Waves and Optics Alıştırma Soruları — JAMB UTME — Sayfa 4

Soru 61Soru →

A progressive wave traveling along a stretched string has a frequency of $250\text{ Hz}$ and a speed of $300\text{ m/s}$ . What is the minimum distance, in meters, between two points on the string that differ in phase by $\frac{\pi}{3}\text{ rad}$ ?

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Cevap: 0.2

Cevap

The minimum distance between the two points is

0.2\text{ m}

.

The wavelength is found using

\lambda = \frac{v}{f} = \frac{300}{250} = 1.2\text{ m}

. Substituting

\lambda = 1.2\text{ m}

and

\Delta \phi = \frac{\pi}{3}\text{ rad}

into the phase difference formula

\Delta \phi = \frac{2\pi \Delta x}{\lambda}

gives

\Delta x = \frac{(\pi / 3) \times 1.2}{2\pi} = 0.2\text{ m}

.

Adım Adım Çözüm

1

Calculate the wavelength ($\lambda$) from wave speed ($v$) and frequency ($f$)

$\lambda = \frac{300}{250} = 1.2\text{ m}$

The fundamental wave equation relates wave speed, frequency, and wavelength as $v = f\lambda$.

2

Apply the phase difference formula to solve for spatial separation ($\Delta x$)

$\Delta x = \frac{\Delta \phi \cdot \lambda}{2\pi} = \frac{(\pi / 3) \cdot 1.2}{2\pi} = 0.2\text{ m}$

A full cycle of $2\pi\text{ radians}$ corresponds to a spatial displacement of one wavelength ($\lambda$).

Anahtar Kavram

Relationship between phase difference and spatial displacement

Soru 62Soru →

A progressive transverse wave traveling in Medium 1 is described by the equation $y = 0.04 \sin(200\pi t - 5\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. If the wave enters Medium 2 where its propagation speed drops to $20\text{ m/s}$ , what is the wavelength of the wave in Medium 2?

0.40\text{ m}

0.20\text{ m}

0.10\text{ m}

0.80\text{ m}

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Cevap: $0.20\text{ m}$

Cevap

0.20\text{ m}

When a wave propagates from one medium into another, its frequency remains unchanged because frequency is determined solely by the periodic source. From the wave equation in Medium 1, the angular frequency

\omega = 200\pi\text{ rad/s}

, which corresponds to a frequency

f = \frac{\omega}{2\pi} = 100\text{ Hz}

. Using

v = f\lambda

in Medium 2 with

v = 20\text{ m/s}

, the new wavelength is

\lambda = \frac{20}{100} = 0.20\text{ m}

.

Adım Adım Çözüm

1

Extract the angular frequency from the wave equation.

Comparing

y = 0.04 \sin(200\pi t - 5\pi x)

to

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

gives

\omega = 200\pi\text{ rad/s}

.

The coefficient of

t

inside the sine term represents the angular frequency

\omega

.

2

Calculate the wave frequency.

f = \frac{\omega}{2\pi} = \frac{200\pi}{2\pi} = 100\text{ Hz}

.

Frequency is related to angular frequency by

\omega = 2\pi f

.

3

Apply the boundary condition for wave refraction.

Frequency in Medium 2 is

f = 100\text{ Hz}

.

Frequency is determined strictly by the wave source and remains invariant when passing between different media.

4

Calculate the wavelength in Medium 2 using the wave speed equation.

\lambda_2 = \frac{v_2}{f} = \frac{20\text{ m/s}}{100\text{ Hz}} = 0.20\text{ m}.

The fundamental wave equation

v = f\lambda

allows finding wavelength from speed and frequency.

Anahtar Kavram

Frequency Invariance Across Media Boundaries

Soru 63Soru →

A transverse progressive wave traveling along a stretched string is represented by the mathematical wave equation $y = 0.05 \sin(160\pi t - 8\pi x)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. What is the speed of propagation of the wave in $\text{m/s}$ ?

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Cevap: 20

Cevap

The speed of propagation of the wave is

20\text{ m/s}

.

Comparing the given equation

y = 0.05 \sin(160\pi t - 8\pi x)

with the standard wave equation

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

gives the angular frequency

\omega = 160\pi\text{ rad/s}

and wave number

k = 8\pi\text{ rad/m}

. Substituting these into

v = \frac{\omega}{k}

yields

v = \frac{160\pi}{8\pi} = 20\text{ m/s}

.

Adım Adım Çözüm

1

Compare the given wave equation with the standard progressive wave equation.

The standard form is

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

. Comparing parameters yields

\omega = 160\pi\text{ rad/s}

and

k = 8\pi\text{ rad/m}

.

Matching coefficients allows direct extraction of angular frequency and wave number.

2

Calculate the wave speed using the relation between angular frequency and wave number.

v = \frac{\omega}{k} = \frac{160\pi\text{ rad/s}}{8\pi\text{ rad/m}} = 20\text{ m/s}

.

Wave speed is defined as the ratio of angular frequency to wave number (

v = \lambda f = \frac{\omega}{k}

).

Anahtar Kavram

Wave Equation Parameter Extraction and Wave Speed Calculation

Soru 64Soru →

A progressive transverse wave propagating along a taut string is governed by the equation $y = 0.03 \sin(80\pi t - 5\pi x)$ , where $x$ and $y$ are in meters and $t$ is in seconds. What is the phase difference in radians between two points on the string separated by a distance of $0.1\text{ m}$ ?

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

\pi\text{ rad}

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

8\pi\text{ rad}

Cevabı ve açıklamayı göster

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

Cevap

The phase difference between the two points is

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

.

Comparing the given equation

y = 0.03 \sin(80\pi t - 5\pi x)

with the standard form

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

shows that

k = 5\pi\text{ rad/m}

. Substituting this wave number and the distance separation

\Delta x = 0.1\text{ m}

into

\Delta \phi = k \Delta x

yields

\Delta \phi = 5\pi \times 0.1 = 0.5\pi\text{ rad}

, which is equal to

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

.

Adım Adım Çözüm

1

Identify the wave number

k

from the given wave equation.

k = 5\pi\text{ rad/m}

The standard progressive wave equation is

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

, where the coefficient of

x

is

k

.

2

Calculate the phase difference

\Delta \phi

for a spatial separation

\Delta x = 0.1\text{ m}

.

\Delta \phi = k \Delta x = 5\pi \times 0.1 = 0.5\pi\text{ rad} = \frac{\pi}{2}\text{ rad}

Phase difference is directly proportional to spatial separation via

\Delta \phi = k \Delta x = \frac{2\pi}{\lambda}\Delta x

.

Anahtar Kavram

Phase Difference in Progressive Waves

Soru 65Soru →

In a Young's double-slit experiment, the separation between two narrow slits is $0.40\text{ mm}$ and the interference pattern is observed on a screen placed $1.20\text{ m}$ away from the slits. If the distance between consecutive bright fringes on the screen is $1.80\text{ mm}$ , what is the wavelength of the light used in nanometers ( $\text{nm}$ )?

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Cevap: 600

Cevap

The wavelength of the light used is

600\text{ nm}

.

Using the Young's double-slit fringe spacing relation

\beta = \frac{\lambda D}{d}

, rearranging yields

\lambda = \frac{\beta d}{D}

. Substituting

\beta = 1.80 \times 10^{-3}\text{ m}

,

d = 4.0 \times 10^{-4}\text{ m}

, and

D = 1.20\text{ m}

gives

\lambda = 6.00 \times 10^{-7}\text{ m}

, which corresponds to

600\text{ nm}

.

Adım Adım Çözüm

1

Convert given physical quantities into standard SI units (meters).

Slit separation

d = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}

, distance to screen

D = 1.20\text{ m}

, and fringe spacing

\beta = 1.80\text{ mm} = 1.80 \times 10^{-3}\text{ m}

.

Standard SI units ensure accuracy when applying wave speed and distance equations.

2

Write the Young's double-slit formula relating fringe width to wavelength.

\beta = \frac{\lambda D}{d}

This relationship defines the spatial period of interference fringes on a screen.

3

Rearrange the equation to isolate the wavelength

\lambda

.

\lambda = \frac{\beta d}{D}

The unknown parameter to solve for is the wavelength of the monochromatic source.

4

Substitute the numerical values and convert the final result to nanometers.

$\lambda = \frac{1.80 \times 10^{-3}\text{ m} \times 4.0 \times 10^{-4}\text{ m}}{1.20\text{ m}} = 6.00 \times 10^{-7}\text{ m} = 600\text{ nm}$

Multiply meters by

10^9

to express the wavelength in nanometers.

Anahtar Kavram

Young's Double-Slit Interference Fringe Spacing

Soru 66Soru →

A light ray travels from Medium X into Medium Y. The speed of light in Medium X is $1.50 \times 10^8\text{ m/s}$ and the speed of light in Medium Y is $2.50 \times 10^8\text{ m/s}$ . What is the sine of the critical angle for total internal reflection at the boundary between these two media?

0.60

1.67

0.50

0.40

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Cevap: $0.60$

Cevap

The sine of the critical angle for total internal reflection at the boundary is

0.60

.

Total internal reflection occurs when light moves from an optically denser medium to an optically less dense medium. By Snell's law,

n_X \sin\theta_c = n_Y \sin 90^\circ

, which gives

\sin\theta_c = n_Y / n_X

. Since refractive index is inversely proportional to speed (

n = c/v

),

\sin\theta_c = v_X / v_Y = (1.50 \times 10^8) / (2.50 \times 10^8) = 0.60

.

Adım Adım Çözüm

1

Relate speed of light in each medium to their respective refractive indices

n_X = \frac{c}{v_X}

and

n_Y = \frac{c}{v_Y}

Refractive index is defined as the ratio of light speed in vacuum to light speed in the medium.

2

Apply Snell's Law for the critical angle condition

\sin\theta_c = \frac{n_Y}{n_X}

At the critical angle, the angle of refraction in the less dense medium is

90^\circ

(

\sin 90^\circ = 1

).

3

Substitute light speeds into the critical angle formula and calculate

\sin\theta_c = \frac{v_X}{v_Y} = \frac{1.50 \times 10^8\text{ m/s}}{2.50 \times 10^8\text{ m/s}} = 0.60

Substituting the expressions for

n_X

and

n_Y

simplifies the refractive index ratio to the direct ratio of speeds

v_X / v_Y

.

Anahtar Kavram

Critical angle and total internal reflection relation to speed of light in media

Soru 67Soru →

Monochromatic light of wavelength $600\text{ nm}$ is incident normally on a diffraction grating having $500\text{ lines per mm}$ . What is the angle of diffraction, in degrees, for the first-order principal maximum?

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Cevap: 17.5

Cevap

The angle of diffraction for the first-order principal maximum is

17.5^\circ

.

Using the grating equation

d \sin \theta = n \lambda

, the slit separation is

d = \frac{10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}

. For

n = 1

and

\lambda = 6.00 \times 10^{-7}\text{ m}

, we get

\sin \theta = \frac{6.00 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.30

. Taking

\arcsin(0.30)

gives approximately

17.5^\circ

.

Adım Adım Çözüm

1

Calculate the grating element (slit spacing)

d

d = 2.00 \times 10^{-6}\text{ m}

Grating spacing

d

is the reciprocal of the line density

N = 500\text{ lines/mm} = 500,000\text{ lines/m}

.

2

Apply the diffraction grating equation

d \sin \theta = n \lambda

\sin \theta = 0.30

For the first-order maximum (

n = 1

),

\sin \theta = \frac{1 \times 600 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.30

.

3

Find the angle

\theta

by taking the inverse sine

\theta = 17.5^\circ

\arcsin(0.30) \approx 17.46^\circ

, which rounds to

17.5^\circ

.

Anahtar Kavram

Diffraction Grating Equation for Principal Maxima

Tahmini Süre:1m 30s

Soru 68Soru →

An echo sounder on a fishing boat emits an acoustic pulse vertically downward toward the seabed and detects the reflected signal $0.80\text{ s}$ later. If the speed of sound in seawater is $1450\text{ m/s}$ , what is the depth of the sea in meters at this location?

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Cevap: 580

Cevap

The depth of the sea is

580\text{ m}

.

An echo signal travels to the reflecting surface and back, covering twice the depth (

2d = v \times t

). Substituting

v = 1450\text{ m/s}

and

t = 0.80\text{ s}

gives

2d = 1160\text{ m}

, so the seabed depth

d = 580\text{ m}

.

Adım Adım Çözüm

1

Identify the given physical quantities

Total travel time

t = 0.80\text{ s}

and speed of sound

v = 1450\text{ m/s}

.

An echo involves sound traveling to the seabed and back, so the recorded time represents a two-way journey.

2

Set up the distance equation for an echo

Total distance traveled by the sound pulse is

2d = v \times t

.

Sound travels to the sea floor and reflects back to the ship, covering a total distance equal to twice the depth.

3

Calculate the depth

d

d = \frac{1450 \times 0.80}{2} = 580\text{ m}

.

Dividing the total round-trip distance by 2 yields the one-way depth.

Anahtar Kavram

Calculation of distance using sound echoes in a medium

Soru 69Soru →

If a light ray undergoes successive reflections from two plane mirrors inclined at an angle $\theta$ to each other in a single plane, the total angle of deviation produced in the ray is independent of the initial angle of incidence at the first mirror.

Doğru

Yanlış

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Cevap: True

Cevap

The statement is TRUE. The total angle of deviation experienced by a ray after two successive reflections at inclined plane mirrors is

\delta = 360^\circ - 2\theta

, which depends only on the angle of inclination

\theta

and is independent of the initial angle of incidence.

The net angular deviation for a ray undergoing two successive reflections at plane mirrors inclined at angle

\theta

is

\delta = 360^\circ - 2\theta

. Since the initial angle of incidence

i_1

cancels out during geometric summation, the total deviation is completely independent of the angle of incidence.

Adım Adım Çözüm

1

Determine the deviation at the first mirror

\delta_1 = 180^\circ - 2i_1

For a single reflection at a plane mirror with angle of incidence

i_1

, the angle of deviation is

\delta_1 = 180^\circ - 2i_1

.

2

Express the angle of incidence at the second mirror in terms of inclination angle

\theta

i_2 = \theta - i_1

From the geometric construction of the ray path inside the triangle formed by the two mirror surfaces, the interior angle relationship gives

i_1 + i_2 = \theta

.

3

Calculate the total deviation after both reflections

\delta = \delta_1 + \delta_2 = (180^\circ - 2i_1) + (180^\circ - 2i_2) = 360^\circ - 2(i_1 + i_2) = 360^\circ - 2\theta

Summing the individual deviations eliminates the variable

i_1

, demonstrating that the total deviation depends only on the inclination angle

\theta

.

Anahtar Kavram

Total Angle of Deviation for Inclined Plane Mirrors

Soru 70Soru →

Two sound waves of frequencies $440\text{ Hz}$ and $445\text{ Hz}$ travel through air and superpose to produce beats. What is the resulting beat frequency, in hertz?

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Cevap: 5

Cevap

The beat frequency produced by the superposition of the two sound waves is

5\text{ Hz}

.

When two waves of slightly different frequencies interfere, periodic variations in sound intensity called beats occur. The number of beats heard per second is equal to the absolute difference between the frequencies of the two superposing waves:

f_{\text{beat}} = |f_2 - f_1| = |445\text{ Hz} - 440\text{ Hz}| = 5\text{ Hz}

.

Adım Adım Çözüm

1

Identify the frequencies of the interfering waves

f_1 = 440\text{ Hz}

and

f_2 = 445\text{ Hz}

Beat frequency is determined by the absolute difference between the individual wave frequencies.

2

Subtract the lower frequency from the higher frequency to find the beat frequency

f_{\text{beat}} = |445 - 440| = 5\text{ Hz}

The rate of periodic intensity variation (beat frequency) is governed by

f_{\text{beat}} = |f_2 - f_1|

.

Anahtar Kavram

Beat Frequency and Wave Superposition

Soru 71Soru →

A convex mirror forms an upright image that is $\frac{1}{3}$ the size of an object. When the object is moved $20\text{ cm}$ further away from the mirror, the size of the image becomes $\frac{1}{5}$ the size of the object. What is the radius of curvature of the mirror?

20\text{ cm}

10\text{ cm}

40\text{ cm}

5\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $20\text{ cm}$

Cevap

The radius of curvature of the convex mirror is

20\text{ cm}

.

For a convex mirror, the focal length is negative (

f = -f_0

). The magnification formula

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

for a virtual upright image gives

m = \frac{-f_0}{-f_0 - u} = \frac{f_0}{f_0 + u}

. For

m_1 = \frac{1}{3}

, we get

u_1 = 2f_0

. For

m_2 = \frac{1}{5}

, we get

u_2 = 4f_0

. The object displacement is

u_2 - u_1 = 2f_0 = 20\text{ cm}

, which gives

f_0 = 10\text{ cm}

. The radius of curvature is

R = 2f_0 = 20\text{ cm}

, making the choice stating

20\text{ cm}

correct.

Adım Adım Çözüm

1

Apply the magnification formula and sign convention for a convex mirror at the first position.

For a convex mirror,

f = -f_0

and the image is virtual (

v_1 = -v_{01}

). Magnification

m_1 = -\frac{v_1}{u_1} = \frac{v_{01}}{u_1} = \frac{1}{3}

, so

v_{01} = \frac{u_1}{3}

.

Convex mirrors always form virtual, upright, and diminished images.

2

Substitute

v_1

into the mirror equation for the first position to express

u_1

in terms of focal length magnitude

f_0

.

\frac{1}{-f_0} = \frac{1}{u_1} + \frac{1}{-v_{01}} = \frac{1}{u_1} - \frac{3}{u_1} = -\frac{2}{u_1} \implies u_1 = 2f_0

.

The mirror formula is

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

with cartesian sign conventions.

3

Repeat the mirror equation calculation for the second object position.

For

m_2 = \frac{1}{5}

,

v_{02} = \frac{u_2}{5}

. Substituting gives

\frac{1}{-f_0} = \frac{1}{u_2} - \frac{5}{u_2} = -\frac{4}{u_2} \implies u_2 = 4f_0

.

The second object position gives a magnification of

\frac{1}{5}

.

4

Use the known object shift distance to solve for

f_0

and radius of curvature

R

.

u_2 - u_1 = 20\text{ cm} \implies 4f_0 - 2f_0 = 20\text{ cm} \implies 2f_0 = 20\text{ cm} \implies f_0 = 10\text{ cm}

. Since

R = 2f_0

,

R = 20\text{ cm}

.

The distance between the two object positions is

20\text{ cm}

, and the radius of curvature of a spherical mirror is twice its focal length.

Anahtar Kavram

Spherical Mirror Formula and Sign Conventions for Convex Mirrors

Soru 72Soru →

An optical communication sensor operates using an electromagnetic wave with a frequency of $1.5 \times 10^{14}\text{ Hz}$ in a vacuum. Given that the speed of light in vacuum is $3.0 \times 10^8\text{ m/s}$ , what is the wavelength of this electromagnetic radiation?

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

5.0 \times 10^5\text{ m}

4.5 \times 10^{22}\text{ m}

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

Cevabı ve açıklamayı göster

Cevap: $2.0 \times 10^{-6}\text{ m}$

Cevap

The wavelength of the electromagnetic wave is

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

.

Using the electromagnetic wave relation

c = f\lambda

, dividing the speed of light (

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

) by the given frequency (

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

) correctly gives

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

.

Adım Adım Çözüm

1

Identify given quantities and formula

Speed of light

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

, frequency

f = 1.5 \times 10^{14}\text{ Hz}

, wave equation

c = f \lambda

The fundamental wave equation relates wave speed, frequency, and wavelength for all electromagnetic waves.

2

Rearrange formula to solve for wavelength

\lambda

\lambda = \frac{c}{f}

Isolating the target unknown variable allows direct calculation.

3

Substitute values and compute

$\lambda = \frac{3.0 \times 10^8}{1.5 \times 10^{14}} = 2.0 \times 10^{-6}\text{ m}$

Dividing the coefficients (

3.0 / 1.5 = 2.0

) and subtracting the powers of ten (

8 - 14 = -6

) yields the accurate wavelength.

Anahtar Kavram

Wave Equation for Electromagnetic Waves

Soru 73Soru →

Arrange the following types of electromagnetic radiation in order of increasing frequency (from lowest frequency to highest frequency).

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct sequence of electromagnetic radiations in order of increasing frequency is: Microwaves, Infrared radiation, Ultraviolet radiation, and Gamma rays.

Electromagnetic waves propagate at the constant speed

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

in a vacuum. Frequency increases as wavelength decreases across the spectrum. Microwaves have the longest wavelength and lowest frequency among the choices, followed by infrared radiation, ultraviolet radiation, and finally gamma rays, which possess the shortest wavelength and highest frequency.

Adım Adım Çözüm

1

Recall the arrangement of the electromagnetic spectrum in terms of frequency and wavelength.

In the electromagnetic spectrum, frequency increases in the order: Radio waves

\rightarrow

Microwaves

\rightarrow

Infrared

\rightarrow

Visible light

\rightarrow

Ultraviolet

\rightarrow

X-rays

\rightarrow

Gamma rays.

Electromagnetic wave energy

E = hf

and frequency

f = \frac{c}{\lambda}

increase as wavelength decreases.

2

Identify the relative position of each given radiation type along the frequency scale.

Microwaves (

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

) < Infrared (

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

) < Ultraviolet (

10^{15} - 10^{16}\text{ Hz}

) < Gamma rays (

>10^{19}\text{ Hz}

).

Comparing their characteristic frequency ranges determines their exact position in the sequence.

3

Order the items from lowest to highest frequency.

1st: Microwaves, 2nd: Infrared radiation, 3rd: Ultraviolet radiation, 4th: Gamma rays.

This sequence satisfies the requirement of strictly increasing frequency.

Anahtar Kavram

Electromagnetic Spectrum Ordering by Frequency and Wavelength

Soru 74Soru →

An FM radio station transmits electromagnetic waves at a frequency of $1.0 \times 10^8\text{ Hz}$ in a vacuum. Given that the speed of light in a vacuum is $3.0 \times 10^8\text{ m/s}$ , what is the wavelength of these radio waves in meters?

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Cevap: 3

Cevap

The wavelength of the radio waves is

3.0\text{ m}

.

Applying the wave equation

c = f\lambda

, rearranging to

\lambda = \frac{c}{f}

, and substituting

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

and

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

gives a wavelength of

3.0\text{ m}

.

Adım Adım Çözüm

1

Identify given parameters and key formula

Speed of light

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

, frequency

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

, wave equation

c = f\lambda

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

2

Rearrange for wavelength and substitute given values

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

Dividing the speed of propagation by the wave frequency yields the spatial wavelength.

Anahtar Kavram

Wave equation relating speed, frequency, and wavelength of electromagnetic waves

Soru 75Soru →

A monochromatic beam of light with a wavelength of $600\text{ nm}$ in air enters a glass block of refractive index $1.50$ . What is the wavelength of the light inside the glass block in nanometers ( $\text{nm}$ )?

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Cevap: 400

Cevap

The wavelength of the light inside the glass block is

400\text{ nm}

.

The speed and wavelength of light both decrease by a factor of the refractive index

n

when entering a medium from air, while the frequency stays constant (

v = f\lambda

). Thus,

\lambda = \frac{\lambda_0}{n} = \frac{600}{1.50} = 400\text{ nm}

.

Adım Adım Çözüm

1

Identify the relevant formula connecting refractive index and wavelength

\lambda = \frac{\lambda_0}{n}

When light passes from air into a medium, its frequency remains unchanged while its speed and wavelength decrease proportionally by a factor of the refractive index

n

.

2

Substitute the given values into the equation

\lambda = \frac{600\text{ nm}}{1.50}

The wavelength in air is

600\text{ nm}

and the refractive index of glass is

1.50

.

3

Perform the calculation

\lambda = 400\text{ nm}

Dividing

600

by

1.50

yields

400

.

Anahtar Kavram

Refraction and Wavelength Change in a Medium

Soru 76Soru →

A progressive wave propagating through an initial elastic medium is represented by the equation $y = 0.08 \sin(120\pi t - 4\pi x)$ , where $x$ and $y$ are measured in meters and $t$ is in seconds. Upon crossing a boundary into a second medium, the wave speed decreases by $25\%$ . What is the wavelength of the wave in the second medium?

0.375\text{ m}

0.667\text{ m}

0.125\text{ m}

0.625\text{ m}

Cevabı ve açıklamayı göster

Cevap: $0.375\text{ m}$

Cevap

0.375\text{ m}

Comparing

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

with the standard wave equation

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

yields an angular frequency

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

and wave number

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

. The frequency is

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

, and the initial speed is

v_1 = \frac{120\pi}{4\pi} = 30\text{ m/s}

. When passing into a new medium, the frequency remains constant at

60\text{ Hz}

. A

25\%

reduction in speed gives

v_2 = 0.75 \times 30\text{ m/s} = 22.5\text{ m/s}

. The new wavelength is therefore

\lambda_2 = \frac{22.5\text{ m/s}}{60\text{ Hz}} = 0.375\text{ m}

.

Adım Adım Çözüm

1

Extract angular frequency and wave number from the wave equation

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

and

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

Comparing the given equation

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

with standard form

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

gives the values for

\omega

and

k

.

2

Calculate the frequency and initial wave speed in medium 1

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

and

v_1 = \frac{\omega}{k} = \frac{120\pi}{4\pi} = 30\text{ m/s}

Frequency is related to angular frequency by

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

, and wave velocity is

v = \frac{\omega}{k}

.

3

Determine the speed in medium 2 and apply the principle of constant frequency

v_2 = 30 \times 0.75 = 22.5\text{ m/s}

and

f_2 = f_1 = 60\text{ Hz}

Wave frequency depends solely on the source and remains unchanged across boundary refraction, whereas speed decreases by

25\%

.

4

Calculate the new wavelength in medium 2

\lambda_2 = \frac{v_2}{f} = \frac{22.5}{60} = 0.375\text{ m}

Using the wave equation relationship

\lambda = \frac{v}{f}

.

Anahtar Kavram

Wave refraction across media boundaries and mathematical wave equation parameter matching

Tahmini Süre:2m 0s

Soru 77Soru →

Match each physical wave propagation scenario on the left with its correct classification and particle vibration characteristics on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Oscillating electric charge in a vacuum matches non-mechanical wave with perpendicular field oscillations; pressure pulse in argon gas matches mechanical longitudinal wave with parallel oscillations; shear pulse on a stretched string matches mechanical transverse wave with perpendicular particle oscillations; water ripple on deep water matches mechanical surface wave with combined circular motion.

The correct pairings align each physical wave scenario with its underlying propagation mechanics. The oscillating charge producing field variations in a vacuum represents a non-mechanical transverse electromagnetic wave. The pressure disturbance in argon represents a longitudinal mechanical wave because fluids transmit energy through density compressions. The pulse on a stretched string is a transverse mechanical wave enabled by string tension. Surface water ripples are two-dimensional interface waves exhibiting combined longitudinal and transverse particle movement.

Adım Adım Çözüm

1

Classify the disturbance in a vacuum (oscillating charge).

Identified as electromagnetic radiation, which is non-mechanical and transverse.

Electromagnetic waves do not require a material medium and involve mutually perpendicular field vectors perpendicular to propagation.

2

Analyze wave propagation through a gaseous medium (argon gas).

Identified as a purely longitudinal mechanical wave.

Gases lack shear strength and can only transmit mechanical energy through volume variations (compressions and rarefactions) parallel to the propagation vector.

3

Evaluate wave propagation along a stretched 1D solid string.

Identified as a transverse mechanical wave.

Tension in the solid string restores lateral shear displacements, causing medium elements to vibrate perpendicular to the string length.

4

Determine the nature of surface waves on liquid interfaces.

Identified as a mechanical surface wave with orbital motion.

Interface boundaries support gravity and surface tension restoring forces, producing elliptical/circular particle trajectories combining transverse and longitudinal components.

Anahtar Kavram

Classification of waves by medium requirement (mechanical vs. electromagnetic) and directional displacement mode (transverse, longitudinal, and surface orbital motion).

Tahmini Süre:2m 0s

Soru 78Soru →

Which of the following types of waves requires a physical material medium for its propagation?

Sound waves

Radio waves

Infrared rays

Gamma rays

Cevabı ve açıklamayı göster

Cevap: Sound waves

Cevap

Sound waves are mechanical waves and require a material medium to propagate.

Sound waves are mechanical longitudinal waves. They propagate through compressions and rarefactions of particles in a medium, meaning they cannot travel through a vacuum.

Adım Adım Çözüm

1

Classify waves based on their propagation requirements.

Waves are classified broadly into mechanical waves (which require a medium) and electromagnetic waves (which do not require a medium).

Understanding the fundamental distinction between mechanical and electromagnetic propagation.

2

Identify the nature of each given wave option.

Radio waves, infrared rays, and gamma rays are electromagnetic waves. Sound waves are mechanical waves.

Mechanical waves depend on the elastic interactions of physical particles to transmit energy.

Anahtar Kavram

Classification of waves into mechanical and electromagnetic based on medium requirement

Tahmini Süre:45s

Soru 79Soru →

A transverse mechanical wave propagates along a stretched string. If the maximum speed of an oscillating particle of the string is equal to the speed of wave propagation, what is the ratio of the wave's amplitude to its wavelength?

2\pi

\frac{1}{2\pi}

\frac{1}{\pi}

1

Cevabı ve açıklamayı göster

Cevap: $\frac{1}{2\pi}$

Cevap

The ratio of the wave's amplitude to its wavelength is

\frac{1}{2\pi}

.

For a transverse wave, the maximum speed of particle oscillation is

v_{p,\text{max}} = A\omega

, while the propagation speed of the wave pattern through the medium is

v_w = \frac{\omega}{k}

. Setting these two speeds equal gives

A\omega = \frac{\omega}{k}

, which simplifies to

A k = 1

. Substituting

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

yields

A \left(\frac{2\pi}{\lambda}\right) = 1

, giving the ratio

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

.

Adım Adım Çözüm

1

Express the maximum speed of a particle in simple harmonic oscillation along a transverse wave.

The maximum particle speed is given by

v_{p,\text{max}} = A\omega

, where

A

is the amplitude and

\omega

is the angular frequency.

Particles in a transverse mechanical wave oscillate perpendicularly to wave propagation in simple harmonic motion.

2

Express the wave propagation speed in terms of angular frequency and wave number.

The wave propagation speed is

v_w = \frac{\omega}{k}

, where

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

is the angular wave number.

Wave propagation speed reflects the rate at which the wave energy and profile move through the medium.

3

Equate the maximum particle speed to the wave propagation speed and solve for

\frac{A}{\lambda}

.

A\omega = \frac{\omega}{k} \implies A k = 1 \implies A \left(\frac{2\pi}{\lambda}\right) = 1 \implies \frac{A}{\lambda} = \frac{1}{2\pi}

.

Setting

v_{p,\text{max}} = v_w

allows the angular frequency

\omega

to cancel out, directly linking amplitude and wavelength.

Anahtar Kavram

Transverse Wave Particle Oscillation vs. Wave Propagation Speed

Tahmini Süre:2m 0s

Soru 80Soru →

A signal generator positioned inside a transparent, sealed glass chamber produces both sound waves and radio waves simultaneously. As an air pump gradually evacuates the chamber to create a vacuum, which of the following best describes what happens to the propagation of these two waves?

Sound waves cease to propagate, whereas radio waves continue to propagate through the chamber.

Radio waves cease to propagate, whereas sound waves continue to propagate through the chamber.

Both sound waves and radio waves cease to propagate completely once a vacuum is achieved.

Both sound waves and radio waves propagate faster as the density of the air medium decreases.

Cevabı ve açıklamayı göster

Cevap: Sound waves cease to propagate, whereas radio waves continue to propagate through the chamber.

Cevap

Sound waves cease to propagate because they are mechanical waves requiring a physical medium, whereas radio waves continue to propagate because they are electromagnetic waves capable of traveling through a vacuum.

Sound waves are mechanical waves that rely on the vibration of medium particles (such as air molecules) to propagate energy. In contrast, radio waves are electromagnetic waves composed of mutually perpendicular oscillating electric and magnetic fields, allowing them to propagate efficiently through both matter and vacuum. Thus, evacuating air stops sound transmission while radio transmission continues.

Adım Adım Çözüm

1

Classify the wave types produced by the generator

Sound waves are mechanical waves (longitudinal), while radio waves are electromagnetic waves (transverse).

Wave propagation mechanisms depend fundamental on whether a wave is mechanical or electromagnetic.

2

Determine medium requirements for each wave class

Mechanical waves require elastic material particles to transfer vibrational energy; electromagnetic waves consist of oscillating electric and magnetic fields which travel without a physical medium.

In a vacuum, particle density approaches zero, preventing mechanical wave transmission.

3

Evaluate the effect of evacuating the glass chamber

As air is removed, sound wave propagation dies out, whereas radio wave propagation remains unimpeded.

The absence of matter blocks sound propagation while permitting electromagnetic wave transmission.

Anahtar Kavram

Classification of Waves by Medium Requirement (Mechanical vs. Electromagnetic Waves)

Waves and Optics

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