PHYSICS:F/SCI.+ENGRS.,V.1
PHYSICS:F/SCI.+ENGRS.,V.1
10th Edition
ISBN: 9781337553575
Author: SERWAY
Publisher: CENGAGE L
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Question
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Chapter 34, Problem 12P

(a)

To determine

The angle of refraction for the sound wave.

(a)

Expert Solution
Check Mark

Answer to Problem 12P

The angle of refraction for sound wave is 87.43° .

Explanation of Solution

Given info: The wavelength of sound wave is 589nm and angle of incidence is 13.0° .

The speed of sound in air at 20°C is 343.216m/s and speed of sound in water at 25°C is 1531m/s .

The expression for the Snell’s law is,

μ1sinθ1=μ2sinθ2

Here,

μ1 is the refractive index of sound in air.

θ1 is angle of incidence.

μ2 is refractive index of sound in water.

θ2 is the angle of refraction.

Rearrange the above formula to find θ2 .

μ1sinθ1=μ2sinθ2sinθ2=μ1μ2sinθ1θ2=sin1(μ1μ2sinθ1) (1)

The formula to calculate speed of sound in water is,

μ1μ2=v2v1 (2)

Here,

μ1 is the refractive index of sound in air.

μ2 is the refractive index of sound in water.

v1 is the speed of sound in air.

v2 is the speed of sound in water.

Substitute v2v1 for μ1μ2 in formula (1) as,

θ2=sin1(v2v1sinθ1)

Substitute 343.216m/s for v1 , 1531m/s for v2 , 13.0° for θ1 in the above formula as,

θ2=sin1(v2v1sinθ1)=sin1(1531m/s343.216m/ssin13°)=sin1(4.460×0.224)=87.43°

Conclusion:

Therefore, the angle of refraction for the sound wave is 87.43°

(b)

To determine

The wavelength of sound in water.

(b)

Expert Solution
Check Mark

Answer to Problem 12P

The wavelength of sound in water is 2627.38nm .

Explanation of Solution

Given info: The wavelength of sound wave is 589nm and angle of incidence is 13.0° .

The formula to calculate the wavelength is,

v1λ1=v2λ2

Here,

v1 is the speed of sound wave in air.

v2 is the speed of sound wave in water.

λ1 is the wavelength of sound wave in air.

λ2 is the wavelength of sound in water.

Rearrange the above formula to find λ2 as,

v1λ1=v2λ2λ2=v2v1λ1

Substitute 343.216m/s for v1 , 1531m/s for v2 , 589nm for λ1 in the above formula as,

λ2=v2v1λ1=1531m/s343.216m/s(589nm)=2627.38nm

Conclusion:

Therefore, the wavelength of sound in water is 2627.38nm

(c)

To determine

The angle of refraction.

(c)

Expert Solution
Check Mark

Answer to Problem 12P

The angle of refraction is 9.67° .

Explanation of Solution

Given info: The wavelength of sodium yellow light is 589nm and angle of incidence is 13.0° .

The formula to calculate the Snell’s law is,

μ1sinθ1=μ2sinθ2

Here,

μ1 is the refractive index for air.

θ1 is angle of incidence.

μ2 is refractive index for water.

θ2 is the angle of refraction.

Rearrange the above formula to find θ2 as,

μ1sinθ1=μ2sinθ2sinθ2=μ1μ2sinθ1θ2=sin1(μ1μ2sinθ1)

Substitute 1 for μ1 , 1.33 for μ2 , 13° for θ1 in the above formula as,

θ2=sin1(μ1μ2sinθ1)=sin1(11.33sin13°)=sin1(0.751×0.224)=9.67°

Conclusion:

Therefore, the angle of refraction is 9.67° .

(d)

To determine

The wavelength of light in water.

(d)

Expert Solution
Check Mark

Answer to Problem 12P

The wavelength of light in water is 442.85nm .

Explanation of Solution

Given info: The wavelength of sodium yellow light is 589nm and angle of incidence is 13.0° .

The formula to calculate the wavelength is,

μ2μ1=λ1λ2

Here,

μ1 is the refractive index of light in air.

μ2 is the refractive index of light in water.

λ1 is the wavelength of light in air.

λ2 is the wavelength of light in water.

Rearrange the above formula to find λ2 as,

μ2μ1=λ1λ2λ2=μ1μ2λ1

Substitute 1 for μ1 , 1.33 for μ2 , 589nm for λ1 in the above formula as,

λ2=μ1μ2λ1=11.33(589nm)=442.85nm

Conclusion:

Therefore, the wavelength of light in water is 442.85nm .

(e)

To determine

The behavior of sound and light waves.

(e)

Expert Solution
Check Mark

Answer to Problem 12P

The behavior of sound and light waves is that the sound waves speeds up when travelling from rarer medium to denser medium and light rays slows down.

Explanation of Solution

Given info: The wavelength of sodium yellow light is 589nm and angle of incidence is 13.0° .

The medium that has low refractive index with respect to another medium is called rarer medium and the medium that has high refractive index with respect to another medium is called denser medium.

From part (b) the wavelength of sound wave in water is larger than the wavelength of sound in air and from part (d) the wavelength of light in water is less than the wavelength of light in air.

The sound waves travelling from rarer to denser medium then the refracted sound waves bend away from the normal that the reason there is an increase in wavelength of sound waves.

The ray of light travelling from rarer to denser medium then the refracted rays bends towards the normal. So that’s the reason there is a decrease in the wavelength of light in water.

Conclusion:

Therefore, The behavior of sound and light waves is that the sound waves speeds up when travelling from rarer medium to denser medium and light rays slows down.

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Chapter 34 Solutions

PHYSICS:F/SCI.+ENGRS.,V.1

Ch. 34 - Prob. 6PCh. 34 - Prob. 7PCh. 34 - Two flat, rectangular mirrors, both perpendicular...Ch. 34 - Prob. 9PCh. 34 - A ray of light strikes a flat block of glass (n =...Ch. 34 - Prob. 11PCh. 34 - Prob. 12PCh. 34 - A laser beam is incident at an angle of 30.0 from...Ch. 34 - A ray of light strikes the midpoint of one face of...Ch. 34 - When you look through a window, by what time...Ch. 34 - Light passes from air into flint glass at a...Ch. 34 - You have just installed a new bathroom in your...Ch. 34 - A triangular glass prism with apex angle 60.0 has...Ch. 34 - You are working at your university swimming...Ch. 34 - Prob. 20PCh. 34 - Prob. 21PCh. 34 - A submarine is 300 m horizontally from the shore...Ch. 34 - Prob. 23PCh. 34 - A light beam containing red and violet wavelengths...Ch. 34 - Prob. 25PCh. 34 - The speed of a water wave is described by v=gd,...Ch. 34 - For 589-nm light, calculate the critical angle for...Ch. 34 - Prob. 28PCh. 34 - A room contains air in which the speed of sound is...Ch. 34 - Prob. 30PCh. 34 - An optical fiber has an index of refraction n and...Ch. 34 - Consider a horizontal interface between air above...Ch. 34 - How many times will the incident beam in Figure...Ch. 34 - Consider a beam of light from the left entering a...Ch. 34 - Why is the following situation impossible? While...Ch. 34 - Prob. 36APCh. 34 - When light is incident normally on the interface...Ch. 34 - Refer to Problem 37 for its description of the...Ch. 34 - A light ray enters the atmosphere of the Earth and...Ch. 34 - A light ray enters the atmosphere of a planet and...Ch. 34 - Prob. 41APCh. 34 - Prob. 42APCh. 34 - Prob. 43APCh. 34 - Prob. 44APCh. 34 - Prob. 45APCh. 34 - As sunlight enters the Earths atmosphere, it...Ch. 34 - A ray of light passes from air into water. For its...Ch. 34 - Prob. 48APCh. 34 - Prob. 49APCh. 34 - Figure P34.50 shows a top view of a square...Ch. 34 - Prob. 51APCh. 34 - Prob. 52CPCh. 34 - Prob. 53CPCh. 34 - Pierre de Fermat (16011665) showed that whenever...Ch. 34 - Prob. 55CPCh. 34 - Suppose a luminous sphere of radius R1 (such as...Ch. 34 - Prob. 57CP
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