Question

A bat emits an ultrasonic sound of frequency $ 100\;kHz $ in the air. If this sound meets a water surface, the wavelength of the reflected and transmitted sound are …… (Speed of sound in air = $ 340m{s^{ - 1}} $ and in water = $ 1500m{s^{ - 1}} $ )
(A) $ 3.4mm,\;30mm $
(B) $ 6.8mm,\;15mm $
(C) $ 3.4mm,\;15mm $
(D) $ 6.8mm,\;30mm $

Answer 1

Hint :The wavelength of the sound can be calculated by dividing the speed of sound in the particular medium by the frequency of the sound. For the given question, reflected sound is traveling through air and the transmitted sound is traveling through water.

Complete Step By Step Answer:
 To begin, let us note down the given data in the question;
Frequency of the sound emitted $ f = 100kHz $
 $ \therefore f = 10000Hz = {10^5}Hz $
Speed of sound in air $ {v_a} = 340m{s^{ - 1}} $
Speed of sound in water $ {v_w} = 1500m{s^{ - 1}} $
Wavelength of reflected sound (in air) $ {\lambda _a} = ? $
Wavelength of transmitted sound (in water) $ {\lambda _w} = ? $
Now, we know that wavelength is defined as the distance traveled by the wave to complete one cycle or the length of one complete wave.
Frequency is defined as the number of waves passing through a point in one second or the number of cycles completed per second.
Hence, if we take the product of the wavelength and the frequency i.e. multiplying distance traveled by wave in one cycle with the number of cycles in one second, we can get the total distance traveled by the wave in one second.
The distance traveled in one second is defined as the speed of an object.
Hence, the product of wavelength of sound and frequency of sound gives us the speed of sound in a particular medium.
Hence, the wavelength can be defined as the ratio of the speed of sound to the frequency of sound.
For the reflected sound, the waves travel through the air
 $ \therefore {\lambda _a} = \dfrac{{{v_a}}}{f} $
Substituting the given values,
 $ \therefore {\lambda _a} = \dfrac{{340m{s^{ - 1}}}}{{{{10}^5}Hz}} $
The unit of frequency can also be written as inverse of time i.e. $ {s^{ - 1}} $
 $ \therefore {\lambda _a} = \dfrac{{340m{s^{ - 1}}}}{{{{10}^5}{s^{ - 1}}}} $
Shifting the power to the numerator and writing the resultant unit
 $ \therefore {\lambda _a} = 340 \times {10^{ - 5}}m $
 $ \therefore {\lambda _a} = 3.4 \times {10^{ - 3}}m $
Now, we know that $ {10^{ - 3}}m = 1mm $
 $ \therefore {\lambda _a} = 3.4mm $
Now, for the transmitted sound, the waves travel through water
 $ \therefore {\lambda _w} = \dfrac{{{v_w}}}{f} $
Substituting the given values,
 $ \therefore {\lambda _w} = \dfrac{{1500m{s^{ - 1}}}}{{{{10}^5}{s^{ - 1}}}} $
Shifting the power to the numerator and writing the resultant unit
 $ \therefore {\lambda _w} = 1500 \times {10^{ - 5}}m $
 $ \therefore {\lambda _w} = 15 \times {10^{ - 3}}m $
Now, we know that $ {10^{ - 3}}m = 1mm $
 $ \therefore {\lambda _w} = 15mm $
Hence, the correct answer is Option $ (C) $ .

Additional Information:
As frequency is the number of waves passing through a point in one second, it only depends on the source of sound. Hence, the frequency remains unaffected when sound travels from one medium to another.

Note :
If a wave travels from a rarer medium to a denser medium its wavelength increases because a denser medium implies more density which further implies more elasticity. Hence, in a denser medium, sound can create larger compressions and rarefactions, which implies an increase in wavelength. Hence, for the given question, the wavelength of the transmitted sound that is traveling through the water should be greater than the wavelength in air.