Problem 7 - Master Resource Book in JEE Main Physics

Question

An object is placed \(30 \mathrm{em}\) to the left of a diverging lens whose focal length is of magnitude \(20 \mathrm{~cm}\). Which one of the following correctly states the nature and position of the virtual image formed? \(\begin{array}{ll}\text { Nature of image } & \text { Distance from lens }\end{array}\) (a) inverted enlarged \(60 \mathrm{~cm}\) to the right (b) erect, diminished \(12 \mathrm{~cm}\) to the left (c) irverted, enlarged \(60 \mathrm{~cm}\) to the left \(\begin{array}{ll}\text { (d) erect, diminished } & 12 \mathrm{~cm} \text { to the right }\end{array}\) (c) imverted, enlarged \(12 \mathrm{~cm}\) to the left

Step-by-Step Solution

Answer

The image is erect and diminished, 60 cm to the left of the lens.

Step 1: Identify Given Values

The object distance from the lens is given as \(u = -30\, \text{cm}\). The focal length of the diverging lens is \(f = -20\, \text{cm}\) (focal length is negative for diverging lenses). We need to find the nature and position of the virtual image.

Step 2: Use Lens Formula

The lens formula relates the object distance \(u\), image distance \(v\), and focal length \(f\) as: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]Plug in the given values: \[ \frac{1}{-20} = \frac{1}{v} + \frac{1}{-30} \]

Step 3: Solve for Image Distance

Rearrange the lens formula to solve for \(v\):\[ \frac{1}{v} = \frac{1}{-20} + \frac{1}{30} = -\frac{3}{60} + \frac{2}{60} = -\frac{1}{60} \]Thus, the image distance is \(v = -60\, \text{cm}\), meaning the image is formed 60 cm to the left of the lens.

Step 4: Determine Image Nature

The image distance \(v = -60\, \text{cm}\) indicates the image is on the same side as the object, implying it is virtual. Since it is formed by a diverging lens, it will be erect and diminished.

Key Concepts

Lens Formula

The lens formula is a fundamental equation used to determine the relationship between the object distance, the image distance, and the focal length of a lens. It is written as: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \]where:

• \(f\) is the focal length of the lens,
• \(v\) is the image distance,
• \(u\) is the object distance.

Using this formula, we can find unknown values when we know the other two. For diverging lenses, the focal length \(f\) is negative, which affects the signs and subsequently the calculations. Always double-check the placements of negative signs, especially in problems involving diverging lenses. More complex problems may require rearranging the formula to solve for \(v\) or \(u\). This formula is crucial because it helps predict where an image will form and whether it will be real or virtual.

Virtual Image

A virtual image is not formed at a place from which light actually comes; rather, it is an image where light rays appear to converge but do not in reality. For diverging lenses, the images formed are typically virtual. This means the image can't be projected onto a screen because it forms on the same side of the lens as the object. Characteristics of a virtual image in the context of a diverging lens include:

• Being upright compared to the object.
• Being smaller than the actual object, hence diminished.
• Occurring on the same side as the object, so it has a negative image distance \(v\).

In practical scenarios, virtual images might commonly be seen in magnifying glasses or certain camera viewfinders, where only the observer experiences the light's pathway, making the image seem projected behind the lens.

Focal Length of Lenses

The focal length of a lens is a critical measure indicating how strongly it converges or diverges light. For diverging lenses, like the one in our exercise, the focal length is always negative, reflecting the lens's behavior of spreading out light rays. It is represented in centimeters or meters depending on the context. Understanding the focal length helps identify:

• The overall strength of the lens - shorter focal lengths indicate stronger divergence.
• The characteristics of the images that form - diverging lenses create smaller, virtual images when objects are placed at standard distances.
• How the lens affects light paths in optical devices, contributing to the design of eyeglasses, cameras, and projections systems.

It is essential to remember that in optical equations, sign conventions matter; for example, with the lens formula, incorporating the correct sign for the focal length ensures accurate results when determining image positions.