**Solutions: Medical Option - the EYE**

**Q1**.

A convex lens is placed 0.25 m from an object.

The focused image produced is virtual and is 0.60 m from the lens.

(a) Calculate

(i) the power of the lens,

v = -0.60 m

u = 0.25 m

f = ?

P = 1/f = 1/u + 1/v = 1/0.25 - 1/0.60 = 2.33 D

(ii) the magnification produced.

M = v/u = 0.6/0.25 = 2.4

**(3 marks) **

(b) Draw a ray diagram to show the formation of the image produced by this lens. The diagram does not have to be to scale, but relevant distances must be marked.

###### diagram to show:

- two correct rays to locate image
** **- through pole of lens and parallel to the principal axis and then through focal point
- correct (virtual, upright) image
** **
- two distances (u and v) shown
** **

**(3 marks) **

(c)

(i) What defect of vision is this lens used to correct?

Longsightedness - hypermetropia

(ii) A person has an unaided near point at 0.60 m and an unaided far point at infinity. Calculate the range of vision of the person when using this lens.

v = -0.60 m

u = ?

1/f = 2.33 D

1/f = 1/u + 1/v

1/u = 1/f - 1/v = 2.33 + 1/0.60 = 4.0 u = 0.25 m

The person's range of vision will therefore be from 0.25m to 0.43 m

**(4 marks) **

**(Total 10 marks)**