Answer to Question #269229 in Optics for NICKO

Question

  A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as the object. (a) What is the focal length of the lens? Is the lens converging or diverging? (b) If the object is 8.50 mm tall, how tall is the image? Is it erect or inverted? (c) Draw a principal-ray diagram.

Answer 1

(a) Since image distance, v is less than the object distance, u and since both object and image are on the same side, the lens is DIVERGING.

so v is taken as negative, and f will be negative too

$\frac{1}{f} = \frac{1}{u} + \frac{1}{v} \\ f = \frac{uv }{ v+u} \\ f = \frac{ 16 \times ( -12) }{ -12 + 16 } \\ f = \frac{-192 }{ 4} \\ f = - 48 \;cm$

negative indicates negative focal length

(b) magnification $= \frac{v}{u}$ = image height / object height

$\frac{-12 }{ 16} = \frac{image \; height }{ 8.5}$

image height $= \frac{ -12 \times 8.5 }{ 16}$

image height $= \frac{-102 }{ 16}$

image height = -6.375 cm

It is erect.

(c)