Answer:
The height of the image of the candle is 20 cm.
Explanation:
Given that,
Size of the candle, h = 12 cm
Object distance from the candle, u = -6 cm
Focal length of converging lens, f = 15 cm
To find,
The height of the image of the candle.
Solution,
Firstly, we will find the image distance of the candle. Let it is equal to v. Using lens formula to find the image distance.
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}[/tex]
v is image distance
[tex]\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{f}+\dfrac{1}{u}\\\\\dfrac{1}{v}=\dfrac{1}{15}+\dfrac{1}{(-6)}\\\\v=-10\ cm[/tex]
If h' is the height of the image. Magnification is given by :
[tex]m=\dfrac{h'}{h}=\dfrac{v}{u}[/tex]
[tex]h'=\dfrac{vh}{u}\\\\h'=\dfrac{-10\times 12}{-6}\\\\h'=20\ cm[/tex]
So, the height of the image of the candle is 20 cm.
