Question

Determine the power of corrective contact lenses for a) a hyperopic eye with near point at 50 cm, b) a myopic eye with far point at 50 cm

Answer 1

Answer:

Part a)

$P =+2 Dioptre$

Part b)

$P = - 2 Dioptre$

Explanation:

Part a)

As we know that Far sighted person has near point shifted to 50 cm distance

so he is able to see the object 50 cm

and the person want to see the objects at distance 25 cm

so here the image distance from lens is 50 cm and the object distance from lens is 25 cm

now from lens formula we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$-\frac{1}{50} + \frac{1}{25} = \frac{1}{f}$

$f = +50 cm$

Now we know that power of lens is given as

$P = \frac{1}{f}$

$P = \frac{1}{0.50} = +2 Dioptre$

Part b)

As we know that near sighted person has far point shifted to 50 cm distance

so he is able to see the object 50 cm

and the person want to see the objects at large distance so here the image distance from lens is 50 cm and the object distance from lens is infinite large

now from lens formula we have

$\frac{1}{d_i} + \frac{1}{d_0} = \frac{1}{f}$

$-\frac{1}{50} + \frac{1}{infinite} = \frac{1}{f}$

$f = -50 cm$

Now we know that power of lens is given as

$P = \frac{1}{f}$

$P = -\frac{1}{0.50} = -2 Dioptre$