Waves and Optics Alıştırma Soruları — JAMB UTME — Sayfa 8

Soru 141Soru →

Match each optical instrument listed on the left with its corresponding lens configuration and image characteristics on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Astronomical Telescope pairs with the configuration having

f_o > f_e

forming an image at infinity; Simple Microscope pairs with a single converging lens forming an erect virtual image; Compound Microscope pairs with two converging lenses having

f_o < f_e

; and Projection Lantern pairs with a converging lens forming a real, inverted image on a screen.

Each instrument matches its distinct optical construction: telescopes use

f_o > f_e

for distant viewing at infinity, simple microscopes use a single convex lens for virtual magnifying, compound microscopes use

f_o < f_e

for double magnification of tiny objects, and projectors use a single convex lens to cast real images onto a distant surface.

Adım Adım Çözüm

1

Determine the lens setup and final image position of an astronomical telescope in normal adjustment.

The objective has a larger focal length than the eyepiece (

f_o > f_e

), and the final image is formed at infinity.

Telescopes gather light from distant objects, requiring a larger objective focal length for high angular magnification and comfortable viewing at infinity.

2

Determine the configuration of a simple microscope.

It consists of a single convex lens producing an erect, virtual, and magnified image.

When an object is placed within the focal length of a single convex lens, it acts as a magnifying glass.

3

Determine the focal length relationship of a compound microscope.

It uses two convex lenses where the objective focal length is shorter than the eyepiece focal length (

f_o < f_e

).

A very short objective focal length maximizes linear magnification of small, near objects before the eyepiece further magnifies the intermediate image.

4

Determine the type of image produced by a projection lantern (slide projector).

It forms a real, inverted, and magnified image on a screen.

Projecting images onto a screen requires a real image formed by a converging lens.

Anahtar Kavram

Optical Instrument Lens Configurations and Image Properties

Soru 142Soru →

An object is placed at a distance $u$ in front of a concave mirror of focal length $12\text{ cm}$ . A plane mirror is placed perpendicular to the principal axis at a distance of $32\text{ cm}$ in front of the concave mirror, between the object and the concave mirror. If the real image formed by the concave mirror coincides in space with the virtual image formed by the plane mirror, what is the value of $u$ in centimeters?

Cevabı ve açıklamayı göster

Cevap: 48

Cevap

The correct object distance

u

is

48\text{ cm}

.

The object is located at distance

u

from the concave mirror. With the plane mirror at

32\text{ cm}

from the concave mirror, the object distance from the plane mirror is

u - 32

. The plane mirror forms an image at distance

u - 32

behind itself, which corresponds to

32 - (u - 32) = 64 - u

from the concave mirror. Setting

v = 64 - u

in the mirror formula

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

gives

u^2 - 64u + 768 = 0

. Factoring yields

u = 48\text{ cm}

or

u = 16\text{ cm}

. Because the plane mirror is between the object and the concave mirror,

u > 32\text{ cm}

, so

u = 48\text{ cm}

.

Adım Adım Çözüm

1

Find the position of the image formed by the plane mirror in terms of

u

.

The object is at a distance

(u - 32)\text{ cm}

in front of the plane mirror. Its virtual image is formed

(u - 32)\text{ cm}

behind the plane mirror, which places it at

32 - (u - 32) = (64 - u)\text{ cm}

in front of the concave mirror.

A plane mirror forms an image behind it at a distance equal to the object distance in front of it.

2

Equate the image distance of the concave mirror

v

to the position of the plane mirror image.

v = 64 - u

The question states that the image formed by the concave mirror coincides in position with the image formed by the plane mirror.

3

Substitute

f = 12\text{ cm}

and

v = 64 - u

into the mirror equation.

\frac{1}{12} = \frac{1}{u} + \frac{1}{64 - u}

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

4

Solve the algebraic equation for

u

.

\frac{1}{12} = \frac{(64 - u) + u}{u(64 - u)} \implies 64u - u^2 = 768 \implies u^2 - 64u + 768 = 0

Combining fractions and multiplying across gives a quadratic equation in standard form.

5

Factor the quadratic equation and select the physical root.

(u - 48)(u - 16) = 0 \implies u = 48\text{ cm}

or

u = 16\text{ cm}

. Since

u > 32\text{ cm}

,

u = 48\text{ cm}

.

The plane mirror is situated between the object and the concave mirror at

32\text{ cm}

, so the object distance

u

must be greater than

32\text{ cm}

.

Anahtar Kavram

Image coincidence in combined plane and curved optical systems

Tahmini Süre:3m 0s

Soru 143Soru →

A pipe closed at one end vibrates in its first overtone. An open pipe vibrating in its fundamental mode has a frequency equal to that of the closed pipe. Neglecting end corrections, what is the ratio of the length of the open pipe to the length of the closed pipe?

2 : 3

3 : 2

4 : 3

1 : 3

Cevabı ve açıklamayı göster

Cevap: $2 : 3$

Cevap

The ratio of the length of the open pipe to the length of the closed pipe is

2 : 3

.

The first overtone of a closed pipe corresponds to its 3rd harmonic, giving a frequency of

f = \frac{3v}{4L_c}

. Equating this to the fundamental frequency of an open pipe (

f = \frac{v}{2L_o}

) yields

\frac{v}{2L_o} = \frac{3v}{4L_c}

, which simplifies to

\frac{L_o}{L_c} = \frac{2}{3}

or

2 : 3

.

Adım Adım Çözüm

1

Write the frequency formula for the first overtone of the closed pipe.

For a closed pipe of length

L_c

, odd harmonics are produced (

n = 1, 3, 5, \dots

). The first overtone is the third harmonic (

n = 3

):

f_{\text{closed}} = \frac{3v}{4L_c}

Closed air columns produce only odd harmonics, where the fundamental is

n=1

and the first overtone is

n=3

.

2

Write the fundamental frequency formula for the open pipe.

For an open pipe of length

L_o

, all harmonics are produced (

m = 1, 2, 3, \dots

). The fundamental frequency (

m = 1

) is:

f_{\text{open}} = \frac{v}{2L_o}

Open air columns have antinodes at both ends, yielding a fundamental wavelength of

\lambda = 2L_o

.

3

Equate the two frequencies and solve for the ratio

\frac{L_o}{L_c}

.

\frac{v}{2L_o} = \frac{3v}{4L_c}

Cancel the speed of sound

v

from both sides:

\frac{1}{2L_o} = \frac{3}{4L_c}

Cross-multiply:

6L_o = 4L_c \implies \frac{L_o}{L_c} = \frac{4}{6} = \frac{2}{3}

The question states that the frequencies of the two pipe configurations are equal.

Anahtar Kavram

Harmonics in Open and Closed Air Columns

Soru 144Soru →

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

+4.0\text{ D}

-4.0\text{ D}

+0.04\text{ D}

+25.0\text{ D}

Cevabı ve açıklamayı göster

Cevap: $+4.0\text{ D}$

Cevap

The optical power of the lens is

+4.0\text{ D}

.

The optical power

P

of a lens in dioptres (

\text{D}

) is calculated using

P = \frac{1}{f}

, where

f

is the focal length in metres. Converting

25\text{ cm}

to metres gives

0.25\text{ m}

. Since the lens is converging, its focal length is positive (

+0.25\text{ m}

). Thus,

P = \frac{1}{+0.25} = +4.0\text{ D}

.

Adım Adım Çözüm

1

Convert the focal length from centimetres to metres

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

The unit of optical power (dioptre,

\text{D}

) requires the focal length to be expressed in metres.

2

Apply the sign convention for a converging lens

f = +0.25\text{ m}

A converging (convex) lens has a real principal focus, so its focal length is positive by sign convention.

3

Calculate the optical power using the formula

P = \frac{1}{f}

P = \frac{1}{+0.25} = +4.0\text{ D}

Optical power is defined as the inverse of the focal length in metres.

Anahtar Kavram

Power of a Thin Lens

Tahmini Süre:45s

Soru 145Soru →

A light ray traveling inside a transparent glass prism of refractive index $1.50$ strikes the boundary with air. What is the sine of the critical angle ( $\sin C$ ) for total internal reflection to occur at this boundary?

0.67

1.50

0.50

0.87

Cevabı ve açıklamayı göster

Cevap: $0.67$

Cevap

The sine of the critical angle (

\sin C

) for total internal reflection at the glass-air boundary is

0.67

.

For light moving from a medium with refractive index

n

into air, total internal reflection occurs when the angle of incidence exceeds the critical angle

C

. The critical angle satisfies

\sin C = \frac{1}{n}

. Substituting

n = 1.50

gives

\sin C = \frac{1}{1.50} = 0.67

.

Adım Adım Çözüm

1

Identify the relationship between the refractive index (

n

) and the critical angle (

C

) when light travels from a medium to air.

\sin C = \frac{1}{n}

By Snell's law, at the critical angle of incidence, the angle of refraction in air is

90^\circ

(so

\sin 90^\circ = 1

).

2

Substitute the given refractive index (

n = 1.50

) into the formula.

\sin C = \frac{1}{1.50} = \frac{2}{3} \approx 0.67

Dividing

1

by

1.50

yields

0.666...

, which rounds to

0.67

.

Anahtar Kavram

Critical Angle and Total Internal Reflection

Tahmini Süre:45s

Soru 146Soru →

A thin converging lens of focal length $15\text{ cm}$ forms an erect image that is magnified three times. What is the distance of the object from the lens?

10\text{ cm}

20\text{ cm}

5\text{ cm}

45\text{ cm}

Cevabı ve açıklamayı göster

Cevap: $10\text{ cm}$

Cevap

10\text{ cm}

An erect image produced by a thin converging lens is virtual, which means the linear magnification is positive (

m = +3

) and the image distance is negative relative to the real object (

v = -3u

). Substituting

f = +15\text{ cm}

and

v = -3u

into the thin lens formula

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

gives

\frac{1}{15} = \frac{1}{u} - \frac{1}{3u} = \frac{2}{3u}

. Solving for

u

yields

u = 10\text{ cm}

.

Adım Adım Çözüm

1

Determine the nature of the image and establish the relationship between image distance and object distance

Since the image formed by a converging lens is erect, it must be virtual. Thus, linear magnification

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

, giving

v = -3u

.

A single convex lens produces an erect image only when the image is virtual, requiring a negative image distance under standard optical sign conventions.

2

Substitute focal length and image distance into the thin lens formula

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \implies \frac{1}{15} = \frac{1}{u} + \frac{1}{-3u}

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

3

Simplify the algebraic expression and solve for object distance u

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

Finding a common denominator allows direct solution for the object distance

u

.

Anahtar Kavram

Thin lens formula and sign conventions for virtual images

Soru 147Soru →

An object is placed $24\text{ cm}$ in front of a concave mirror with a focal length of $8\text{ cm}$ . What is the distance of the image from the mirror in $\text{cm}$ ?

Cevabı ve açıklamayı göster

Cevap: 12

Cevap

The distance of the image from the mirror is

12\text{ cm}

.

Applying the spherical mirror formula

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

with focal length

f = 8\text{ cm}

and object distance

u = 24\text{ cm}

yields

\frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{1}{12}

, giving an image distance of

12\text{ cm}

.

Adım Adım Çözüm

1

Identify the given parameters and select the appropriate relation.

Focal length

f = 8\text{ cm}

, object distance

u = 24\text{ cm}

, using the mirror equation

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

.

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

2

Substitute the given values into the formula and solve for

\frac{1}{v}

.

\frac{1}{8} = \frac{1}{24} + \frac{1}{v} \implies \frac{1}{v} = \frac{1}{8} - \frac{1}{24} = \frac{3-1}{24} = \frac{2}{24} = \frac{1}{12}

Subtracting

\frac{1}{24}

from

\frac{1}{8}

isolates the reciprocal of the image distance.

3

Invert the result to determine the image distance

v

.

v = 12\text{ cm}

Taking the reciprocal of

\frac{1}{12}

gives the image distance in centimeters.

Anahtar Kavram

Mirror Formula for Concave Mirrors

Tahmini Süre:45s

Soru 148Soru →

A light ray enters the first face of a glass prism surrounded by air. The prism has a refracting angle of $75^\circ$ and a refractive index of $\sqrt{2}$ . Inside the prism, the ray strikes the second refracting face at the critical angle for total internal reflection. Calculate the angle of incidence, in degrees, at the first face.

Cevabı ve açıklamayı göster

Cevap: 45

Cevap

The angle of incidence at the first face is

45^\circ

.

To find the angle of incidence at the first face, first determine the critical angle

C

at the second glass-air surface using

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

, which yields

C = 45^\circ

. Since the ray strikes the second face at this critical angle,

r_2 = 45^\circ

. Next, using the geometric relationship for a prism

A = r_1 + r_2

, the angle of refraction at the first surface is

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

. Finally, applying Snell's law at the first face gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \frac{\sqrt{2}}{2}

. Taking the inverse sine yields

i = 45^\circ

.

Adım Adım Çözüm

1

Find the critical angle at the second face

Critical angle

C = 45^\circ

, so

r_2 = 45^\circ

Light travels from glass to air at the critical angle, so

\sin C = \frac{1}{n} = \frac{1}{\sqrt{2}}

.

2

Determine the angle of refraction at the first face

r_1 = 30^\circ

The apex angle of a prism satisfies

A = r_1 + r_2

, hence

r_1 = A - r_2 = 75^\circ - 45^\circ = 30^\circ

.

3

Apply Snell's Law at the entry boundary

\sin i = \frac{\sqrt{2}}{2}

Refraction at the first surface gives

\sin i = n \sin r_1 = \sqrt{2} \sin 30^\circ = \sqrt{2} \times 0.5 = \frac{\sqrt{2}}{2}

.

4

Solve for the incident angle

i

i = 45^\circ

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

.

Anahtar Kavram

Refraction through a prism combined with total internal reflection critical angle condition

Soru 149Soru →

Match each optical instrument component or device on the left with its correct focal length requirement and image formation condition on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Compound Microscope Objective Lens pairs with very short focal length producing a real, inverted, and magnified intermediate image. Astronomical Telescope Objective Lens pairs with long focal length forming a real, inverted, and diminished image of a distant body. Simple Magnifying Glass pairs with single convex lens with object placed within focal length producing an erect, virtual, and magnified image. Slide / Film Projector Lens pairs with convex lens with object positioned between f and 2f producing a real, inverted, and enlarged image on a screen.

The matching correct pairs align each optical device with its precise optical parameter: microscope objectives utilize short focal lengths for high magnification of near objects, telescope objectives utilize long focal lengths for distant objects, simple magnifiers place objects closer than the focal point to form virtual erect images, and projectors place objects between one and two focal lengths to cast real enlarged images on a screen.

Adım Adım Çözüm

1

Analyze the objective lens of a compound microscope

It requires a very short focal length to achieve high linear magnification of near objects.

The specimen is placed just outside the focal point (

f < u < 2f

), producing a real, inverted, magnified image inside the tube.

2

Analyze the objective lens of an astronomical telescope

It requires a long focal length and large aperture.

Distant celestial bodies subtend tiny angles at the eye; a long focal length objective creates a larger real intermediate image for the eyepiece to magnify.

3

Analyze the ray path condition of a simple magnifying glass

The object is held within the principal focus (

u < f

).

Light rays emerging from the lens diverge, forming an enlarged, virtual, and erect image on the same side of the lens as the object.

4

Analyze the image projection setup in a slide projector

The slide is placed between

f

and

2f

of a converging lens.

According to the thin lens formula, placing an object between

f

and

2f

yields a real, inverted, magnified image beyond

2f

.

Anahtar Kavram

Operating principles and ray placement parameters of optical instruments

Soru 150Soru →

An astronomical telescope operating in normal adjustment consists of an objective lens with a focal length of $80\text{ cm}$ and an eyepiece with a focal length of $4\text{ cm}$ . What is the magnitude of the angular magnification produced by the telescope?

Cevabı ve açıklamayı göster

Cevap: 20

Cevap

The magnitude of the angular magnification produced by the telescope is 20.

The angular magnification

M

of an astronomical telescope in normal adjustment is defined as the ratio of the focal length of the objective lens

f_o

to the focal length of the eyepiece lens

f_e

, given by

M = \frac{f_o}{f_e}

. Substituting the given values

f_o = 80\text{ cm}

and

f_e = 4\text{ cm}

yields

M = \frac{80}{4} = 20

.

Adım Adım Çözüm

1

Identify the given optical parameters and formula for angular magnification.

Objective focal length

f_o = 80\text{ cm}

, Eyepiece focal length

f_e = 4\text{ cm}

. Formula:

M = \frac{f_o}{f_e}

.

For an astronomical telescope in normal adjustment, the light rays emerge parallel, and the angular magnification is given by the ratio of the focal length of the objective lens to that of the eyepiece.

2

Calculate the angular magnification value.

M = \frac{80\text{ cm}}{4\text{ cm}} = 20

.

Dividing the focal length of the objective lens by the focal length of the eyepiece yields the dimensionless magnification ratio.

Anahtar Kavram

Angular Magnification of an Astronomical Telescope in Normal Adjustment

Soru 151Soru →

Arrange the following electromagnetic radiation applications in order of increasing photon energy, starting from the radiation with the lowest energy to the one with the highest energy.

Öğeleri doğru sıraya koymak için sürükleyin

Cevabı ve açıklamayı göster

Cevap

The correct sequence from lowest to highest photon energy is: Radar waves (microwaves), radiant heat (infrared), sterilizing radiation (ultraviolet), and nuclear gamma emissions.

The photon energy of electromagnetic radiation is directly proportional to its frequency (

E = hf

). Microwaves have the lowest frequency among the listed types, followed by infrared radiation, then ultraviolet radiation, with gamma rays having the highest frequency and thus the greatest photon energy per photon.

Adım Adım Çözüm

1

Identify the region of the electromagnetic spectrum for each listed application.

Radar waves belong to microwaves; radiant heat corresponds to infrared radiation; sterilization uses ultraviolet light; nuclear emissions are gamma rays.

Connecting applications to their respective spectral regions is required to compare physical properties.

2

Recall the relationship between frequency and photon energy in the electromagnetic spectrum using

E = h f

.

Photon energy is directly proportional to frequency (

E \propto f

), meaning higher frequency waves carry greater energy per photon.

Understanding quantum photon energy helps determine the correct energy ranking.

3

Order the spectral regions from lowest frequency to highest frequency.

The order of increasing frequency (and energy) is: Microwaves < Infrared < Ultraviolet < Gamma rays.

This matches the physical ordering of the electromagnetic spectrum by increasing frequency.

Anahtar Kavram

Photon energy across the electromagnetic spectrum increases with increasing frequency (

E = h f

).

Soru 152Soru →

In a resonance tube experiment using a tuning fork of frequency $340\text{ Hz}$ , the first two consecutive resonance lengths of the air column closed at one end are $22\text{ cm}$ and $72\text{ cm}$ . If a pipe open at both ends with a physical length of $28\text{ cm}$ is operated in the same environment, what is its fundamental frequency when end corrections at both open ends are taken into account?

500\text{ Hz}

250\text{ Hz}

680\text{ Hz}

1000\text{ Hz}

Cevabı ve açıklamayı göster

Cevap: $500\text{ Hz}$

Cevap

The fundamental frequency of the open pipe is

500\text{ Hz}

.

The difference between successive resonant lengths in the closed tube gives half a wavelength (

L_2 - L_1 = 0.50\text{ m} \implies \lambda = 1.00\text{ m}

). Using the tuning fork frequency

340\text{ Hz}

, the speed of sound is

340\text{ m/s}

. The end correction is

e = \lambda / 4 - L_1 = 0.25\text{ m} - 0.22\text{ m} = 0.03\text{ m}

. For a pipe open at both ends, end corrections apply at both openings, making the effective length

L_{\text{eff}} = 0.28\text{ m} + 2(0.03\text{ m}) = 0.34\text{ m}

. Its fundamental frequency is

f_0 = v / (2 L_{\text{eff}}) = 340 / (2 \times 0.34) = 500\text{ Hz}

.

Adım Adım Çözüm

1

Determine the wavelength and speed of sound from the resonance tube data.

\lambda = 2(L_2 - L_1) = 2(0.72\text{ m} - 0.22\text{ m}) = 1.00\text{ m}

. Speed of sound

v = f \lambda = 340\text{ Hz} \times 1.00\text{ m} = 340\text{ m/s}

.

The distance between consecutive resonance positions in a closed pipe is equal to half a wavelength.

2

Calculate the end correction

e

of the tube.

L_1 + e = \frac{\lambda}{4} \implies 0.22\text{ m} + e = 0.25\text{ m} \implies e = 0.03\text{ m} = 3\text{ cm}

.

The first resonance of a pipe closed at one end occurs when the effective length equals one quarter of a wavelength.

3

Calculate the effective length

L_{\text{eff}}

of the open pipe.

L_{\text{eff}} = L + 2e = 28\text{ cm} + 2(3\text{ cm}) = 34\text{ cm} = 0.34\text{ m}

.

A pipe open at both ends requires an end-correction term added at each open boundary.

4

Calculate the fundamental frequency of the open pipe.

f_0 = \frac{v}{2 L_{\text{eff}}} = \frac{340\text{ m/s}}{2 \times 0.34\text{ m}} = 500\text{ Hz}

.

The fundamental wavelength of a pipe open at both ends is twice its effective length.

Anahtar Kavram

Resonance tube end correction and boundary conditions of open vs closed pipes

Tahmini Süre:2m 0s

Soru 153Soru →

A light wave travels through medium A at a speed of $2.25 \times 10^8\text{ m s}^{-1}$ and enters medium B, where its speed decreases to $1.50 \times 10^8\text{ m s}^{-1}$ . What is the value of the sine of the critical angle for total internal reflection between these two media, and in which medium must the light ray originate?

\frac{2}{3}

, originating in medium B

\frac{2}{3}

, originating in medium A

\frac{3}{2}

, originating in medium B

\frac{3}{2}

, originating in medium A

Cevabı ve açıklamayı göster

Cevap: $\frac{2}{3}$ , originating in medium B

Cevap

The sine of the critical angle is

\frac{2}{3}

, and the light ray must originate in medium B.

Total internal reflection occurs when light travels from an optically denser medium to an optically rarer medium. Since the speed of light is lower in medium B (

1.50 \times 10^8\text{ m s}^{-1}

) than in medium A (

2.25 \times 10^8\text{ m s}^{-1}

), medium B is the denser medium. The critical angle

C

satisfies

\sin C = \frac{v_{\text{dense}}}{v_{\text{rare}}} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}

. Therefore, the light must originate in medium B and the sine of the critical angle is

\frac{2}{3}

.

Adım Adım Çözüm

1

Determine the relative optical densities of medium A and medium B from wave speed

Medium B has a lower light speed (

1.50 \times 10^8\text{ m s}^{-1}

) than medium A (

2.25 \times 10^8\text{ m s}^{-1}

), so medium B is optically denser than medium A.

Refractive index is inversely proportional to wave speed (

n \propto \frac{1}{v}

).

2

Identify the required direction of light propagation for total internal reflection

The light ray must originate in medium B and travel toward medium A.

Total internal reflection occurs only when light attempts to pass from a medium of higher refractive index (denser) to a medium of lower refractive index (rarer).

3

Calculate the sine of the critical angle

\sin C = \frac{v_B}{v_A} = \frac{1.50 \times 10^8}{2.25 \times 10^8} = \frac{2}{3}

.

By Snell's law at the critical angle,

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

.

Anahtar Kavram

Conditions for Total Internal Reflection and Critical Angle calculation from wave speeds

Tahmini Süre:1m 30s

Soru 154Soru →

A satellite communication system transmits an ultra-high frequency electromagnetic wave with a wavelength of $0.05\text{ m}$ . If the wave travels at a speed of $3.0 \times 10^{8}\text{ m/s}$ in a vacuum, what is the frequency of the transmission in gigahertz ( $\text{GHz}$ )?

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Cevap: 6

Cevap

The frequency of the electromagnetic transmission is

6\text{ GHz}

.

Using the wave equation

c = f \lambda

, the frequency is

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.05\text{ m}} = 6.0 \times 10^9\text{ Hz}

. Converting to gigahertz (

1\text{ GHz} = 10^9\text{ Hz}

) yields

6\text{ GHz}

.

Adım Adım Çözüm

1

Identify the relationship between electromagnetic wave velocity, frequency, and wavelength.

c = f \lambda

, where

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

and

\lambda = 0.05\text{ m}

.

All electromagnetic waves travel at the speed of light

c

in a vacuum.

2

Rearrange the wave equation to isolate frequency (

f

).

f = \frac{c}{\lambda} = \frac{3.0 \times 10^8\text{ m/s}}{0.05\text{ m}} = 6.0 \times 10^9\text{ Hz}

.

Dividing the wave speed by the wavelength yields the frequency in hertz.

3

Convert the frequency into gigahertz (

\text{GHz}

).

\frac{6.0 \times 10^9\text{ Hz}}{10^9\text{ Hz/GHz}} = 6\text{ GHz}

.

The prefix giga (G) denotes a factor of

10^9

.

Anahtar Kavram

Electromagnetic wave propagation equation (

c = f \lambda

) and unit conversion

Tahmini Süre:1m 15s

Soru 155Soru →

A concave mirror produces a real image that is $3$ times the size of an object placed in front of it. If the distance between the object and its image is $40\text{ cm}$ , what is the focal length of the mirror in $\text{cm}$ ?

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Cevap: 15

Cevap

The focal length of the concave mirror is

15\text{ cm}

.

Using the magnification relation

v = 3u

and the object-image separation of

40\text{ cm}

, we obtain

3u - u = 40\text{ cm}

, which yields

u = 20\text{ cm}

and

v = 60\text{ cm}

. Substituting these distances into the mirror equation

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

gives

\frac{1}{f} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

, so

f = 15\text{ cm}

.

Adım Adım Çözüm

1

Relate the image distance

v

to the object distance

u

using the linear magnification formula

v = 3u

Since the mirror forms a real, magnified image that is 3 times the size of the object, linear magnification

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

.

2

Formulate an equation from the given object-to-image separation distance to solve for

u

and

v

u = 20\text{ cm}

and

v = 60\text{ cm}

The separation distance between the image and object is

v - u = 40\text{ cm}

. Substituting

v = 3u

yields

2u = 40\text{ cm} \implies u = 20\text{ cm}

and

v = 60\text{ cm}

.

3

Substitute the values of

u

and

v

into the mirror formula to compute the focal length

f

f = 15\text{ cm}

Applying

\frac{1}{f} = \frac{1}{u} + \frac{1}{v} = \frac{1}{20} + \frac{1}{60} = \frac{4}{60} = \frac{1}{15}

gives

f = 15\text{ cm}

.

Anahtar Kavram

Linear magnification and mirror formula for concave mirrors

Soru 156Soru →

A light ray strikes a plane mirror. Keeping the direction of the incident ray fixed, the mirror is rotated through an angle of $15^\circ$ about an axis lying in its plane. What is the angle of rotation, in degrees, of the reflected ray?

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Cevap: 30

Cevap

The reflected ray turns through an angle of

30^\circ

.

When a plane mirror is rotated through an angle

\theta

while keeping the incident ray direction constant, the normal turns by

\theta

. This changes the angle of incidence by

\theta

and the angle of reflection by

\theta

, causing the reflected ray to rotate by a total angle of

2\theta

. For a mirror rotation of

15^\circ

, the reflected ray rotates through

2 \times 15^\circ = 30^\circ

.

Adım Adım Çözüm

1

Analyze the effect of mirror rotation on the normal line

When the plane mirror rotates by

15^\circ

, the normal to the mirror surface also rotates by

15^\circ

.

The normal line is always perpendicular to the surface of the mirror.

2

Determine the change in the angle of incidence and reflection

The angle of incidence changes by

15^\circ

, so the angle of reflection relative to the new normal also changes by

15^\circ

.

According to the law of reflection, the angle of incidence equals the angle of reflection (

i = r

).

3

Calculate the total angular deviation of the reflected ray

The total shift of the reflected ray relative to its original path is

15^\circ + 15^\circ = 30^\circ

.

The rotation of the reflected ray is twice the angle of rotation of the mirror.

Anahtar Kavram

Rotation of Reflected Ray by a Plane Mirror

Soru 157Soru →

A glass prism with a refracting angle of $60^\circ$ has a refractive index of $\sqrt{2}$ . What is the angle of minimum deviation, in degrees, experienced by a light ray passing through this prism?

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Cevap: 30

Cevap

The angle of minimum deviation is

30^\circ

.

Using the prism minimum deviation relation

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

, substituting

n = \sqrt{2}

and

A = 60^\circ

yields

\sin\left((60^\circ + D_m)/2\right) = \frac{1}{\sqrt{2}}

. This gives

(60^\circ + D_m)/2 = 45^\circ

, so

D_m = 30^\circ

.

Adım Adım Çözüm

1

Apply the prism minimum deviation equation.

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

This relates refractive index, prism refracting angle, and minimum deviation angle.

2

Substitute the given values

A = 60^\circ

and

n = \sqrt{2}

.

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

Dividing the refracting angle

60^\circ

by 2 gives

30^\circ

for the denominator angle.

3

Calculate the numerator sine term.

\sin\left(\frac{60^\circ + D_m}{2}\right) = \frac{1}{\sqrt{2}}

Multiplying

\sqrt{2}

by

\sin(30^\circ) = 0.5

gives

\frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

.

4

Solve for the angle

D_m

.

D_m = 30^\circ

Since

\arcsin\left(\frac{1}{\sqrt{2}}\right) = 45^\circ

, we have

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

, leading to

60^\circ + D_m = 90^\circ

.

Anahtar Kavram

Minimum Deviation in Triangular Prisms

Soru 158Soru →

A block of transparent polymer with a refractive index of $1.25$ is placed over a small mark on a table. When viewed vertically from directly above, the mark appears to be shifted upward by $3.0\text{ cm}$ . What is the real thickness of the polymer block in centimeters?

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Cevap: 15

Cevap

The real thickness of the polymer block is

15.0\text{ cm}

.

Refraction at the interface produces an upward shift

s = d\left(1 - \frac{1}{n}\right)

. Substituting

s = 3.0\text{ cm}

and

n = 1.25

gives

3.0 = d(1 - 0.80) = 0.20d

, which solves to a real thickness

d = 15.0\text{ cm}

.

Adım Adım Çözüm

1

Relate apparent depth

d'

to real depth

d

using the refractive index.

d' = \frac{d}{n}

Light refracting at the surface of a denser medium causes the object to appear at a shallower depth.

2

Express the apparent displacement (upward shift)

s

in terms of

d

and

n

.

s = d - d' = d\left(1 - \frac{1}{n}\right)

The upward shift is the vertical distance between the true position and the apparent position.

3

Substitute

s = 3.0\text{ cm}

and

n = 1.25

into the equation and solve for

d

.

3.0 = d\left(1 - \frac{1}{1.25}\right) = d(1 - 0.80) = 0.20 d \implies d = 15.0\text{ cm}

Algebraic evaluation yields the exact real thickness of the polymer block.

Anahtar Kavram

Apparent depth and upward displacement due to refraction

Soru 159Soru →

An optical fiber consists of a core with a refractive index of $1.50$ surrounded by a cladding with a refractive index of $1.20$ . Which of the following correctly describes the direction a light ray must travel for total internal reflection to take place at the boundary, as well as the sine of the critical angle?

From core to cladding with

\sin C = 0.80

From cladding to core with

\sin C = 0.80

From core to cladding with

\sin C = 1.25

From cladding to core with

\sin C = 1.25

Cevabı ve açıklamayı göster

Cevap: From core to cladding with $\sin C = 0.80$

Cevap

Light must travel from the core to the cladding with

\sin C = 0.80

Total internal reflection takes place only when light travels from an optically denser medium (

n_1 = 1.50

) to an optically less dense medium (

n_2 = 1.20

). Applying Snell's law at the critical boundary gives

\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.50} = 0.80

.

Adım Adım Çözüm

1

Determine the direction requirement for Total Internal Reflection (TIR)

Light must travel from the denser medium (core,

n_1 = 1.50

) toward the less dense medium (cladding,

n_2 = 1.20

).

TIR only occurs when light attempts to pass into a medium of lower optical density so that the refracted ray bends away from the normal.

2

Calculate the sine of the critical angle

C

\sin C = \frac{n_2}{n_1} = \frac{1.20}{1.50} = 0.80

By Snell's law at the critical angle,

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

.

Anahtar Kavram

Total Internal Reflection and Critical Angle

Tahmini Süre:1m 30s

Soru 160Soru →

A thin converging lens with a focal length of $20\text{ cm}$ is placed in direct contact with a thin diverging lens with a focal length of $50\text{ cm}$ . What is the combined optical power of this two-lens system in dioptres?

+3.0\text{ D}

+7.0\text{ D}

-3.0\text{ D}

+0.03\text{ D}

Cevabı ve açıklamayı göster

Cevap: $+3.0\text{ D}$

Cevap

The combined optical power of the lens system is

+3.0\text{ D}

.

The individual power of the converging lens is

+5.0\text{ D}

and the power of the diverging lens is

-2.0\text{ D}

. Summing these values algebraically yields a net power of

+3.0\text{ D}

.

Adım Adım Çözüm

1

Convert the focal length of each lens from centimeters to meters and apply sign conventions.

Converging lens focal length

f_1 = +0.20\text{ m}

; Diverging lens focal length

f_2 = -0.50\text{ m}

.

Optical power in dioptres requires focal length to be expressed in meters, with converging lenses being positive and diverging lenses being negative.

2

Calculate the individual power of each lens.

P_1 = \frac{1}{+0.20} = +5.0\text{ D}

and

P_2 = \frac{1}{-0.50} = -2.0\text{ D}

.

Power of a thin lens is defined as

P = \frac{1}{f}

where

f

is in meters.

3

Sum the individual powers to find the total power of thin lenses in contact.

P = P_1 + P_2 = +5.0\text{ D} + (-2.0\text{ D}) = +3.0\text{ D}

.

When thin lenses are in contact, their net optical power is equal to the algebraic sum of their individual powers.

Anahtar Kavram

Power of Thin Lenses in Contact

Tahmini Süre:1m 15s

Waves and Optics

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