vacuum. Given that the speed of light in a vacuum is $c = 3.0 \times 10^8\text{ m/s}$ , what is the wavelength of this radiation, and to which part of the electromagnetic spectrum does it belong?

5.0 \times 10^{-11}\text{ m}

, X-rays

5.0 \times 10^{-11}\text{ m}

, Ultraviolet rays

2.0 \times 10^{10}\text{ m}

, X-rays

1.8 \times 10^{27}\text{ m}

, Gamma rays

Show answer & explanation

Answer: $5.0 \times 10^{-11}\text{ m}$ , X-rays

Answer

The wavelength of the radiation is

5.0 \times 10^{-11}\text{ m}

, which corresponds to X-rays.

Using the wave equation

\lambda = \frac{c}{f}

, dividing

3.0 \times 10^8\text{ m/s}

by

6.0 \times 10^{18}\text{ Hz}

yields

\lambda = 5.0 \times 10^{-11}\text{ m}

. Radiation with a wavelength of

5.0 \times 10^{-11}\text{ m}

(frequency

6.0 \times 10^{18}\text{ Hz}

) falls in the X-ray portion of the electromagnetic spectrum.

Step-by-Step Solution

1

Identify the given values and the wave equation relating speed, frequency, and wavelength.

Speed of light

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

, frequency

f = 6.0 \times 10^{18}\text{ Hz}

. Equation:

c = f \lambda \implies \lambda = \frac{c}{f}

.

Electromagnetic waves propagate in a vacuum at speed

c

.

2

Calculate the wavelength

\lambda

.

\lambda = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{18}\text{ s}^{-1}} = 0.5 \times 10^{-10}\text{ m} = 5.0 \times 10^{-11}\text{ m}.

Proper division of scientific notation terms.

3

Classify the electromagnetic spectral region based on the calculated wavelength.

Wavelengths in the range

10^{-11}\text{ m}

to

10^{-8}\text{ m}

correspond to X-rays.

X-rays typically span frequencies between

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

and

3 \times 10^{19}\text{ Hz}

.

Key Concept

Wave equation for electromagnetic waves and spectral region identification

Question 58Question →

A longitudinal mechanical wave propagates through a gas at a speed of $340\text{ m/s}$ . If a particle in the gas completes $50$ full oscillations in $0.10\text{ s}$ , what is the distance between a compression and the immediate next rarefaction?

0.34\text{ m}

0.68\text{ m}

1.70\text{ m}

1.36\text{ m}

Show answer & explanation

Answer: $0.34\text{ m}$

Answer

The distance between a compression and the adjacent rarefaction is

0.34\text{ m}

.

The frequency of the particle oscillation is

f = 50 / 0.10 = 500\text{ Hz}

. Using the wave equation

v = f\lambda

, the wavelength

\lambda = 340 / 500 = 0.68\text{ m}

. Because a compression and the consecutive rarefaction represent half of a full wave cycle, the distance between them is

\lambda / 2 = 0.34\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of oscillation of the wave

f = \frac{\text{Number of oscillations}}{\text{Time interval}} = \frac{50}{0.10\text{ s}} = 500\text{ Hz}

Frequency is defined as the number of complete vibrations per unit time.

2

Calculate the wavelength using the wave equation

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

The fundamental wave equation relates speed, frequency, and wavelength.

3

Determine the distance between consecutive compression and rarefaction

d = \frac{\lambda}{2} = \frac{0.68\text{ m}}{2} = 0.34\text{ m}

In a longitudinal wave, a compression and the nearest rarefaction are separated by half a wavelength.

Key Concept

Relation between frequency, wavelength, wave speed, and structural intervals in longitudinal waves

Question 59Question →

Match each wave phenomenon on the left with its accurate wave classification and propagation mechanism on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Sound waves in air correspond to longitudinal mechanical waves requiring a material medium. Light waves traveling through a vacuum correspond to transverse electromagnetic waves needing no material medium. Ripples on a stretched string correspond to transverse mechanical waves with perpendicular particle vibration. Primary seismic P-waves correspond to longitudinal body waves with parallel particle movement.

Sound in air and seismic P-waves are longitudinal mechanical waves because medium particles vibrate parallel to the direction of propagation. Light is a transverse electromagnetic wave because it propagates via perpendicular field oscillations without needing a material medium. Waves on a string are transverse mechanical waves because string segments vibrate perpendicular to the length of the string.

Step-by-Step Solution

1

Analyze medium dependence

Sound, string waves, and seismic P-waves require a material medium (mechanical waves), whereas light waves travel through a vacuum (electromagnetic waves).

Waves are broadly categorized as mechanical or non-mechanical (electromagnetic) based on whether they require a physical medium for propagation.

2

Analyze particle motion relative to wave propagation

Sound and P-waves oscillate parallel to propagation (longitudinal); light waves and string vibrations oscillate perpendicular to propagation (transverse).

Direction of medium displacement relative to energy transmission determines whether a wave is transverse or longitudinal.

3

Match each physical scenario to its unique classification pair

Match sound wave to longitudinal mechanical, light to transverse electromagnetic, string wave to transverse mechanical, and seismic P-wave to longitudinal body wave.

Synthesize medium dependence and displacement direction for each specific physical system.

Key Concept

Classification of waves based on material medium dependence (mechanical vs. electromagnetic) and particle oscillation direction (transverse vs. longitudinal).

Question 60Question →

A sound wave traveling through air at a speed of $340\text{ m/s}$ has a wavelength of $0.85\text{ m}$ . Upon entering a second gaseous medium, its speed decreases to $272\text{ m/s}$ . What is the wavelength of the wave in the second medium?

0.68\text{ m}

1.06\text{ m}

0.85\text{ m}

0.54\text{ m}

Show answer & explanation

Answer: $0.68\text{ m}$

Answer

The wavelength of the wave in the second medium is

0.68\text{ m}

.

Because wave frequency is determined by the source, it stays constant at

400\text{ Hz}

across both media. Using

\lambda_2 = \frac{v_2}{f}

, the wavelength in the second medium is

\frac{272\text{ m/s}}{400\text{ Hz}} = 0.68\text{ m}

.

Step-by-Step Solution

1

Calculate the frequency of the wave in the first medium using the wave equation

v = f \lambda

.

f = \frac{v_1}{\lambda_1} = \frac{340\text{ m/s}}{0.85\text{ m}} = 400\text{ Hz}

.

The source determines the wave frequency, which remains unchanged when passing into a new medium.

2

Apply the wave equation with the constant frequency to find the new wavelength in the second medium.

\lambda_2 = \frac{v_2}{f} = \frac{272\text{ m/s}}{400\text{ Hz}} = 0.68\text{ m}

.

Wave speed changes in a new medium lead directly to proportional changes in wavelength since frequency is constant.

Key Concept

Invariance of wave frequency across media boundaries and application of the wave equation

v = f \lambda

.

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