Question

Goal Apply geometric optics to correct nearsightedness. Problem A particular nearsighted patient can't see objects clearly when they are beyond 24 cm (the far point of the eye). (a) What focal length should the prescribed contact lens have to correct this problem? (b) Find the power of the lens, in diopters. Neglect the distance between the eye and the corrective lens. Strategy The purpose of the lens in this instance is to take objects at infinity and create an image of them at the patient's far point. Apply the thin-lens equation. Solution (a) Find the focal length of the corrective lens. Apply the thin-lens equation for an object at infinity 11 1 and image at 24.0 cm. cm (b) Find the power of the lens in diopters. diopters Remarks The focal length is negative, consistent with a diverging lens. Notice that the power is also negative and has the same numeric value as the sum on the left side of the thin-lens equation. Exercise 25.2 Hints: Getting Started | I'm Stuck (a) What power lens would you prescribe for a patient with a far point of 30.7 cm? diopters (b) Repeat, assuming an eye-corrective lens distance of 2.00 cm. diopters

Answer 1

Nearsightedness means object can see near object and can't see far object.

Here  q is the maximum distance patient can see.

a) As you know 1/f =1/p +1/q

here q=-24 cm , p=$\infty$

Put in above equation

1/f =1/$\infty$ +1/-24

=0+1/-24

f =-24 cm

b) f=-24 cm =-0.24 m

p=1/f =1/-0.24

p=-4.17 diopters

Exercise 25.2

a) here p=$\infty$ ,q=-30.7cm

1/f =1/$\infty$ +1/-30.7

f=-30.7 cm

p=1/-0.307

p=-3.257 diopters

b)here p =$\infty$ , q=-(30.7-2) =-28.7 cm

similarly f=-28.7 cm

p=-1/0.287

p=-3.484 diopters.

Answer 2

n a) focal Janoth vequind to pre scarked (onac+ Jans have to orsod this poblam which sheusd be giuan using dhin dens yormulaj Saun do di f-29 cm