91Ó°ÊÓ

Candle Distance from wall = 2m = 200cm

Focal length of lens = 32cm

Flame height = 2cm

Applying Lens formula

$$\begin{gathered} \frac{1}{v}-\frac{1}{u}=\frac{1}{f} \\ ⇒\frac{u-v}{vu}=\frac{1}{32} \end{gathered}$$

Applying sign convention we know value of 'u' will be negative

$$\begin{gathered} ⇒\frac{-u-v}{uv}=\frac{1}{32} \\ ⇒\frac{-(u+v)}{uv}=\frac{1}{32} \\ ⇒\frac{-200}{uv}=\frac{1}{32}\{u+v=200cm \\ ⇒uv=-6400 \end{gathered}$$

Let's call this equation 1,

We also know that sum of object distance and image distance will be equal to distance of candle from wall.

i.e. u + v = 200cm.

Let's call this equation 2.

Now we have two equations with two variables, so this can be solved.

Substituting v = u - 200 from equation 2 to equation 1.

$$\begin{gathered} u(u-200)=-6400 \\ ⇒u^2-200u+6400=0 \\ Solveforu \\ u=160,40 \end{gathered}$$

So we can put the lens either at 160 cm away from candle or 40 cm away from candle in order to obtain a sharp image on wall.

Now we know $m=\frac{v}{u}=\frac{Heightofimage}{Heightofobject}$

For u = 40 (v = 160)

$$\begin{gathered} m=\frac{160cm}{-40cm}=\frac{Heightofimage}{2cm} \\ ⇒-4×2cm=heightofimage \\ ⇒heightofimage=-8cm \end{gathered}$$

Image is real and inverted.

For u = 160(v = 40)

$$\begin{gathered} m=\frac{40}{-160}=\frac{heightofimage}{2cm} \\ ⇒\frac{-1×2cm}{4}=heightofimage \\ ⇒-0.5cm=heightofimage \end{gathered}$$

Image is real and inverted.