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Q38(iii):
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 under: Position of candle = 12.0 cm, Position of convex lens = 50.0 cm, Position of the screen = 88.0 cm. (iii) What will be the nature of the image formed if he further shifts the candle towards the lens?

Solution :

Step 1: Identifying Initial Optical Parameters

First, we calculate the object distance ($u$) and image distance ($v$) from the lens position ($P_l = 50.0\text{ cm}$):

  • Position of candle ($P_c$) = $12.0\text{ cm}$
  • Position of lens ($P_l$) = $50.0\text{ cm}$
  • Position of screen ($P_s$) = $88.0\text{ cm}$

Object distance ($u$) = $P_c - P_l = 12.0 - 50.0 = -38.0\text{ cm}$ [By sign convention, light travels left to right; distances measured opposite to incident light are negative].

Image distance ($v$) = $P_s - P_l = 88.0 - 50.0 = +38.0\text{ cm}$.

Step 2: Calculating the Focal Length ($f$)

Using the Lens Formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$

$\frac{1}{f} = \frac{1}{38} - \frac{1}{-38} = \frac{1}{38} + \frac{1}{38} = \frac{2}{38} = \frac{1}{19}$

$f = 19.0\text{ cm}$

[The object is placed at $2f$ ($2 \times 19 = 38\text{ cm}$), which explains why the image is formed at $2f$ on the other side].

Step 3: Analyzing the Shift of the Candle

The student shifts the candle towards the lens. If the candle moves from $12.0\text{ cm}$ toward $50.0\text{ cm}$, the new object distance $u'$ becomes less than $38.0\text{ cm}$. Specifically, $19.0\text{ cm} < |u'| < 38.0\text{ cm}$ [When the object is placed between $F$ and $2F$].

Lens F 2F Object

Step 4: Determining the Nature of the Image

Based on the lens theory for a convex lens: [When an object is placed between $F$ and $2F$, the image formed is real, inverted, and magnified].

Object Position Image Position Nature of Image
$u = 2f$ $v = 2f$ Real, Inverted, Same size
$f < u < 2f$ $v > 2f$ Real, Inverted, Magnified

As the candle is shifted towards the lens (reducing $u$ to a value between $f$ and $2f$), the image moves beyond the initial $2f$ position and increases in size.

Final Solution: The image formed will be real, inverted, and magnified.


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