questions for corrective lens word

د. تيماء نوري Questions for corrective lens 5/5/2016

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 D
57.7 = 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 vision
1/f=1/P +1/Q
1/f = 1/ ȹ +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- 2 D= 52 D
52 = 1/Po + 1/0.02

Po = 0.5m his near point without glass.

Question for Doppler effect
Q. 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=zero).

Δf (Doppler shift frequency) is

Δ f = 2 f v cosθ
c
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 * co s(0)}/1540

v= { (0.1 *10 3 ) *1540}/2* 3.5* 10 6

v= 22*10 -3 m/sec

Q. Calculate the ratios of pressure amplitude and the intensities of the reflected
and transmitted sound waves from air to muscle.
Acoustic impedance of air = 430 kg/m^3. sec
Acoustic impedance of muscle=1.64 x 10^6 kg/m^3. sec
R = Z2 – Z1
A Z1 +Z2
R = 1.64 x 10^6 – 430 = 0.9995
A 1.64 x 10^6 + 430


T = 2 Z2
A Z1+Z2

T = 2(1.64 x 10^6 ) = 1.9995

A 1.64 x 10^6 + 430