A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as given below,
Position of candle = $12 cm$
Position of convex lens = $50 cm$

Position of the screen = $88 cm$

The image will be formed if he shifts the candle towards the lens to a position of $31 cm$ . The image formed if he further shifts the candle towards the lens will be virtual, erect and[[1]].

Question

A student focussed the image of a candle flame on a white screen using a convex lens. He noted down the position of the candle, screen and the lens as given below,
Position of candle = $12 cm$
Position of convex lens = $50 cm$

Position of the screen = $88 cm$

The image will be formed if he shifts the candle towards the lens to a position of $31 cm$ . The image formed if he further shifts the candle towards the lens will be virtual, erect and[[1]].

Accepted Answer

A student focussed the image of a candle flame on a white screen using a convex lens. The image formed if he further shifts the candle towards the lens will be virtual, erect and magnified.
Given,

u

is the distance of candle from the lens

\Rightarrow 50 - 12 cm = 38 cm

According to sign convention,
u =

- 38 cm

Now,

v

is the distance of image from the lens

\Rightarrow 88 - 50 = 38 cm

As per sign convention,
v = +

38 cm

Using lens formula,

\frac{1}{f} = \frac{1}{v} - \frac{1}{u}

\frac{1}{f} = \frac{1}{38} + \frac{1}{38}

f = \frac{38}{2}

= 19 cm

If the candle is further shifted towards the lens, the candle lies between the optical center and focus of the lens.
Thus a virtual, erect and magnified image is formed on the same side as the object.

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