Question #172823

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.


1
Expert's answer
2021-03-18T14:01:48-0400

Given,

Far point of the person = 30cm

Distance between eye and eyeglass frame =1.5 cm

So, image distance (v)=(30cm1.5cm)=28.5cm(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)=(u)=\infty

We know that,

1f=1v1u\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

Now, substituting the values,

1f=128.51\Rightarrow \frac{1}{f}=\frac{1}{-28.5}-\frac{1}{\infty}


1f=128.5\Rightarrow \frac{1}{f}=\frac{1}{-28.5}


We know that,

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


P=10028.5D\Rightarrow P=\frac{100}{-28.5}D


P=3.5D\Rightarrow P =-3.5D

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


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