Waves and Optics Practice Questions — JAMB UTME — Page 7

Question 121Question →

A convex spherical mirror produces an image that is one-third the size of an object placed in front of it. If the distance of the object from the mirror is doubled, what is the new linear magnification of the image?

\frac{1}{5}

\frac{1}{4}

\frac{1}{6}

\frac{1}{2}

Show answer & explanation

Answer: $\frac{1}{5}$

Answer

The new linear magnification of the image is

\frac{1}{5}

.

For a convex mirror, the linear magnification

m

relates object distance

u

and focal magnitude

f

by

m = \frac{f}{f + u}

. Given

m = \frac{1}{3}

, solving

\frac{1}{3} = \frac{f}{f + u}

yields

u = 2f

. When the object distance is doubled to

u' = 4f

, the new magnification becomes

m' = \frac{f}{f + 4f} = \frac{1}{5}

.

Step-by-Step Solution

1

Express linear magnification in terms of object distance and focal length for a convex mirror

m = \frac{f}{f + u}

For a convex mirror, the focal length is negative under the Cartesian sign convention, making the virtual image distance

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

, so magnification

m = -\frac{v}{u} = \frac{f}{f + u}

.

2

Substitute the initial magnification

m = \frac{1}{3}

to express initial object distance

u

in terms of focal length

f

\frac{1}{3} = \frac{f}{f + u} \implies f + u = 3f \implies u = 2f

This establishes that the object was originally located at a distance equal to twice the focal length of the mirror.

3

Calculate the new object distance

u'

when distance is doubled

u' = 2u = 2(2f) = 4f

The problem states the object distance from the mirror is doubled.

4

Compute the new linear magnification

m'

m' = \frac{f}{f + u'} = \frac{f}{f + 4f} = \frac{f}{5f} = \frac{1}{5}

Substituting

u' = 4f

into the magnification formula yields the final reduced magnification.

Key Concept

Linear magnification and sign convention for convex mirrors

Estimated Time:2m 0s

Question 122Question →

Which type of electromagnetic radiation is primarily detected using a thermopile or a bolometer due to its heating effect?

Infrared radiation

Ultraviolet radiation

Gamma radiation

Radio waves

Show answer & explanation

Answer: Infrared radiation

Answer

Infrared radiation is the component of the electromagnetic spectrum primarily detected by a thermopile or bolometer.

Infrared radiation is absorbed by matter primarily as heat, raising the temperature of the absorber. Thermopiles measure this temperature rise by generating a small voltage across thermocouple junctions, making infrared radiation the primary band detected by thermopiles.

Step-by-Step Solution

1

Identify the characteristic mechanism of detection for thermopiles

Thermopiles convert thermal energy (heat) into an electrical voltage via the thermoelectric effect.

Understanding the working principle of the detector points to the radiation band with prominent heating properties.

2

Match the detector to the corresponding electromagnetic band

Infrared radiation causes significant thermal excitation and temperature rise upon absorption, making thermopiles ideal detectors.

Infrared rays are also known as heat waves because they transfer thermal energy efficiently.

Key Concept

Detection mechanisms of electromagnetic waves

Estimated Time:45s

Question 123Question →

An electromagnetic wave propagating in a vacuum has a frequency of $6.0 \times 10^{14}\text{ Hz}$ . It passes from the vacuum into a dense glass block with a refractive index of $1.50$ . 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 inside the glass block, and to which region of the electromagnetic spectrum does the wave belong based on its vacuum properties?

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

, Visible light

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

, Visible light

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

, Infrared

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

, Ultraviolet

Show answer & explanation

Answer: $3.33 \times 10^{-7}\text{ m}$ , Visible light

Answer

The wavelength of the wave inside the glass block is

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

, and the wave belongs to the Visible light region.

The vacuum wavelength of the wave is

\lambda_0 = c/f = 5.0 \times 10^{-7}\text{ m}

, placing it in the visible light spectrum. Upon entering the glass medium (

n = 1.50

), the wave frequency remains unchanged while its wavelength is compressed by the factor

n

, giving

\lambda = (5.0 \times 10^{-7})/1.50 = 3.33 \times 10^{-7}\text{ m}

.

Step-by-Step Solution

1

Calculate the vacuum wavelength of the electromagnetic wave

\lambda_0 = \frac{c}{f} = \frac{3.0 \times 10^8\text{ m/s}}{6.0 \times 10^{14}\text{ Hz}} = 5.0 \times 10^{-7}\text{ m}

The wave equation in vacuum relates wave speed, frequency, and wavelength by

c = f \lambda_0

.

2

Classify the electromagnetic spectral region

Visible light region

A vacuum wavelength of

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

(

500\text{ nm}

) falls within the visible light band (

400\text{ nm} - 700\text{ nm}

).

3

Calculate the wavelength inside the glass block

\lambda = \frac{\lambda_0}{n} = \frac{5.0 \times 10^{-7}\text{ m}}{1.50} = 3.33 \times 10^{-7}\text{ m}

When passing into a medium with refractive index

n

, frequency remains constant while the wave speed and wavelength are reduced by a factor of

n

.

Key Concept

Electromagnetic Wave Refraction and Spectrum Classification

Question 124Question →

A converging lens of focal length $20\text{ cm}$ is placed in thin coaxial contact with a diverging lens of focal length $50\text{ cm}$ . An object is placed $30\text{ cm}$ in front of this lens combination. What is the position and nature of the final image formed?

300\text{ cm}

in front of the combination (virtual image)

60\text{ cm}

behind the combination (real image)

27.3\text{ cm}

behind the combination (real image)

15.8\text{ cm}

behind the combination (real image)

Show answer & explanation

Answer: $300\text{ cm}$ in front of the combination (virtual image)

Answer

The image is virtual and formed

300\text{ cm}

in front of the lens combination.

Combining a converging lens (

f = +20\text{ cm}

) and a diverging lens (

f = -50\text{ cm}

) yields an effective focal length of

F = +\frac{100}{3}\text{ cm}

. Using the lens formula

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

with object distance

u = 30\text{ cm}

gives

\frac{1}{v} = \frac{3}{100} - \frac{1}{30} = -\frac{1}{300}\text{ cm}^{-1}

, resulting in

v = -300\text{ cm}

. The negative sign confirms the image is virtual and located

300\text{ cm}

in front of the combination.

Step-by-Step Solution

1

Calculate the effective focal length (

F

) of the two lenses in contact.

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = \frac{1}{20\text{ cm}} + \frac{1}{-50\text{ cm}} = \frac{5 - 2}{100\text{ cm}} = \frac{3}{100}\text{ cm}^{-1}

, so

F = +\frac{100}{3}\text{ cm}

.

Thin lenses in contact combine algebraically according to their optical powers, taking signs into account (positive for converging, negative for diverging).

2

Apply the lens formula to find the image distance (

v

).

\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{30} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{1}{30} = \frac{9 - 10}{300} = -\frac{1}{300}\text{ cm}^{-1}

.

Rearranging the thin lens equation allows us to solve for the image distance

v

given object distance

u = 30\text{ cm}

.

3

Interpret the sign and magnitude of

v

.

v = -300\text{ cm}

, which signifies a virtual image located

300\text{ cm}

in front of the lens combination (on the object side).

A negative image distance in the standard real-is-positive convention denotes a virtual image.

Key Concept

Combination of thin lenses in contact and lens sign conventions

Estimated Time:2m 0s

Question 125Question →

A myopic person has a far point of $52\text{ cm}$ from the eye. A corrective lens is placed $2\text{ cm}$ in front of the eye to enable the person to see distant objects clearly. A second thin converging lens of focal length $+20\text{ cm}$ is then placed in thin coaxial contact with this corrective lens. If an object is placed $50\text{ cm}$ in front of this combined lens system, what is the position and nature of the final image formed?

100\text{ cm}

behind the combined lens (real)

20\text{ cm}

behind the combined lens (real)

100\text{ cm}

in front of the combined lens (virtual)

20\text{ cm}

in front of the combined lens (virtual)

Show answer & explanation

Answer: $100\text{ cm}$ behind the combined lens (real)

Answer

100\text{ cm}

behind the combined lens (real image)

The corrective lens for short-sightedness (myopia) must be a diverging lens with a negative focal length. Accounting for the

2\text{ cm}

distance between the eye and the lens, the far point relative to the lens is

50\text{ cm}

, giving

f_1 = -50\text{ cm}

. Combining this lens with the converging lens (

f_2 = +20\text{ cm}

) yields a net power

\frac{1}{F} = -\frac{1}{50} + \frac{1}{20} = +\frac{3}{100}\text{ cm}^{-1}

, or

F = +\frac{100}{3}\text{ cm}

. Placing an object at

u = 50\text{ cm}

gives

\frac{1}{v} = \frac{3}{100} - \frac{1}{50} = +\frac{1}{100}\text{ cm}^{-1}

, which yields

v = +100\text{ cm}

. The positive sign confirms a real image formed

100\text{ cm}

behind the lens system.

Step-by-Step Solution

1

Calculate the focal length

f_1

of the corrective lens required for myopia.

f_1 = -50\text{ cm}

The far point distance from the lens is

52\text{ cm} - 2\text{ cm} = 50\text{ cm}

. A diverging (concave) lens is needed to form a virtual image of distant objects (

u = \infty

) at the far point (

v = -50\text{ cm}

).

2

Determine the focal length

F

of the two lenses in thin contact.

F = +\frac{100}{3}\text{ cm}

Using the combined focal length equation

\frac{1}{F} = \frac{1}{f_1} + \frac{1}{f_2} = -\frac{1}{50} + \frac{1}{20} = \frac{-2 + 5}{100} = +\frac{3}{100}\text{ cm}^{-1}

.

3

Apply the thin lens formula to locate the final image distance

v

for

u = 50\text{ cm}

.

v = +100\text{ cm}

Using

\frac{1}{F} = \frac{1}{u} + \frac{1}{v} \implies \frac{3}{100} = \frac{1}{50} + \frac{1}{v} \implies \frac{1}{v} = \frac{3}{100} - \frac{2}{100} = \frac{1}{100}\text{ cm}^{-1}

.

4

Determine the nature of the image from the sign of

v

.

Real image formed

100\text{ cm}

behind the combined lens system.

A positive value of image distance (

v > 0

) indicates a real image formed on the opposite side (behind) the lens system.

Key Concept

Thin lens combination and sight defect correction sign conventions

Estimated Time:3m 0s

Question 126Question →

A short-sighted person cannot see objects clearly beyond a distance of $80\text{ cm}$ . What type of lens, of what focal length and power, is required to correct this vision defect so that the person can view distant objects clearly?

Diverging lens of focal length

80\text{ cm}

and power

-1.25\text{ D}

Converging lens of focal length

80\text{ cm}

and power

+1.25\text{ D}

Converging lens of focal length

36.4\text{ cm}

and power

+2.75\text{ D}

Diverging lens of focal length

80\text{ cm}

and power

-0.0125\text{ D}

Show answer & explanation

Answer: Diverging lens of focal length $80\text{ cm}$ and power $-1.25\text{ D}$

Answer

Diverging lens of focal length

80\text{ cm}

and power

-1.25\text{ D}

For a myopic person with a far point at

80\text{ cm}

, parallel rays from a distant object (

u = \infty

) must be diverged so they appear to come from the far point (

v = -80\text{ cm} = -0.8\text{ m}

). Using

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

, we obtain

f = -0.8\text{ m} = -80\text{ cm}

. The negative focal length corresponds to a diverging lens, and its power is

P = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}

.

Step-by-Step Solution

1

Identify the optical requirements for correcting myopia (short-sightedness).

For a distant object (

u = \infty

), the corrective lens must form a virtual image at the eye's far point (

v = -80\text{ cm} = -0.8\text{ m}

).

Myopic eyes focus rays from infinity in front of the retina; placing a virtual image at the far point allows the eye lens to focus it correctly onto the retina.

2

Apply the thin lens formula to calculate the required focal length.

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{\infty} + \frac{1}{-0.8\text{ m}} = 0 - 1.25\text{ m}^{-1} \implies f = -0.8\text{ m} = -80\text{ cm}$.

The negative sign indicates that a concave (diverging) lens is required.

3

Calculate the power of the corrective lens in dioptres.

$P = \frac{1}{f\text{ (in metres)}} = \frac{1}{-0.8\text{ m}} = -1.25\text{ D}$.

Lens power in dioptres (D) is the reciprocal of the focal length expressed in meters.

Key Concept

Correction of Myopia (Short-Sightedness) using Diverging Lenses

Question 127Question →

Arrange the following regions of the electromagnetic spectrum in order of decreasing wavelength (from longest wavelength to shortest wavelength).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence from longest to shortest wavelength is: Microwaves, Infrared radiation, Ultraviolet radiation, and Gamma rays.

The electromagnetic spectrum ordered by decreasing wavelength (longest to shortest) follows the sequence: radio waves/microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. Thus, microwaves come first with the longest wavelength, followed by infrared, ultraviolet, and finally gamma rays with the shortest wavelength.

Step-by-Step Solution

1

Recall the wave equation

c = f \lambda

connecting frequency (

f

) and wavelength (

\lambda

) for electromagnetic waves in a vacuum.

Wavelength is inversely proportional to frequency and photon energy.

Since the speed of light

c

is constant, waves with lower frequencies have longer wavelengths.

2

Identify the relative wavelengths of each specified region of the electromagnetic spectrum.

Microwaves (

10^{-3}\text{ m} - 10^{-1}\text{ m}

) > Infrared (

7 \times 10^{-7}\text{ m} - 10^{-3}\text{ m}

) > Ultraviolet (

10^{-8}\text{ m} - 4 \times 10^{-7}\text{ m}

) > Gamma rays (

< 10^{-11}\text{ m}

).

Microwaves sit near the radio end of the spectrum, while gamma rays lie at the extreme high-energy end.

3

Sequence the items from longest wavelength to shortest wavelength.

Microwaves

\rightarrow

Infrared radiation

\rightarrow

Ultraviolet radiation

\rightarrow

Gamma rays.

This arranges the waves in strict order of decreasing wavelength.

Key Concept

Electromagnetic spectrum wavelength and frequency hierarchy

Question 128Question →

An electromagnetic microwave signal used in telecommunication has a wavelength of $0.02\text{ m}$ in a vacuum. Given that the speed of light in a vacuum is $3.0 \times 10^8\text{ m/s}$ , calculate the frequency of the signal in gigahertz ( $\text{GHz}$ ).

Show answer & explanation

Answer: 15

Answer

The frequency of the microwave signal is

15\text{ GHz}

.

Using the electromagnetic wave equation

c = f \lambda

, the frequency in Hz is calculated as

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.02\text{ m}} = 1.5 \times 10^{10}\text{ Hz}

. Dividing by

10^9

to convert into gigahertz gives

15\text{ GHz}

.

Step-by-Step Solution

1

Identify the given parameters and formula.

Speed of light

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

, wavelength

\lambda = 0.02\text{ m}

, and wave equation

c = f \lambda

.

Electromagnetic waves propagate at speed

c

in a vacuum, relating frequency and wavelength.

2

Calculate the frequency in Hertz (Hz).

f = \frac{3.0 \times 10^8}{0.02} = 1.5 \times 10^{10}\text{ Hz}

.

Dividing the wave speed by the wavelength yields the frequency.

3

Convert the unit from Hz to GHz.

\frac{1.5 \times 10^{10}\text{ Hz}}{10^9\text{ Hz/GHz}} = 15\text{ GHz}

.

One gigahertz (

1\text{ GHz}

) equals

10^9\text{ Hz}

.

Key Concept

Relationship between speed of light, frequency, and wavelength (

c = f \lambda

) for electromagnetic radiation.

Question 129Question →

A thin converging lens forms a real image of an object on a screen placed $60\text{ cm}$ from the lens. If the object is located $30\text{ cm}$ in front of the lens, what is the focal length of the lens in centimeters ( $\text{cm}$ )?

Show answer & explanation

Answer: 20

Answer

The focal length of the converging lens is

20\text{ cm}

.

Using the thin lens formula

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

with an object distance

u = 30\text{ cm}

and a real image distance

v = 60\text{ cm}

yields

\frac{1}{f} = \frac{1}{30} + \frac{1}{60} = \frac{1}{20}

, giving

f = 20\text{ cm}

.

Step-by-Step Solution

1

Identify given parameters and apply correct sign conventions

u = +30\text{ cm}

(real object) and

v = +60\text{ cm}

(real image on screen)

In thin lens calculations for real objects and images formed on screens, both distances are positive.

2

Substitute parameters into the thin lens formula

\frac{1}{f} = \frac{1}{30} + \frac{1}{60}

The thin lens equation relates focal length

f

, object distance

u

, and image distance

v

.

3

Perform fraction addition and solve for focal length

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

Taking the common denominator gives

\frac{1}{20}\text{ cm}^{-1}

, which yields

f = 20\text{ cm}

.

Key Concept

Thin Lens Formula for Real Image Formation

Question 130Question →

The electromagnetic spectrum consists of waves with varying frequencies, wavelengths, and photon energies, each associated with distinct physical detection mechanisms and applications. Consider the following types of electromagnetic radiation:

I. Radiation emitted by warm bodies, primarily detected using a thermopile.
II. Radiation utilized in radar systems and satellite communications.
III. Radiation emitted during nuclear decay processes, detected using a Geiger-Müller counter.
IV. Radiation responsible for sun tanning and detected by its ability to induce fluorescence on zinc sulfide screens.

Arrange these four types of electromagnetic radiation in order of increasing photon energy (from lowest photon energy to highest photon energy).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence in order of increasing photon energy is: Radiation utilized in radar systems (Microwaves) < Radiation emitted by warm bodies (Infrared) < Radiation causing sun tanning (Ultraviolet) < Radiation emitted during nuclear decay (Gamma rays).

Microwaves possess the lowest frequency among the four types, followed by infrared radiation, then ultraviolet radiation, and finally gamma rays which possess the highest frequency and photon energy.

Step-by-Step Solution

1

Identify the region of the electromagnetic spectrum corresponding to each property and detector described.

Item I corresponds to Infrared radiation; Item II corresponds to Microwaves; Item III corresponds to Gamma rays; Item IV corresponds to Ultraviolet radiation.

Thermopiles detect thermal radiation (IR); radar uses microwaves; Geiger-Müller counters detect nuclear ionizing radiation (Gamma rays); fluorescence on ZnS is caused by UV light.

2

Relate photon energy

E

to frequency

f

and wavelength

\lambda

using Planck's relation

E = hf = \frac{hc}{\lambda}

.

Photon energy is directly proportional to frequency (

E \propto f

) and inversely proportional to wavelength (

E \propto \frac{1}{\lambda}

).

Higher frequency radiation consists of more energetic individual photons.

3

Sequence the identified electromagnetic waves from lowest frequency to highest frequency.

Microwaves (

f \approx 10^9 - 10^{11}\text{ Hz}

) < Infrared (

f \approx 10^{11} - 4 \times 10^{14}\text{ Hz}

) < Ultraviolet (

f \approx 7.5 \times 10^{14} - 3 \times 10^{16}\text{ Hz}

) < Gamma rays (

f > 10^{19}\text{ Hz}

).

This sequence reflects the fundamental order of increasing photon energy across the spectrum.

Key Concept

Electromagnetic Spectrum Spectral Regions, Detection Devices, and Photon Energy Ordering

Question 131Question →

Match each visual defect or optical condition listed in Column A with its corresponding cause and corrective lens in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Myopia pairs with rays focusing in front of the retina (diverging lens correction); Hypermetropia pairs with rays focusing behind the retina (converging lens correction); Astigmatism pairs with uneven corneal curvature (cylindrical lens correction); Presbyopia pairs with age-related loss of accommodation (bifocal lens correction).

Each defect of vision is accurately matched to its optical cause and standard corrective device: Myopia uses diverging lenses to push the image focal plane onto the retina, Hypermetropia uses converging lenses to pull the image forward onto the retina, Astigmatism uses cylindrical lenses for non-spherical corneal curves, and Presbyopia uses bifocal/converging lenses to correct age-related accommodation loss.

Step-by-Step Solution

1

Analyze Myopia

Myopia causes distant rays to focus before reaching the retina because the eye lens is overly converging or the eye focal length is too short; a diverging (concave) lens spreads rays to push the focal point back onto the retina.

Identify optical cause and lens remedy for short-sightedness.

2

Analyze Hypermetropia

Hypermetropia causes near rays to focus behind the retina; a converging (convex) lens bends incoming light rays inwards to bring the focal point onto the retina.

Identify optical cause and lens remedy for long-sightedness.

3

Analyze Astigmatism

Astigmatism arises from non-uniform curvature of the refracting surfaces, requiring a cylindrical lens with differential curvature along different planes.

Identify refractive error causing multiple focal planes.

4

Analyze Presbyopia

Presbyopia is due to age-induced stiffening of the eye lens and loss of ciliary accommodation power, which is managed using bifocal or converging lenses.

Distinguish physiological aging effects on focal accommodation.

Key Concept

Defects of Vision and Corrective Lenses

Estimated Time:1m 30s

Question 132Question →

Match each defect of vision on the left with its corresponding corrective optical lens on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Myopia matches with Concave (diverging) lens, Hypermetropia matches with Convex (converging) lens, Presbyopia matches with Bifocal lens, and Astigmatism matches with Cylindrical lens.

Each eye condition is paired with its specific optical correction: Myopia requires a concave lens to diverge light, Hypermetropia requires a convex lens to converge light, Presbyopia uses a bifocal lens to assist diminished accommodation, and Astigmatism relies on a cylindrical lens to correct asymmetrical curvature.

Step-by-Step Solution

1

Analyze Myopia (Short-sightedness)

Parallel rays focus in front of the retina due to an elongated eyeball or over-refractive lens.

Diverging (concave) lenses spread out incoming rays slightly before entering the eye so the focal point shifts back onto the retina.

2

Analyze Hypermetropia (Long-sightedness)

Light rays focus behind the retina due to a shortened eyeball or insufficient focal power.

Converging (convex) lenses provide additional converging power to focus rays directly on the retina.

3

Analyze Presbyopia

The eye lens loses elasticity with age, reducing its power to accommodate both near and far objects.

Bifocal lenses have two distinct focal lengths in a single glass unit to assist with both near and distant vision.

4

Analyze Astigmatism

Cornea or crystalline lens curvature is uneven along different axes, producing distorted vision.

Cylindrical lenses correct uneven refractive power by bending light along one axis without affecting the orthogonal axis.

Key Concept

Defects of Vision and Corrective Lenses

Estimated Time:45s

Question 133Question →

An object is placed at a distance of $15\text{ cm}$ in front of a concave mirror with a focal length of $10\text{ cm}$ . What is the distance of the image formed from the mirror?

30\text{ cm}

6\text{ cm}

25\text{ cm}

-30\text{ cm}

Show answer & explanation

Answer: $30\text{ cm}$

Answer

The distance of the image formed from the mirror is

30\text{ cm}

.

Applying the mirror formula

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

with

f = 10\text{ cm}

and

u = 15\text{ cm}

gives

\frac{1}{v} = \frac{1}{10} - \frac{1}{15} = \frac{1}{30}

, yielding an image distance of

30\text{ cm}

.

Step-by-Step Solution

1

Identify given quantities and signs

Focal length

f = 10\text{ cm}

and object distance

u = 15\text{ cm}

.

For a concave mirror, the real focus and real object distances are both positive.

2

Set up the mirror formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{10} = \frac{1}{15} + \frac{1}{v}

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

3

Solve for the image distance

v

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

Subtracting the reciprocals and taking the inverse gives the image position.

Key Concept

Concave Mirror Formula

Question 134Question →

A physics laboratory utilizes four specialized instruments to detect different regions of the electromagnetic spectrum: an aerial antenna, a thermopile, a photographic plate sensitive to sun-tanning radiation, and a Geiger-Müller tube.

Arrange these detectors in order of INCREASING frequency of the electromagnetic radiation they are primarily designed to detect (from lowest frequency to highest frequency).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order from lowest frequency to highest frequency is: Aerial antenna (Radio waves) → Thermopile (Infrared) → Photographic plate sensitive to sun-tanning radiation (Ultraviolet) → Geiger-Müller tube (Gamma rays).

The correct sequence arranges the instruments according to the increasing frequency of the radiation they detect. Radio waves (detected by an aerial antenna) have the lowest frequency, followed by infrared radiation (detected by a thermopile), ultraviolet radiation (detected by photographic plates sensitive to sun-tanning rays), and gamma rays (detected by a Geiger-Müller tube) which have the highest frequency.

Step-by-Step Solution

1

Identify the type of electromagnetic radiation detected by each device.

Aerial antenna detects radio waves; Thermopile detects infrared radiation; Photographic plate for tanning radiation detects ultraviolet radiation; Geiger-Müller tube detects gamma rays.

Each detector operates on specific physical properties characteristic of a particular band of the electromagnetic spectrum.

2

Recall the order of the electromagnetic spectrum in terms of frequency (

f

).

Radio waves (

< 10^9\text{ Hz}

) < Infrared (

10^{11} - 10^{14}\text{ Hz}

) < Ultraviolet (

10^{15} - 10^{17}\text{ Hz}

) < Gamma rays (

> 10^{19}\text{ Hz}

).

Frequency increases continuously across the spectrum from radio waves to gamma rays.

3

Sequence the detectors based on their associated radiation frequencies from lowest to highest.

Aerial antenna

\rightarrow

Thermopile

\rightarrow

Photographic plate sensitive to sun-tanning radiation

\rightarrow

Geiger-Müller tube.

This directly matches the increasing frequency order of radio waves, infrared, ultraviolet, and gamma rays.

Key Concept

Detection mechanisms and frequency distribution across the electromagnetic spectrum

Estimated Time:2m 0s

Question 135Question →

A swimming pool has an apparent depth of $1.8\text{ m}$ when viewed vertically from directly above. If the refractive index of water relative to air is $\frac{4}{3}$ , what is the real depth of the pool in meters?

Show answer & explanation

Answer: 2.4

Answer

The real depth of the pool is

2.4\text{ m}

.

The refractive index

n

of a medium is defined as the ratio of the real depth to the apparent depth (

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

). Multiplying the observed apparent depth of

1.8\text{ m}

by the refractive index

\frac{4}{3}

gives the true real depth of

2.4\text{ m}

.

Step-by-Step Solution

1

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

Refractive index

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

.

Light bending at the boundary causes an object submerged in a denser medium to appear closer to the surface.

2

Rearrange the equation to make Real Depth the subject.

\text{Real Depth} = n \times \text{Apparent Depth}.

To calculate the true depth from the observed apparent depth and the optical density of water.

3

Substitute the given values into the formula.

\text{Real Depth} = \frac{4}{3} \times 1.8\text{ m} = 2.4\text{ m}.

Multiplying the apparent depth by the refractive index yields the actual physical depth.

Key Concept

Refraction of Light and Real/Apparent Depth

Question 136Question →

An object is placed $15.0\text{ cm}$ in front of a thin diverging lens with a focal length of $10.0\text{ cm}$ . What is the image distance formed by the lens?

-6.0\text{ cm}

(

6.0\text{ cm}

on the same side as the object)

+30.0\text{ cm}

(

30.0\text{ cm}

on the opposite side of the object)

+6.0\text{ cm}

(

6.0\text{ cm}

on the opposite side of the object)

-30.0\text{ cm}

(

30.0\text{ cm}

on the same side as the object)

Show answer & explanation

Answer: $-6.0\text{ cm}$ ( $6.0\text{ cm}$ on the same side as the object)

Answer

The image distance is

-6.0\text{ cm}

, indicating a virtual image located

6.0\text{ cm}

in front of the lens on the same side as the object.

Using the thin lens equation

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

with

u = +15.0\text{ cm}

and

f = -10.0\text{ cm}

for the diverging lens gives

\frac{1}{v} = -\frac{1}{10} - \frac{1}{15} = -\frac{1}{6}

, resulting in

v = -6.0\text{ cm}

. The negative sign confirms the image is virtual and formed on the same side as the object.

Step-by-Step Solution

1

Identify the given optical parameters and apply the proper sign convention.

Object distance

u = +15.0\text{ cm}

; Focal length for a diverging lens

f = -10.0\text{ cm}

.

By the real-is-positive sign convention, real object distance

u

is positive, while the focal length

f

of a concave/diverging lens is strictly negative.

2

Set up the thin lens formula

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

and solve for

\frac{1}{v}

.

\frac{1}{-10.0} = \frac{1}{15.0} + \frac{1}{v} \implies \frac{1}{v} = -\frac{1}{10.0} - \frac{1}{15.0}

Isolating the reciprocal image distance term requires subtracting

\frac{1}{u}

from both sides.

3

Calculate the common denominator and evaluate

v

.

\frac{1}{v} = \frac{-3 - 2}{30.0} = -\frac{5.0}{30.0} = -\frac{1}{6.0} \implies v = -6.0\text{ cm}

Inverting the reciprocal yields the final signed image distance.

Key Concept

Thin Lens Formula and Sign Conventions for Diverging Lenses

Estimated Time:1m 15s

Question 137Question →

Match each physical modification of a vibrating string or air pipe system on the left with its corresponding effect on the system's frequency on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Quadrupling tension corresponds to increasing frequency by a factor of 2; quadrupling linear mass density corresponds to reducing frequency to half; doubling tension corresponds to increasing frequency by a factor of

\sqrt{2}

; transitioning a closed pipe from fundamental mode to first overtone corresponds to increasing frequency by a factor of 3.

Each physical modification correctly maps to its quantitative outcome based on wave mechanics: string frequency scales with

\sqrt{T}

and

1/\sqrt{\mu}

, while closed pipe overtones follow odd harmonic multipliers (

1, 3, 5, \dots

).

Step-by-Step Solution

1

Examine the fundamental frequency formula for a stretched string under tension:

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

.

Frequency is directly proportional to

\sqrt{T}

and inversely proportional to

\sqrt{\mu}

.

This establishes how changes in tension and mass per unit length scale the fundamental frequency.

2

Calculate scaling factors for the string modifications.

Quadrupling

T

multiplies frequency by

\sqrt{4} = 2

. Quadrupling

\mu

multiplies frequency by

1/\sqrt{4} = 0.5

. Doubling

T

multiplies frequency by

\sqrt{2}

.

Applying square roots to the parameter change factors gives the resultant frequency change.

3

Analyze harmonic ratios for air columns in pipes closed at one end.

The fundamental mode frequency is

f_1 = \frac{v}{4L}

. The first overtone is the third harmonic (

f_3 = \frac{3v}{4L} = 3f_1

).

Closed air columns produce only odd harmonics (

n = 1, 3, 5, \dots

).

Key Concept

Parameter scaling of transverse waves on stretched strings and harmonic modes in closed air columns

Question 138Question →

A concave shaving mirror has a radius of curvature of $60\text{ cm}$ . A person places their face in front of the mirror such that an upright image magnified $3$ times is formed. What is the distance of the face from the mirror, in centimeters?

Show answer & explanation

Answer: 20

Answer

The distance of the person's face from the mirror is

20\text{ cm}

.

For a concave mirror with a radius of curvature of

60\text{ cm}

, the focal length is

f = +30\text{ cm}

. An upright image is virtual, corresponding to a positive magnification

m = +3

. Since

m = -\frac{v}{u}

, the image distance is

v = -3u

. Substituting these into the mirror formula

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

gives

\frac{1}{30} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

, solving to

u = 20\text{ cm}

.

Step-by-Step Solution

1

Determine the focal length of the concave mirror.

f = 30\text{ cm}

The focal length is half the radius of curvature (

f = \frac{r}{2} = \frac{60\text{ cm}}{2} = 30\text{ cm}

).

2

Express the image distance

v

in terms of the object distance

u

using the magnification relationship.

v = -3u

An upright image produced by a spherical mirror is virtual, so linear magnification

m = +3

. Using

m = -\frac{v}{u} = +3

, we obtain

v = -3u

.

3

Substitute

f

and

v

into the mirror equation to solve for

u

.

u = 20\text{ cm}

Applying the mirror formula

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

.

Key Concept

Mirror equation and sign conventions for virtual images formed by concave mirrors

Question 139Question →

A small object lies at the bottom of a transparent vessel containing two immiscible liquid layers, $A$ and $B$ . Layer $A$ (top) has a real thickness of $7.0\text{ cm}$ and a refractive index of $1.40$ . Layer $B$ (bottom) has a real thickness of $8.0\text{ cm}$ and a refractive index of $1.60$ . Calculate the apparent displacement of the object, in centimeters, when viewed vertically from directly above.

Show answer & explanation

Answer: 5

Answer

The apparent displacement of the object is

5.0\text{ cm}

.

For multiple refractive layers viewed normally, the total apparent depth is the sum of individual layer apparent depths:

\frac{7.0}{1.40} + \frac{8.0}{1.60} = 5.0 + 5.0 = 10.0\text{ cm}

. Subtracting this total apparent depth from the total real depth of

15.0\text{ cm}

gives an apparent vertical shift (displacement) of

5.0\text{ cm}

.

Step-by-Step Solution

1

Calculate the apparent depth of the top liquid layer (Layer A)

Apparent depth of Layer A =

5.0\text{ cm}

Apparent depth for a single medium is obtained by dividing real depth by its refractive index:

\frac{7.0}{1.40} = 5.0\text{ cm}

.

2

Calculate the apparent depth of the bottom liquid layer (Layer B)

Apparent depth of Layer B =

5.0\text{ cm}

Apparent depth for Layer B is

\frac{8.0}{1.60} = 5.0\text{ cm}

.

3

Calculate total real depth and total apparent depth

Total real depth =

15.0\text{ cm}

; Total apparent depth =

10.0\text{ cm}

Depths in composite media are additive.

4

Calculate vertical apparent displacement

Apparent displacement =

5.0\text{ cm}

Displacement is the difference between total real depth and total apparent depth:

15.0\text{ cm} - 10.0\text{ cm} = 5.0\text{ cm}

.

Key Concept

Refraction through composite media and vertical apparent displacement

Question 140Question →

A hypermetropic (far-sighted) person has a near point located at a distance of $100\text{ cm}$ from the eye. What power of spectacle lens, in dioptres, is required to enable this person to read print comfortably held at the standard near point of $25\text{ cm}$ ?

+3.0\text{ D}

-3.0\text{ D}

+5.0\text{ D}

+0.75\text{ D}

Show answer & explanation

Answer: $+3.0\text{ D}$

Answer

+3.0\text{ D}

To correct hypermetropia, a converging (convex) lens is required to bend incoming rays so that an object placed at the standard near point of

25\text{ cm}

(

+0.25\text{ m}

) forms a virtual image at the defective eye's near point of

100\text{ cm}

(

-1.0\text{ m}

). Substituting

u = +0.25\text{ m}

and

v = -1.0\text{ m}

into the power formula

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

yields

P = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

.

Step-by-Step Solution

1

Identify object distance (

u

) and required virtual image distance (

v

)

u = +25\text{ cm} = +0.25\text{ m}

and

v = -100\text{ cm} = -1.0\text{ m}

The lens must create a virtual image (on the same side as the object) at the person's actual near point when an object is placed at the standard reading distance.

2

Apply the thin lens formula to determine lens power

P

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

Power in dioptres is the reciprocal of the focal length in metres.

3

Calculate the numerical value of lens power

P = \frac{1}{0.25\text{ m}} + \frac{1}{-1.0\text{ m}} = +4.0\text{ D} - 1.0\text{ D} = +3.0\text{ D}

Adding the reciprocal quantities yields a positive focal power of

+3.0\text{ D}

.

Key Concept

Correction of Hypermetropia using Converging Lenses

Estimated Time:1m 30s

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