Question

Electromagnetic Wave PenetrationRadio waves and mircowaves are used in therapy to provide "deep heating" of tissue because the waves penetrate beneth the surface of the body and deposit energy. We define the penetration depth as the depth at which the wave intensity has decreased to 37% of its value at the surface. The penetration depth is 15cm for 27MHz radio waves. For radio frequencies such as this, the penetration depth is proportional to (sqrt of lambda), which is the square root of the wavelength. What is the wavelength of 27 MHz radio waves? A. 11 m.B. 9.0 m.C. 0.011 m.D. 0.009 m.If the frequency of the radio waves is increased, the depth of penetration:_____.A. Increases. B. Does not change. C. Decreases. 3. For 27 MHzMHz radio waves, the wave intensity has been reduced by a factor of 3 at a depth of approximately 15 cmcm. At this point in the tissue, the electric field amplitude has decreased by a factor of:____.A. 9 77. B. 3√3. C. 3. D. √3.

Answer 1

Hence the square root of the wavelength. What is the wavelength of 27 MHz radio waves 11 m; option A is the correct answer.

Hence, as the frequency of the radio waves increases the penetration depth decreases.

Option C is the correct answer.

Hence the electric field amplitude has decreased by a factor of √3.

Hence the correct option is D.

Step-by-step explanation:

`c=n\lambda`, where the symbols have their usual meanings.

`\lambda =c/n\\= 3*10^(8)/27*10^(6)\\\\=11.11 m`

Hence, wavelength = 11 m; option A is the correct answer.

As given that, the penetration depth is proportional to the square root of wavelength, and we know that,

wavelength and frequency are inversely proportional to each other.

Hence, as the frequency of the radio waves increases the penetration depth decreases.

Option C is the correct answer.

Electric field amplitude is directly proportional to the square root of the intensity. Therefore, the ratio between the electric field amplitude is,

`(E\sigma_(1) )/(E\sigma_(2) ) =\sqrt{(I_(1) )/(I_(2) ) }`

Here to find the ratio of intensities,

`{(I_(1) )/(I_(2) ) } = 3`

Therefore,

`(E\sigma_(1) )/(E\sigma_(2) ) =√(3 )`

Hence the correct option is D.