General Physics, 2nd Edition
General Physics, 2nd Edition
2nd Edition
ISBN: 9780471522782
Author: Morton M. Sternheim
Publisher: WILEY
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Chapter 22, Problem 63E

(a)

To determine

The percentage of intensity being reflected.

(a)

Expert Solution
Check Mark

Answer to Problem 63E

The percentage of intensity being reflected is 99.89%_.

Explanation of Solution

Write the expression for ratio of reflected intensity to incident intensity.

    IrIi=(ρ1c1ρ2c2ρ1c1+ρ2c2)2        (I)

Here, Ir is the intensity of the reflected wave, Ii is the intensity of the incident wave, ρ1 is the density of air medium, c1 is the speed of the sound wave in air medium 1, ρ2 is the density of water medium, and c2 is the speed of sound in water medium.

Write the expression for intensity reflected.

    p=IrIi×100%        (II)

Here, p is the percentage of intensity reflected.

Conclusion:

Substitute 1.20kg/m3 for ρ1, 344m/s for c1, 998kg/m3 for ρ2, and 1498m/s for c2 in equation (I), to find IrIi.

    IrIi=((1.20kg/m3)(344m/s)(998kg/m3)(1498m/s)(1.20kg/m3)(344m/s)+(998kg/m3)(1498m/s))2=(412.81495004412.8+1495004)=0.998896

Substitute 0.998896 for IrIi in equation (II), to find p.

    p=(0.998896)×100%=99.8896%99.89%

Therefore, the percentage of intensity being reflected is 99.89%_.

(b)

To determine

The percentage of intensity being transmitted.

(b)

Expert Solution
Check Mark

Answer to Problem 63E

The percentage of intensity being transmitted is 11%_.

Explanation of Solution

Write the expression for transmitted intensity.

    It=IiIr

Here, It is the transmitted intensity.

Write the expression for transmitted intensity.

    p=(IiIr)Ii×100%

Here, p is the percentage of transmitted intensity.

Rearrange the above equation.

    p=(IiIiIrIi)100%=(1IrIi)100%        (III)

From subpart (a), the ratio of IrIi is obtained as 0.998896.

Use the above condition in equation (III).

    p=(10.998896)100%=(0.001104)100%=11.04%11%

Conclusion:

Therefore, The percentage of intensity being transmitted is 11%_.

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