Example for Doppler shift frequency
hyperemia
Let us consider afar sighted eye with a near point of 2.0 m. what power lens will let this person read comfortably at 0.25 m?(The distance between the lens and retina is 0.02 m.)
The strength of a good eye focused at 0.25 m is given by 1/F=(1/0.25)+ ( 1/ 0.02)
= 4+ 50 =54D.
An eye focused at 2m has a strength of
1/F= (1/2.0) + (1/ 0.02) =0.5 + 50 =50.5 D
A corrective lens of 54-50.5= +3.5 D would prescribed for his eye.
Myopia
A person who is focusing an 0bject at 1m has a lens strength of1/F =(1/1.0)+(1/0.02) =51 D
An eye able to focus at infinity has a strength of 1/F=(1/∞ )+ (1/0.02) = 50 D
51 D – 50D =1D
Thus a myopic person with a far point 1m has 1 D, and a negative lens of -1.0 D will correct his vision.
Question for corrective lens
Q1.If a myope has a near point of 15 cm without glasses and wears a corrective lens of
-1.0 D, what is her near point wearing glasses?
1/f = 1/P +1/Q
1/f = 1/0.15 + 1/0.02
1/f = 6.6 + 50 = 56.66 = 56.7 D
1/f = 56.7 D without glasses
Correctives lens = - 1 D
With glass the strength = 56.7-(-1)= 57.7 D57.7 = 1/P+ 1/0.02
57.7 = 1/P + 50
1/P =57.7 -50 = 7.7
P= 0.18 m ,her near point with glasses
Q2.If an emmetrope has an accommodation of 3D,what is her near point?
For distance vision1/f=1/P +1/Q
1/f = 1/ ∞ +1/0.02 =50 D
50+3 =53D
53= 1/P + 1/0.02
P= 1/3 m her near point
Q3.If a former emmetrope wears reading glasses
Of +2D to read at a distance of 25 cm ,what is his near point without glasses?
Let the near point be Po without glasses,
and Pg with glass.
At Pg =25 cm,1/f = 1/0.25 +1/0.02 =54 D
Without glass the person would have a maximum of 52 D.54 D -2D= 52 D
52 D =1/Po + 1/0.02
Po =0.5m his near point without glass.
A cardiologist uses an ultrasound scanner with an operating frequency of 3.5MHz that can detect Doppler frequency shifts as small as 0.1kHz . Assume the velocity of ultrasound in soft tissue = 1540 m/s. Calculate the smallest flow velocity detectable in m/s(consider θ=0).
Δf (Doppler shift frequency) is
Δ f = 2 f v cosθc
f= 3.5MHz =3.5*10^6
C = 1540 m/sec
θ =0
Cos(0)=1
0.1 *10^3={ 2* 3.5* 10^6*V*COS(0)}/1540