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Q38(i):
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. (i) What is the focal length of the convex lens?

Solution :

Step 1: Identification of Coordinates and Object-Image Distances

To determine the focal length ($f$), we must first calculate the object distance ($u$) and the image distance ($v$) relative to the optical center of the convex lens, located at $50.0\text{ cm}$.

  • Position of Object (Candle) = $12.0\text{ cm}$
  • Position of Lens = $50.0\text{ cm}$
  • Position of Screen = $88.0\text{ cm}$

Using the sign convention [Cartesian Sign Convention for Spherical Lenses], light travels from left to right. The lens is at the origin of the coordinate system.

Object Distance ($u$): $u = (\text{Position of Candle} - \text{Position of Lens}) = 12.0 - 50.0 = -38.0\text{ cm}$.

Image Distance ($v$): $v = (\text{Position of Screen} - \text{Position of Lens}) = 88.0 - 50.0 = +38.0\text{ cm}$.

Step 2: Theoretical Application of the Lens Formula

The relationship between focal length ($f$), image distance ($v$), and object distance ($u$) is defined by the Thin Lens Equation [Gaussian Form]:

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

Substituting the calculated values:

$\frac{1}{f} = \frac{1}{38.0} - \frac{1}{-38.0}$

$\frac{1}{f} = \frac{1}{38.0} + \frac{1}{38.0}$

$\frac{1}{f} = \frac{2}{38.0}$

Step 3: Calculating Focal Length

Solving for $f$:

$f = \frac{38.0}{2} = 19.0\text{ cm}$

Step 4: Geometric Representation of the Optical System

Candle (12cm) Lens (50cm) Screen (88cm) u = -38cm v = +38cm

Note: Since $|u| = |v|$, the object is placed at $2f$. Given $u = 38\text{ cm}$, $2f = 38\text{ cm}$, which confirms $f = 19\text{ cm}$ [Property of Convex Lens: When an object is at $2f$, the image is formed at $2f$ on the other side].

Final Solution: The focal length of the convex lens is 19.0 cm.


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