Question

A convex lens of focal length of $20\text{ cm}$ is used to form the image of an object placed $30\text{ cm}$ away from the lens. Find the position and nature of the image formed.

Hint:

For a quick mental check during examinations: When an object is placed between $F$ and $2F$ of a convex lens (here, $F=20\text{ cm}$ and $2F=40\text{ cm}$, object is at $30\text{ cm}$), the image must always be formed beyond $2F$ ($>40\text{ cm}$), and it must be real, inverted, and magnified. Our result of $+60\text{ cm}$ aligns perfectly!

Answer 1

Concept: For a thin spherical lens, the relationship between the focal length ($f$), the object distance ($u$), and the image distance ($v$) is governed accurately by the Thin Lens Formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] The nature of the image (whether it is real/inverted or virtual/erect) along with its size amplification can be derived using the linear magnification formula ($m$): \[ m = \frac{v}{u} \] A positive value of $v$ indicates a real image formed on the opposite side of the lens, whereas a negative magnification value confirms that the image is inverted.

Step 1: Identify given values and apply sign conventions. From the question statement, we extract the parameters for a convex lens:
• Focal length of the convex lens, $f = +20\text{ cm}$ (positive for a converging lens)
• Object distance, $u = -30\text{ cm}$ (negative since the object is placed in front of the lens against the direction of incident light) We need to calculate the image distance ($v$) and deduce its nature.

Step 2: Substitute values into the Thin Lens Formula. The lens formula is stated as: \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Isolating the unknown variable term $\frac{1}{v}$: \[ \frac{1}{v} = \frac{1}{f} + \frac{1}{u} \] Substitute the numerical values along with their proper operational signs: \[ \frac{1}{v} = \frac{1}{20} + \frac{1}{-30} \] \[ \frac{1}{v} = \frac{1}{20} - \frac{1}{30} \] To compute the subtraction, find the Least Common Multiple (LCM) of the denominators 20 and 30, which is 60: \[ \frac{1}{v} = \frac{3 \cdot 1 - 2 \cdot 1}{60} \] \[ \frac{1}{v} = \frac{3 - 2}{60} = \frac{1}{60} \] Taking the reciprocal of both sides gives the exact image position: \[ v = +60\text{ cm} \] The positive sign indicates that the image is formed at a distance of $60\text{ cm}$ behind the lens on the other side from where the object is located.

Step 3: Determine the magnification and nature of the image. Let us calculate the linear magnification ($m$) to confidently declare the final orientation and structural characteristics of the image: \[ m = \frac{v}{u} = \frac{60}{-30} = -2 \] Analysis of magnification value:
• The negative sign of the magnification explicitly signifies that the image formed is inverted.
• Since the real light rays actually converge to form an image on the back side of the lens ($v > 0$), the image is real.
• The absolute value $|m| = 2 > 1$ implies that the image is magnified to twice the size of the object. Hence, the image is located at a distance of $60\text{ cm}$ on the other side of the lens, and its nature is real and inverted.