Answer to Question #172823 in Optics for Jessica Boateng

Question

What power of spectacle lens is needed to correct the vision of a nearsighted person whose far point is 30.0cm. Assume the spectacle lens is held 1.50cm away from the eye by eyeglass frames.

Answer 1

Given,

Far point of the person = 30cm

Distance between eye and eyeglass frame =1.5 cm

So, image distance $(v) = -(30cm-1.5cm) =- 28.5cm$

Here the distance is negative because it is in same side of the spectacle as the object.

Object distance $(u)=\infty$

We know that,

$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$

Now, substituting the values,

$\Rightarrow \frac{1}{f}=\frac{1}{-28.5}-\frac{1}{\infty}$

$\Rightarrow \frac{1}{f}=\frac{1}{-28.5}$

We know that,

Power of lens $(P)=\frac{100}{f}$

$\Rightarrow P=\frac{100}{-28.5}D$

$\Rightarrow P =-3.5D$

Here, negative sign representing that the lens should be concave.