Question

A 24 cm candle is placed 6 m in front of a thin diverging lens with a focal length magnitude of 3 m. Find the image's height and classify it as real or virtual, upright or inverted, enlarged or reduced.

Answer 1

Image is virtual and formed on the same side as the object. 2 m from the lens.

The size of the image is 7.97 cm

Image is upright as the magnification is positive and smaller than the object.

Step-by-step explanation:

u = Object distance = 6 m

v = Image distance

f = Focal length = -3 m (concave lens)

`h_u`= Object height = 24 cm

Lens Equation

`(1)/(f)=(1)/(u)+(1)/(v)\\\Rightarrow (1)/(f)-(1)/(u)=(1)/(v)\\\Rightarrow (1)/(v)=(1)/(-3)-(1)/(6)\\\Rightarrow (1)/(v)=(-1)/(2) \\\Rightarrow v=(-2)/(1)=-2\ m`

Image is virtual and formed on the same side as the object. 2 m from the lens.

Magnification

`m=-(v)/(u)\\\Rightarrow m=-(-2)/(6)\\\Rightarrow m=0.33`

`m=(h_v)/(h_u)\\\Rightarrow 0.33=(h_v)/(0.24)\\\Rightarrow h_v=0.33* 0.24=0.0797\ m`

The size of the image is 7.97 cm

Image is upright as the magnification is positive and smaller than the object.