1. Example for Doppler shift frequency

3. 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)
= =54D.
An eye focused at 2m has a strength of
1/F= (1/2.0) + (1/ 0.02) = =50.5 D
A corrective lens of = +3.5 D would prescribed for his eye.

4. Myopia A person who is focusing an 0bject at 1m has a lens strength of
1/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.

5. 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/ /0.02
1/f = = = 56.7 D
1/f = 56.7 D without glasses

6. Correctives lens = - 1 D
With glass the strength = 56.7-(-1)= 57.7 D
57.7 = 1/P+ 1/0.02
57.7 = 1/P + 50
1/P = = 7.7
P= 0.18 m ,her near point with glasses

7. Q2.If an emmetrope has an accommodation of 3D,what is her near point?
For distance vision
1/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

8. 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.

9. At Pg =25 cm,1/f = 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.

10. A cardiologist uses an ultrasound scanner with an operating frequency of 3.5MHz that can detect Doppler frequency shifts as small as kHz . Assume the velocity of ultrasound in soft tissue = 1540 m/s. Calculate the smallest flow velocity detectable in m/s(consider θ=0).

11. c Δf (Doppler shift frequency) is Δ f = 2 f v cosθ f= 3.5MHz =3.5*10^6
C = 1540 m/sec
θ =0
Cos(0)=1

12. 0.1 *10^3={ 2* 3.5* 10^6*V*COS(0)}/1540
V= { (0.1 *10^3)*1540}/2* 3.5* 10^6
V= 22*10^-3 m/sec