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The lens formula is a crucial tool for understanding how light interacts with lenses. It relates the object distance, image distance, and focal length of the lens. This formula is expressed as:\[ \frac{1}{f} = \frac{1}{d_o} + \frac{1}{d_i} \]Here, \(f\) is the focal length (positive for converging lenses), \(d_o\) is the distance from the object to the lens, and \(d_i\) is the distance from the image to the lens. The lens formula helps determine the position of the image formed by a lens when the object distance and focal length are known.
In our exercise, understanding the lens formula is the first step in finding the image distance for a converging lens. By rearranging and solving, we can calculate \(d_i\), the position where the image forms, which is critical for determining the nature of the image.

Image distance, denoted as \(d_i\), describes how far the formed image is from the lens. When solving optical problems with lenses, determining \(d_i\) gives insight into the physical layout of the imaging system.
For a converging lens:

• If \(d_i\) is positive, the image forms on the opposite side of the object, signifying a real image.
• If \(d_i\) is negative, the image appears on the same side as the object, indicating a virtual image.

In our specific exercise:- With an object distance \(d_o = 150\text{ cm}\), \(d_i = 300\text{ cm}\), showing a real image.- For \(d_o = 75\text{ cm}\), \(d_i = -300\text{ cm}\), demonstrating a virtual image.
This essential measurement helps classify the type and orientation of the image.

Transverse magnification (\(m\)) describes how much larger or smaller the image is compared to the object. It is defined mathematically as the ratio of the image distance to the object distance, with a negative sign:\[ m = -\frac{d_i}{d_o} \]The sign of \(m\) provides additional insights:

• A negative magnification indicates that the image is inverted compared to the object.
• A positive magnification suggests that the image is upright.

For the given exercise:- Case (a): With a magnification of \(-2\), the image is inverted and twice the size of the object.- Case (b): With a magnification of \(4\), the image is upright and four times larger than the object.
Understanding magnification is essential for predicting how the image will appear in relation to the original object.

A lens can form images that are either real or virtual, two distinct types each with its characteristics.
**Real Images:**

• Form on the opposite side of the lens from the object.
• Can be projected onto a screen.
• Typically inverted from the object.

**Virtual Images:**

• Appear on the same side as the object relative to the lens.
• Cannot be projected; only visible when looking through the lens.
• Generally upright.

In the specific cases of our exercise:- When the object is placed at \(150\text{ cm}\) from the lens, a real image is formed at \(300\text{ cm}\) on the opposite side.- At an object distance of \(75\text{ cm}\), a virtual image appears at \(-300\text{ cm}\) on the same side.
Knowing whether an image is real or virtual is important in applications ranging from everyday lenses like eyeglasses to complex optical instruments.