We are back with some of the basic physics concepts .We have already discussed about two major eye sight problems Myopia and Hypermetropia and how these can be cured . Other topics covered in previous posts include Difference between Myopia and Hypermetropia and what are the causes for these eye defects . In this post we are going to cover some of the basic numericals based on the Power of lens and which lens is used for which type of eye defect .

# Numericals based on Power of Lens

There are some important things one should kept in mind before doing the numericals based on power of lens are that for Hypermetropia , * we always use convex lens as a correcting lens and for myopia , concave lens is used* .The formula for both cases is different .

### Lens Formula in case of Hypermetropia is

**(1/ v ) – (1/ u ) = 1/ f**

By this above formula we calculate focal length and hence Power of the lens . Here u and v are

**u = object distance** = normal near point ( taken in negative )

**v = image distance** = Defective near Point

so the formula becomes

**(1/ v ) + (1/ u ) = 1/ f**

u and v are usually in cm

**Power = 1 / f ( in m )**

### Hypermetropia Numerical # 1

For an eye , the defective near point is 150 cm . Calculate power of correcting convex lens to correct this vision defect .

**Solution # 1**

We are given the following values

Hypermetropic Near Point = 150 cm

Therefore Image distance = v = -150

Object distance = normal near point = -25 cm

**(1/ v ) + (1/ u ) = 1/ f**

(1/ v ) + (1/ 25 ) = 1/ f

(1/ -150 ) + (1/ u ) = 1/ f

f = 150/5 cm = 0.3 m

**Power = 1 / f = 3.3 D ( diopter )**

It means Convex lens of power 3.3 Diopter is required to correct the vision .

### Myopia Numerical # 2

The far point of myopic person is 80 cm in front of an eye . Describe the nature of lens and the power required to correct the vision .

**Solution # 2**

Given –

Corrective lens required is Concave

Distance of far object = 80 cm

To view the distance correctly , the focal length needed is = -80 cm

Power of lens = 1 / f = 1/ -.80 = – -1. 25 D

**Note = Power of corrective convex lens is positive and convex lens is negative**

### Hypermetropia Numerical # 3

The near point of eye suffering from hypermetropia is 1 m . Calculate the Power that is needed to correct this eye defect . In this case we are assuming the near point as 25 cm .

**Solution # 3**

Here we know the values of u and v .Put these values in the formula given above and final f i.e focal length .

Value of u = -1 m = -100 cm

Value of v = -25 cm

Use the Lens formula

(1/ v ) – (1/ u ) = 1/ f

value of f = 3.33cm

Power of lens = 1/ f = 3 D

### Lens Formula bases numerical # 4

If a person can’t see the object that is placed at a distance less than 50 cm , then what type of defect he is suffering from ? Also tell the nature of lens and calculate the power of the lens so that he can clearly see the object that is placed at 25 cm from the eye .

**Numerical Solution # 4**

Defect from which the person suffering = Long sightedness

Given

Value of u = -25 cm

Value of v = -50 cm

Use the lens formula

(1/ v ) + (1/ u ) = 1/ f

Put the values of u and v in the above given formula

(1/ -50 ) + (1/ -25 ) = 1/ f

f = 50 cm = .5 m

Power of lens = 2D

To correct this defect , convex lens with power 2 D is required to be used .

These were some of the basic numericals with solutions based on Myopia and Hypermetropia .For more solves problems on topics such as Lens power or eye defects correcting lens , stay tuned .