Question

What magnification will be produced by a lens of power −4.10 D (such as might be used to correct myopia) if an object is held 22.5 cm away?

Answer 1

Answer:

The magnification = 0.52

Step-by-step explanation:

Power, P = -4.10D

Object distance, u = 22.5 cm

Focal length, f = 100/power

f = 100/-4.10

f = -24.39 cm

Using the formula below, calculate the image distance (v)

`(1)/(f)=(1)/(u)+(1)/(v)`

Substitute u = 22.5, f = -24.39, and solve for v

`\begin{gathered} (1)/(-24.39)=(1)/(22.5)+(1)/(v) \\ \\ (1)/(v)=-(1)/(24.39)-(1)/(22.5) \\ \\ (1)/(v)=-0.041-0.044 \\ \\ (1)/(v)=-0.085 \\ \\ v=-(1)/(0.085) \\ \\ v=-11.8 \end{gathered}`

The magnification, M = |v/u|

M = |-11.8/22.5|

M = |-0.52|

M = 0.52

The magnification = 0.52