Problem 54 - Master Resource Book in JEE Main Physics

Question

In order of obtain a real image of magnification 2 , using a converging lens of focal length \(20 \mathrm{~cm}\), where should an object be placed (a) \(50 \mathrm{~cm}\) (b) \(30 \mathrm{~cm}\) (c) \(-50 \mathrm{~cm}\) (d) \(-30 \mathrm{~cm}\)

Step-by-Step Solution

Answer

None of the given distances work; revisiting other possible distances is needed for verification.

Step 1: Understanding the Problem

We want to find the object distance for a converging lens that creates a real image with a magnification of 2. The focal length given is 20 cm. We need to check different object positions to see which allows this condition to be met.

Step 2: Lens Formula and Magnification Formula

To solve this problem, we will use the lens formula and magnification formula:1. Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( f \) is the focal length, \( v \) is the image distance, and \( u \) is the object distance.2. Magnification \( m = \frac{v}{u} \). Given \( m = 2 \), we have \( \frac{v}{u} = 2 \), thus, \( v = 2u \).

Step 3: Calculate using Option (a) \( u = 50 \mathrm{~cm} \)

Substitute \( v = 2u = 100 \mathrm{~cm} \) and \( u = 50 \mathrm{~cm} \) into the lens formula:\[ \frac{1}{20} = \frac{1}{100} - \frac{1}{50} \]\[ \frac{1}{20} eq \frac{1}{100} - \frac{1}{50} \]This leads to a contradiction, so \( u = 50 \mathrm{~cm} \) isn't the correct answer.

Step 4: Calculate using Option (b) \( u = 30 \mathrm{~cm} \)

Substitute \( v = 2u = 60 \mathrm{~cm} \) and \( u = 30 \mathrm{~cm} \) into the lens formula:\[ \frac{1}{20} = \frac{1}{60} - \frac{1}{30} \]\[ \frac{1}{20} = \frac{1}{60} + \frac{1}{60} = \frac{2}{60} = \frac{1}{30} \]This is incorrect because it leads to contradiction, so \( u = 30 \mathrm{~cm} \) isn't the correct answer.

Step 5: Calculate using Option (c) \( u = -50 \mathrm{~cm} \)

Here, \( u \) would imply the object is virtual, which isn't relevant for a real image formation in this context, so we don't use negative object distances here.

Step 6: Calculate using Option (d) \( u = -30 \mathrm{~cm} \)

Similarly to (c), using \( u = -30 \mathrm{~cm} \) also implies a virtual object, thus falling outside the relevance for creating a real image. Negative values are not pertinent for this problem as we deal explicitly with real images.

Key Concepts

Lens Formula

The lens formula is essential for understanding how lenses work, particularly when talking about converging lenses. In a simple converging lens system, the lens formula is expressed as:

• \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)

Here, \(f\) is the lens's focal length, \(v\) represents the image distance, and \(u\) stands for the object distance. This formula helps in calculating where an object should be placed to form an image of a certain quality or size. It reflects the relationship between these distances, so you can predict how a change in one will affect the other.
Understanding this relationship is crucial because it lets us determine how each element of a lens setup interacts. In essence, it's the mathematical backbone behind lens design and usage.
By applying this formula, we can solve problems involving image and object distances effectively.

Magnification

Magnification is another vital concept when it comes to lenses and optics in general. Magnification tells us how much larger or smaller an image is compared to the object itself. For lenses, the magnification (\(m\)) formula is:

• \( m = \frac{v}{u} \)

where \(v\) is the image distance and \(u\) is the object distance. A magnification of 2, as seen in the given exercise, means the image is twice the size of the object.
When dealing with real images produced by converging lenses, positive magnification indicates the image is upright, while a negative value would suggest an inverted image. The exercise asked for a real image with positive magnification.
By ensuring \( m = \frac{v}{u} = 2 \), you can substitute \( v = 2u \) into the lens formula to solve for specific object and image distances, making this a key calculation step in optics problems.

Focal Length

The focal length is a fundamental parameter in optics that characterizes lenses. It is the distance from the center of the lens to the focal point, where parallel rays of light converge. For a converging lens, the focal length is positive and pivotal in determining the overall behavior of the lens.

• Focal length, denoted as \(f\), plays a critical role in forming images.
• It is the main driver behind a lens's ability to converge or diverge light.

For a converging lens with a focal length of \(20\) cm, as indicated in our problem, it controls how rays bend to meet at a point, thus forming a real image.
Knowing the focal length allows us to use the lens formula effectively, as seen in the provided exercise. It enables us to find the right object distances needed to create specific types of images.
Mastering this concept is crucial for any learner aiming to understand or design optical systems.