Question

A concave lens with focal length-15.5 cm creates a virtual image-13.5 cm from the lens. If theimage is 6.25 cm tall, what is theobject height?(Mind your minus signs.)(Unit = cm)

Answer 1

Answer:

Object height = 48.45 cm

Step-by-step explanation:

Focal length, f = -15.5 cm

Image distance, v = -13.5 cm

Object distance, u = ?

Relationship between image distance, object distance, and focal length is:

`(1)/(f)=(1)/(u)+(1)/(v)`
`\begin{gathered} (-1)/(15.5)=(1)/(u)-(1)/(13.5) \\ \\ (1)/(u)=(1)/(13.5)-(1)/(15.5) \\ \\ (1)/(u)=0.00955794504 \\ \\ u=(1)/(0.00955794504) \\ \\ u=104.625\text{ cm} \\ \\ \\ \end{gathered}`

Magnification = |v/u|

`\begin{gathered} M=|(v)/(u)| \\ \\ M=|(-13.5)/(104.625)| \\ \\ M=0.129 \end{gathered}`

Image height, H = 6.25 cm

Object height, h = ?

`\begin{gathered} M=(H)/(h) \\ \\ 0.129=(6.25)/(h) \\ \\ h=(6.25)/(0.129) \\ \\ h=48.45\text{ cm} \end{gathered}`

Object height = 48.45 cm