Examkin

Wave Phenomena: Reflection, Refraction, Interference, Diffraction, and Polarization

16 soru

Tüm türler16Çoktan seçmeli8Sayısal8
Soru 1Soru →

Monochromatic light of wavelength 500 nm500\text{ nm} is incident normally on a plane diffraction grating. If the second-order principal maximum is observed at an angle of 30∘30^\circ to the normal, calculate the number of lines per millimeter ruled on the grating.

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Cevap: 500

Cevap

500 lines/mm
Using the diffraction grating equation dsin⁡θ=nλd \sin\theta = n\lambda with n=2n = 2, λ=5.0×10−7 m\lambda = 5.0 \times 10^{-7}\text{ m}, and sin⁡(30∘)=0.5\sin(30^\circ) = 0.5 gives a slit separation of d=2.0×10−6 md = 2.0 \times 10^{-6}\text{ m}. Converting to grating ruling density per millimeter yields 10−3 m2.0×10−6 m=500 lines/mm\frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}.

Adım Adım Çözüm

1
Identify the given physical parameters and convert wavelength to standard SI meters.
Order n=2n = 2, diffraction angle θ=30∘\theta = 30^\circ, wavelength λ=500×10−9 m=5.0×10−7 m\lambda = 500 \times 10^{-9}\text{ m} = 5.0 \times 10^{-7}\text{ m}.
Standard SI unit conversion is required to perform wave equation calculations accurately.
2
Apply the diffraction grating equation dsin⁡θ=nλd \sin\theta = n\lambda to compute the grating spacing dd.
d=2×5.0×10−7 msin⁡30∘=1.0×10−60.5=2.0×10−6 md = \frac{2 \times 5.0 \times 10^{-7}\text{ m}}{\sin 30^\circ} = \frac{1.0 \times 10^{-6}}{0.5} = 2.0 \times 10^{-6}\text{ m}.
The condition for principal constructive interference maxima is dsin⁡θ=nλd \sin\theta = n\lambda.
3
Calculate the number of lines per millimeter by dividing 1 mm1\text{ mm} (10−3 m10^{-3}\text{ m}) by the grating spacing dd.
Nmm=10−3 m2.0×10−6 m=500 lines/mmN_{\text{mm}} = \frac{10^{-3}\text{ m}}{2.0 \times 10^{-6}\text{ m}} = 500\text{ lines/mm}.
Grating density (lines per unit length) is the reciprocal of the slit separation dd.

Anahtar Kavram

Diffraction Grating Principal Maxima Condition
Tahmini Süre:2m 0s
Soru 2Soru →

A light wave of frequency 5.00×1014 Hz5.00 \times 10^{14}\text{ Hz} travels from air into a glass medium with a refractive index of 1.501.50. What is the frequency of the light wave inside the glass medium?

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Cevap: 5.00×1014 Hz5.00 \times 10^{14}\text{ Hz}

Cevap

The frequency of the light wave inside the glass medium remains 5.00×1014 Hz5.00 \times 10^{14}\text{ Hz}.
When light passes from one medium into another, its speed and wavelength change, but its frequency remains constant because it depends solely on the source producing the wave.

Adım Adım Çözüm

1
Identify the given physical parameters.
Initial frequency f=5.00×1014 Hzf = 5.00 \times 10^{14}\text{ Hz} and refractive index n=1.50n = 1.50.
Understanding provided quantities is the starting point for wave refraction analysis.
2
Apply the principle of frequency conservation across optical boundaries.
Frequency in glass medium fglass=fair=5.00×1014 Hzf_{\text{glass}} = f_{\text{air}} = 5.00 \times 10^{14}\text{ Hz}.
When a wave passes into a different medium, its speed and wavelength alter in proportion to the refractive index, but its frequency is determined solely by the periodic source and remains constant.

Anahtar Kavram

Invariance of Wave Frequency across Refracting Media
Soru 3Soru →

In a Young's double-slit experiment, the separation between two narrow slits is 0.40 mm0.40\text{ mm} and the interference pattern is observed on a screen placed 1.20 m1.20\text{ m} away from the slits. If the distance between consecutive bright fringes on the screen is 1.80 mm1.80\text{ mm}, what is the wavelength of the light used in nanometers (nm\text{nm})?

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Cevap: 600

Cevap

The wavelength of the light used is 600 nm600\text{ nm}.
Using the Young's double-slit fringe spacing relation β=λDd\beta = \frac{\lambda D}{d}, rearranging yields λ=βdD\lambda = \frac{\beta d}{D}. Substituting β=1.80×10−3 m\beta = 1.80 \times 10^{-3}\text{ m}, d=4.0×10−4 md = 4.0 \times 10^{-4}\text{ m}, and D=1.20 mD = 1.20\text{ m} gives λ=6.00×10−7 m\lambda = 6.00 \times 10^{-7}\text{ m}, which corresponds to 600 nm600\text{ nm}.

Adım Adım Çözüm

1
Convert given physical quantities into standard SI units (meters).
Slit separation d=0.40 mm=4.0×10−4 md = 0.40\text{ mm} = 4.0 \times 10^{-4}\text{ m}, distance to screen D=1.20 mD = 1.20\text{ m}, and fringe spacing β=1.80 mm=1.80×10−3 m\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 λ\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 10910^9 to express the wavelength in nanometers.

Anahtar Kavram

Young's Double-Slit Interference Fringe Spacing
Soru 4Soru →

A light ray travels from Medium X into Medium Y. The speed of light in Medium X is 1.50×108 m/s1.50 \times 10^8\text{ m/s} and the speed of light in Medium Y is 2.50×108 m/s2.50 \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?

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Cevap: 0.600.60

Cevap

The sine of the critical angle for total internal reflection at the boundary is 0.600.60.
Total internal reflection occurs when light moves from an optically denser medium to an optically less dense medium. By Snell's law, nXsin⁡θc=nYsin⁡90∘n_X \sin\theta_c = n_Y \sin 90^\circ, which gives sin⁡θc=nY/nX\sin\theta_c = n_Y / n_X. Since refractive index is inversely proportional to speed (n=c/vn = c/v), sin⁡θc=vX/vY=(1.50×108)/(2.50×108)=0.60\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
nX=cvXn_X = \frac{c}{v_X} and nY=cvYn_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⁡θc=nYnX\sin\theta_c = \frac{n_Y}{n_X}
At the critical angle, the angle of refraction in the less dense medium is 90∘90^\circ (sin⁡90∘=1\sin 90^\circ = 1).
3
Substitute light speeds into the critical angle formula and calculate
sin⁡θc=vXvY=1.50×108 m/s2.50×108 m/s=0.60\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 nXn_X and nYn_Y simplifies the refractive index ratio to the direct ratio of speeds vX/vYv_X / v_Y.

Anahtar Kavram

Critical angle and total internal reflection relation to speed of light in media
Soru 5Soru →

Monochromatic light of wavelength 600 nm600\text{ nm} is incident normally on a diffraction grating having 500 lines per mm500\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∘17.5^\circ.
Using the grating equation dsin⁡θ=nλd \sin \theta = n \lambda, the slit separation is d=10−3 m500=2.00×10−6 md = \frac{10^{-3}\text{ m}}{500} = 2.00 \times 10^{-6}\text{ m}. For n=1n = 1 and λ=6.00×10−7 m\lambda = 6.00 \times 10^{-7}\text{ m}, we get sin⁡θ=6.00×10−72.00×10−6=0.30\sin \theta = \frac{6.00 \times 10^{-7}}{2.00 \times 10^{-6}} = 0.30. Taking arcsin⁡(0.30)\arcsin(0.30) gives approximately 17.5∘17.5^\circ.

Adım Adım Çözüm

1
Calculate the grating element (slit spacing) dd
d=2.00×10−6 md = 2.00 \times 10^{-6}\text{ m}
Grating spacing dd is the reciprocal of the line density N=500 lines/mm=500,000 lines/mN = 500\text{ lines/mm} = 500,000\text{ lines/m}.
2
Apply the diffraction grating equation dsin⁡θ=nλd \sin \theta = n \lambda
sin⁡θ=0.30\sin \theta = 0.30
For the first-order maximum (n=1n = 1), sin⁡θ=1×600×10−9 m2.00×10−6 m=0.30\sin \theta = \frac{1 \times 600 \times 10^{-9}\text{ m}}{2.00 \times 10^{-6}\text{ m}} = 0.30.
3
Find the angle θ\theta by taking the inverse sine
θ=17.5∘\theta = 17.5^\circ
arcsin⁡(0.30)≈17.46∘\arcsin(0.30) \approx 17.46^\circ, which rounds to 17.5∘17.5^\circ.

Anahtar Kavram

Diffraction Grating Equation for Principal Maxima
Tahmini Süre:1m 30s
Soru 6Soru →

Two sound waves of frequencies 440 Hz440\text{ Hz} and 445 Hz445\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 Hz5\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: fbeat=∣f2−f1∣=∣445 Hz−440 Hz∣=5 Hzf_{\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
f1=440 Hzf_1 = 440\text{ Hz} and f2=445 Hzf_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
fbeat=∣445−440∣=5 Hzf_{\text{beat}} = |445 - 440| = 5\text{ Hz}
The rate of periodic intensity variation (beat frequency) is governed by fbeat=∣f2−f1∣f_{\text{beat}} = |f_2 - f_1|.

Anahtar Kavram

Beat Frequency and Wave Superposition
Soru 7Soru →

A monochromatic beam of light with a wavelength of 600 nm600\text{ nm} in air enters a glass block of refractive index 1.501.50. What is the wavelength of the light inside the glass block in nanometers (nm\text{nm})?

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Cevap: 400

Cevap

The wavelength of the light inside the glass block is 400 nm400\text{ nm}.
The speed and wavelength of light both decrease by a factor of the refractive index nn when entering a medium from air, while the frequency stays constant (v=fλv = f\lambda). Thus, λ=λ0n=6001.50=400 nm\lambda = \frac{\lambda_0}{n} = \frac{600}{1.50} = 400\text{ nm}.

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1
Identify the relevant formula connecting refractive index and wavelength
λ=λ0n\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 nn.
2
Substitute the given values into the equation
λ=600 nm1.50\lambda = \frac{600\text{ nm}}{1.50}
The wavelength in air is 600 nm600\text{ nm} and the refractive index of glass is 1.501.50.
3
Perform the calculation
λ=400 nm\lambda = 400\text{ nm}
Dividing 600600 by 1.501.50 yields 400400.

Anahtar Kavram

Refraction and Wavelength Change in a Medium
Soru 8Soru →

Under which of the following conditions will total internal reflection occur when light encounters the boundary between two transparent media?

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Cevap: When light travels from a medium of higher refractive index to one of lower refractive index at an angle of incidence greater than the critical angle.

Cevap

Total internal reflection occurs when light travels from a medium of higher refractive index to a medium of lower refractive index at an angle of incidence greater than the critical angle.
Total internal reflection takes place only when light moves from an optically denser medium (higher refractive index) into an optically less dense medium (lower refractive index), and the angle of incidence at the interface is greater than the critical angle for the two media.

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1
Identify the optical density requirement for total internal reflection
Light must travel from an optically denser medium (higher refractive index n1n_1) toward an optically rarer medium (lower refractive index n2n_2).
This condition ensures that the light refracts away from the normal into the second medium.
2
Identify the angular requirement at the interface
The angle of incidence ii must be strictly greater than the critical angle θc\theta_c (sin⁡θc=n2n1\sin\theta_c = \frac{n_2}{n_1}).
When i>θci > \theta_c, no light can refract into the second medium, resulting in total reflection back into the initial medium.

Anahtar Kavram

Conditions for Total Internal Reflection
Soru 9Soru →

A ray of light traveling inside a transparent medium strikes the boundary with air. If the critical angle for total internal reflection at this boundary is 30∘30^\circ, what is the refractive index of the medium?

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Cevap: 2.002.00

Cevap

The refractive index of the medium is 2.002.00.
For light traveling from a medium into air, the critical angle CC is related to the refractive index nn by n=1sin⁡Cn = \frac{1}{\sin C}. Substituting C=30∘C = 30^\circ gives n=1sin⁡30∘=10.5=2.00n = \frac{1}{\sin 30^\circ} = \frac{1}{0.5} = 2.00.

Adım Adım Çözüm

1
Identify the relationship between critical angle CC and refractive index nn
The formula for light passing into air is n=1sin⁡Cn = \frac{1}{\sin C}.
Total internal reflection occurs when light travels from a denser medium to a less dense medium (air, nair=1n_{air} = 1) at an angle greater than the critical angle.
2
Substitute the given critical angle C=30∘C = 30^\circ into the formula
sin⁡30∘=0.5\sin 30^\circ = 0.5, so n=10.5=2.00n = \frac{1}{0.5} = 2.00.
Taking the reciprocal of sin⁡30∘\sin 30^\circ gives the correct refractive index.

Anahtar Kavram

Critical Angle and Refractive Index
Soru 10Soru →

In a double-slit experiment, monochromatic light of wavelength 500 nm500\text{ nm} is incident normally on two narrow slits. If the third-order (m=3m = 3) bright fringe is observed at an angle of 30∘30^\circ from the central maximum, what is the slit separation, dd, in micrometers (μm\mu\text{m})?

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Cevap: 3

Cevap

The slit separation is 3.0 μm3.0\text{ }\mu\text{m}.
For bright fringes in a double-slit setup, constructive interference occurs when dsin⁡θ=mλd \sin\theta = m\lambda. Rearranging for slit separation yields d=mλsin⁡θd = \frac{m\lambda}{\sin\theta}. Substituting m=3m = 3, λ=0.50 μm\lambda = 0.50\text{ }\mu\text{m}, and θ=30∘\theta = 30^\circ gives d=3×0.50 μmsin⁡30∘=1.50 μm0.50=3.0 μmd = \frac{3 \times 0.50\text{ }\mu\text{m}}{\sin 30^\circ} = \frac{1.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}.

Adım Adım Çözüm

1
Identify the constructive interference equation for Young's double-slit experiment.
dsin⁡θ=mλd \sin\theta = m\lambda
Bright fringes occur where waves from the two slits interfere constructively, which corresponds to path differences equal to integer multiples of the wavelength.
2
Convert the given wavelength to micrometers and substitute known values.
λ=0.50 μm\lambda = 0.50\text{ }\mu\text{m}, m=3m = 3, sin⁡(30∘)=0.50\sin(30^\circ) = 0.50
Converting units to micrometers early simplifies direct calculation of dd in μm\mu\text{m}.
3
Rearrange the equation and evaluate for dd.
d=3×0.50 μm0.50=3.0 μmd = \frac{3 \times 0.50\text{ }\mu\text{m}}{0.50} = 3.0\text{ }\mu\text{m}
Dividing the numerator by sin⁡(30∘)=0.50\sin(30^\circ) = 0.50 doubles the value of mλm\lambda.

Anahtar Kavram

Angular position condition for constructive interference in double-slit diffraction
Soru 11Soru →

An unpolarized light beam with an initial intensity of 80 W/m280\text{ W/m}^2 passes through an ideal linear polarizer. What is the intensity of the transmitted light in W/m2\text{W/m}^2?

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Cevap: 40

Cevap

The intensity of the transmitted light is 40 W/m240\text{ W/m}^2.
When unpolarized light of initial intensity I0I_0 encounters an ideal linear polarizing filter, the transmitted intensity II is always equal to half of the incident intensity (I=12I0I = \frac{1}{2}I_0). Substituting 80 W/m280\text{ W/m}^2 gives I=40 W/m2I = 40\text{ W/m}^2.

Adım Adım Çözüm

1
Determine the fraction of unpolarized light intensity transmitted by a polarizer.
Transmitted intensity formula I=12I0I = \frac{1}{2} I_0.
Unpolarized light consists of randomly oriented electric field vectors, resulting in an average transmission factor of one-half.
2
Substitute the incident intensity value into the formula.
I=802=40 W/m2I = \frac{80}{2} = 40\text{ W/m}^2.
Direct mathematical calculation.

Anahtar Kavram

Polarization and intensity reduction of unpolarized light upon passing through a linear polarizer.
Soru 12Soru →

A beam of light traveling in air is incident on a transparent liquid at the polarizing angle (Brewster's angle) of 53.1∘53.1^\circ, where tan⁡53.1∘=1.33\tan 53.1^\circ = 1.33. What is the critical angle for total internal reflection when light travels from this liquid into air?

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Cevap: sin⁡−1(0.75)\sin^{-1}(0.75)

Cevap

The critical angle for total internal reflection at the liquid-air boundary is sin⁡−1(0.75)\sin^{-1}(0.75).
According to Brewster's law, the refractive index of the liquid is given by n=tan⁡(53.1∘)=1.33=43n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}. When light travels from the denser liquid medium to the rarer air medium, the critical angle θc\theta_c for total internal reflection satisfies sin⁡θc=1n\sin\theta_c = \frac{1}{n}. Substituting n=43n = \frac{4}{3} gives sin⁡θc=34=0.75\sin\theta_c = \frac{3}{4} = 0.75, so θc=sin⁡−1(0.75)\theta_c = \sin^{-1}(0.75).

Adım Adım Çözüm

1
Determine the refractive index of the liquid using Brewster's law.
n=tan⁡(53.1∘)=1.33=43n = \tan(53.1^\circ) = 1.33 = \frac{4}{3}.
Brewster's law states that when light in air (n1=1n_1 = 1) is incident at the polarizing angle θB\theta_B on a medium of index nn, tan⁡θB=n\tan\theta_B = n.
2
Apply the total internal reflection condition for light passing from liquid to air.
sin⁡θc=1n=14/3=34=0.75\sin\theta_c = \frac{1}{n} = \frac{1}{4/3} = \frac{3}{4} = 0.75.
Total internal reflection occurs at an interface when light travels from a denser medium (nn) to a less dense medium (11) at an angle greater than θc\theta_c, where sin⁡θc=1n\sin\theta_c = \frac{1}{n}.
3
Solve for the critical angle θc\theta_c.
θc=sin⁡−1(0.75)\theta_c = \sin^{-1}(0.75).
Taking the inverse sine of 0.750.75 yields the critical angle.

Anahtar Kavram

Synthesizing Brewster's law of polarization with total internal reflection critical angle
Soru 13Soru →

A ray of light travels from air into a liquid with a refractive index of 1.331.33. If the sine of the angle of incidence in air is 0.800.80, what is the sine of the angle of refraction in the liquid?

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Cevap: 0.6

Cevap

The sine of the angle of refraction in the liquid is 0.60.
According to Snell's law for light passing from air into a medium, the refractive index nn is given by n=sin⁡isin⁡rn = \frac{\sin i}{\sin r}. Rearranging this equation to solve for the sine of the angle of refraction yields sin⁡r=sin⁡in\sin r = \frac{\sin i}{n}. Substituting sin⁡i=0.80\sin i = 0.80 and n=1.33n = 1.33 (or 43\frac{4}{3}) gives sin⁡r=0.804/3=0.60\sin r = \frac{0.80}{4/3} = 0.60.

Adım Adım Çözüm

1
Identify the given physical parameters and state Snell's law
Refractive index n=1.33n = 1.33 (or 43\frac{4}{3}), sin⁡i=0.80\sin i = 0.80. Snell's law: n=sin⁡isin⁡rn = \frac{\sin i}{\sin r}
Snell's law relates the ratio of the sines of the angles of incidence and refraction to the refractive index of the medium.
2
Rearrange the equation to express the sine of the angle of refraction
sin⁡r=sin⁡in\sin r = \frac{\sin i}{n}
Algebraically isolating sin⁡r\sin r allows direct substitution of the known quantities.
3
Substitute the values and compute the result
\sin r = \frac{0.80}{4/3} = 0.60
Dividing 0.800.80 by 43\frac{4}{3} gives 0.600.60.

Anahtar Kavram

Snell's Law of Refraction

Alternatif Yöntem

Convert decimal numbers into simple fractions: n=43n = \frac{4}{3} and sin⁡i=45\sin i = \frac{4}{5}. Evaluating sin⁡r=4/54/3\sin r = \frac{4/5}{4/3} simplifies directly to 35=0.60\frac{3}{5} = 0.60.
Tahmini Süre:45s
Soru 14Soru →

A ray of light traveling within a dense glass prism of refractive index 1.601.60 strikes the boundary with a surrounding transparent liquid. If total internal reflection just occurs at an angle of incidence of 45.0∘45.0^\circ in the glass, what is the refractive index of the liquid? (Take sin⁡45.0∘=0.707\sin 45.0^\circ = 0.707)

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Cevap: 1.131.13

Cevap

The refractive index of the liquid is 1.131.13.
For light traveling from a denser medium (nglassn_{\text{glass}}) to a rarer medium (nliquidn_{\text{liquid}}), the critical angle θc\theta_c is defined by sin⁡θc=nliquidnglass\sin \theta_c = \frac{n_{\text{liquid}}}{n_{\text{glass}}}. Substituting nglass=1.60n_{\text{glass}} = 1.60 and sin⁡45.0∘=0.707\sin 45.0^\circ = 0.707 gives nliquid=1.60×0.707=1.13n_{\text{liquid}} = 1.60 \times 0.707 = 1.13.

Adım Adım Çözüm

1
Identify the given parameters and formula for total internal reflection
Refractive index of denser medium nglass=1.60n_{\text{glass}} = 1.60, critical angle θc=45.0∘\theta_c = 45.0^\circ, and formula sin⁡θc=nrarerndenser\sin \theta_c = \frac{n_{\text{rarer}}}{n_{\text{denser}}}.
Total internal reflection occurs at the critical angle when light travels from an optically denser medium to a less dense (rarer) medium.
2
Rearrange the equation to solve for the refractive index of the liquid (nliquidn_{\text{liquid}})
nliquid=nglass×sin⁡θcn_{\text{liquid}} = n_{\text{glass}} \times \sin \theta_c.
Multiplying both sides of the critical angle equation by nglassn_{\text{glass}} isolates the target variable.
3
Substitute the values and calculate nliquidn_{\text{liquid}}
nliquid=1.60×0.707=1.1312≈1.13n_{\text{liquid}} = 1.60 \times 0.707 = 1.1312 \approx 1.13.
Carrying out the arithmetic yields the refractive index of the liquid.

Anahtar Kavram

Total Internal Reflection and Critical Angle
Soru 15Soru →

A water wave traveling in deep water has a wavelength of 0.80 m0.80\text{ m} and a speed of 2.4 m/s2.4\text{ m/s}. Upon entering a shallow region, its speed drops to 1.8 m/s1.8\text{ m/s}. What are the frequency and wavelength of the wave in the shallow region?

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Cevap: Frequency = 3.0 Hz3.0\text{ Hz}, Wavelength = 0.60 m0.60\text{ m}

Cevap

Frequency = 3.0 Hz3.0\text{ Hz}, Wavelength = 0.60 m0.60\text{ m}
When a wave passes from deep to shallow water (refraction), its frequency remains unchanged because frequency is fixed by the source. Using v=fλv = f\lambda, the initial frequency is f=2.40.80=3.0 Hzf = \frac{2.4}{0.80} = 3.0\text{ Hz}. In shallow water, the new wavelength is λ′=1.83.0=0.60 m\lambda' = \frac{1.8}{3.0} = 0.60\text{ m}.

Adım Adım Çözüm

1
Calculate the frequency of the wave in deep water using the wave equation v=fλv = f \lambda.
f=vλ=2.4 m/s0.80 m=3.0 Hzf = \frac{v}{\lambda} = \frac{2.4\text{ m/s}}{0.80\text{ m}} = 3.0\text{ Hz}.
The frequency depends on the wave source and can be determined from the given initial speed and wavelength.
2
Apply the boundary condition for wave refraction.
The frequency in shallow water remains f=3.0 Hzf = 3.0\text{ Hz}.
When a wave travels from one medium to another, its frequency remains constant.
3
Calculate the new wavelength in shallow water using λ′=v′f\lambda' = \frac{v'}{f}.
λ′=1.8 m/s3.0 Hz=0.60 m\lambda' = \frac{1.8\text{ m/s}}{3.0\text{ Hz}} = 0.60\text{ m}.
The wavelength changes proportionally with speed when frequency is constant.

Anahtar Kavram

Constancy of wave frequency during refraction across medium boundaries
Soru 16Soru →

A beam of monochromatic light of wavelength 6.25×10−7 m6.25 \times 10^{-7}\text{ m} is incident normally on a plane diffraction grating having 400 lines/mm400\text{ lines/mm}. What is the angle of diffraction for the second-order principal maximum?

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Cevap: 30.0∘30.0^\circ

Cevap

The angle of diffraction for the second-order principal maximum is 30.0∘30.0^\circ.
The grating spacing is d=1400×103 m−1=2.50×10−6 md = \frac{1}{400\times 10^3\text{ m}^{-1}} = 2.50 \times 10^{-6}\text{ m}. Using dsin⁡θ=nλd \sin\theta = n\lambda for n=2n=2 and λ=6.25×10−7 m\lambda = 6.25 \times 10^{-7}\text{ m} gives sin⁡θ=2×6.25×10−72.50×10−6=0.50\sin\theta = \frac{2 \times 6.25 \times 10^{-7}}{2.50 \times 10^{-6}} = 0.50, corresponding to θ=30.0∘\theta = 30.0^\circ.

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1
Calculate the grating spacing (dd).
d=1 mm400=2.50×10−3 mm=2.50×10−6 md = \frac{1\text{ mm}}{400} = 2.50 \times 10^{-3}\text{ mm} = 2.50 \times 10^{-6}\text{ m}.
Grating spacing dd is the reciprocal of the line density NN per unit length.
2
Apply the diffraction grating equation for the second order (n=2n=2).
dsin⁡θ=nλ  ⟹  (2.50×10−6)sin⁡θ=2×(6.25×10−7)=1.25×10−6d \sin\theta = n\lambda \implies (2.50 \times 10^{-6}) \sin\theta = 2 \times (6.25 \times 10^{-7}) = 1.25 \times 10^{-6}.
The principal maximum condition relates slit spacing, diffraction angle, spectral order, and wavelength.
3
Solve for the diffraction angle θ\theta.
sin⁡θ=1.25×10−62.50×10−6=0.50  ⟹  θ=arcsin⁡(0.50)=30.0∘\sin\theta = \frac{1.25 \times 10^{-6}}{2.50 \times 10^{-6}} = 0.50 \implies \theta = \arcsin(0.50) = 30.0^\circ.
Taking the inverse sine of 0.500.50 yields the exact angle of diffraction.

Anahtar Kavram

Diffraction Grating Equation (dsin⁡θ=nλd \sin\theta = n\lambda)