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 2.0 m high. How high is the image formed by the contacts?

• A. 0.1 m
• B. 1.5 m
• C. 0.75 m
• D. 0.50 m

Answer 1

Answer: (D)

0.50 m

Answer 2

: The far point of $6.0$ m tell us that the focal length of the lens is

$\mathrm { h } = 2 \mathrm {~m}$

Using, $\frac { 1 } { \mathrm { f } } = \frac { 1 } { \mathrm { v } } - \frac { 1 } { \mathrm { u } } \Rightarrow \frac { 1 } { \mathrm { v } } = \frac { 1 } { \mathrm { f } } + \frac { 1 } { \mathrm { u } } = \frac { 1 } { - 6.0 } - \frac { 1 } { 18.0 }$

$\therefore$The image size, h’ = h