A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Using lens formula, find the position, size and nature of the image formed.

Using the lens formula and magnification equations, we can determine the position, size, and nature of the image formed by the convex lens.

Given Data:

• Height of the object ($h$) = $+4\text{ cm}$
• Focal length of the convex lens ($f$) = $+24\text{ cm}$ (positive for a convex lens)
• Object distance ($u$) = $-16\text{ cm}$ (always negative according to sign convention)

1. Position of the Image ($v$):

According to the lens formula:

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

Rearranging the formula to solve for $\frac{1}{v}$:

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

Substituting the given values:

$$\frac{1}{v} = \frac{1}{24} + \frac{1}{-16}$$

$$\frac{1}{v} = \frac{1}{24} - \frac{1}{16}$$

To subtract these fractions, find the Least Common Multiple (LCM) of $24$ and $16$, which is $48$:

$$\frac{1}{v} = \frac{2 - 3}{48}$$

$$\frac{1}{v} = -\frac{1}{48}$$

$$v = -48\text{ cm}$$

Position: The image is formed at a distance of $48\text{ cm}$ from the optical centre, on the same side of the lens as the object.

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2. Size of the Image ($h'$):

The magnification ($m$) of a lens is given by:

$$m = \frac{v}{u} = \frac{h'}{h}$$

Substitute the values of $v$, $u$, and $h$:

$$\frac{-48}{-16} = \frac{h'}{4}$$

$$3 = \frac{h'}{4}$$

$$h' = 3 \times 4 = +12\text{ cm}$$

Size: The image is $12\text{ cm}$ tall (magnified to three times the object's size).

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3. Nature of the Image:

• Since the image distance ($v$) is negative, the image is formed on the same side as the object, meaning it is virtual.
• Since the height of the image ($h'$) is positive, the image is erect.

Conclusion: The image is virtual, erect, and magnified.