Question

The far point of a near sighted person is 6.0 m from her eyes, and she wears contacts that enable her to
see distant objects clearly. A tree is 18.0 m away and 2.0 high. How high is the image formed by the contacts?

• A. 1.0 m
• B. 1.5 m
• C. 0.75 m
• D. 0.50 m

Answer 1

The correct option is D 0.50 m

The far point of 6.0 m tell us that focal length of the lens is f = – 6.0 m, u = – 18 m and h = 2 m,
Using, $\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\Rightarrow \frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{-6.0}-\frac{1}{18.0}\Rightarrow v=-4.5m$
$\therefore$ The image size.
$h^{\prime }=h\left(\frac{v}{u}\right)=2\times \left(\frac{-4.5}{-18.0}\right)=0.50m$