Problem 35 - Master Resource Book in JEE Main Physics

Question

A convex lens of focal length, \(f\) is placed somewhere in between an object and a screen. The distance between object and screen is \(x\). If numerical value of magnification produced by lens is \(m\), focal length of lens is (a) \(\frac{m x}{(m+1)^{2}}\) (b) \(\frac{m x}{(m-1)^{2}}\) (c) \(\frac{(m+1]^{2}}{m} x\) (d) \(\frac{(m-1)^{2}}{m} x\)

Step-by-Step Solution

Answer

The correct answer is (b) \( \frac{m x}{(m-1)^{2}} \).

Step 1: Understand the Problem

We are given a convex lens with focal length \( f \), an object, and a screen. The distance between the object and the screen is \( x \). The lens produces a magnification \( m \). We need to find the expression for the focal length of the lens in terms of \( m \) and \( x \).

Step 2: Recall Lens Formula and Magnification Equation

The lens formula is \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \), where \( u \) is the object distance and \( v \) is the image distance. The magnification \( m \) is given by \( m = \frac{v}{u} \).

Step 3: Express Image and Object Distance in Terms of Given Variables

We know the total distance between the object and screen is \( x \). So, \( u + v = x \). Using the magnification equation \( m = \frac{v}{u} \), we can express \( v = m \cdot u \).

Step 4: Substitute into the Equation for v

Using the equation \( u + v = x \) and substituting \( v = m \cdot u \), we get: \( u + m \cdot u = x \). Simplifying, we find \( u(1 + m) = x \), and thus \( u = \frac{x}{1 + m} \).

Step 5: Determine Image Distance v

Substitute \( u \) in \( v = m \cdot u \) to obtain \( v = m \cdot \frac{x}{1 + m} = \frac{mx}{1 + m} \).

Step 6: Use Lens Formula to Find Focal Length

Substitute \( u = \frac{x}{1 + m} \) and \( v = \frac{mx}{1 + m} \) into the lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} = \frac{1+ m}{mx} - \frac{1+ m}{x} \). Simplify to get \( \frac{1}{f} = \frac{(1 + m) - (1 + m)m}{mx} = \frac{(1 + m)(1 - m)}{mx} \).

Step 7: Simplify to Find Expression for f

Simplify the expression \( \frac{(1 + m)(1 - m)}{m} \frac{x}{(1+m)^{2}} = \frac{(1 - m^2)}{m} \frac{x}{(1+m)^{2}} \). This simplifies to \( f = \frac{m x}{(m-1)^2} \).

Step 8: Match with Given Options

Comparing with the given options, the expression \( \frac{m x}{(m-1)^2} \) corresponds to option (b).

Key Concepts

Magnification and Lens Formula

In the study of optics, understanding the magnification and lens formula is essential.
Convex lenses are a common type of lens that converges light to form an image.
Essential to these calculations are two formulas: the lens formula and the magnification equation.The lens formula relates the focal length (\( f \)), the object distance (\( u \)), and the image distance (\( v \)) by the equation:

• \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]

This formula is crucial in determining how a lens will form an image at a specific location.Magnification (\( m \)) is another key concept in lens optics, defining how much larger or smaller an image appears compared to the object itself. The equation for magnification in terms of image and object distance is:

• \[ m = \frac{v}{u} \]

This ratio shows that magnification depends on how close or far the image is compared to the object. Understanding these formulas enables us to solve complex problems related to image formation in lenses, making them a powerful tool in optics.

Image and Object Distance

To determine where an image will form in relation to the lens, it's important to understand the relationship between image and object distance.
These distances are represented as \( v \) (for the image) and \( u \) (for the object).Consider a scenario where a convex lens is positioned between an object and a screen, and the distance between them is \( x \). Knowing that the sum of the image and object distance equals this total distance, we have:

• \[ u + v = x \]

Given the magnification \( m \) equation (\( m = \frac{v}{u} \)), we can express one distance in terms of the other. For instance, if we express the image distance (\( v \)) in terms of the object distance (\( u \)), it becomes:

• \[ v = m \cdot u \]

Substituting back into \( u + v = x \), we can solve for each distance:

• \[ u (1 + m) = x \]
• \[ u = \frac{x}{1 + m} \]
• \[ v = \frac{mx}{1 + m} \]

Knowing these distances allows for accurate placement of the lens to form a sharp image.

Optics Problem Solving

Solving lens-related problems requires a clear understanding of both the equations involved and logical problem-solving skills.
In this exercise, we began by understanding the relationships between the given variables and derived values using fundamental optics formulas.The process involved:

• First, using the constraint of total distance \( u + v = x \), we found expressions for both object and image distances in terms of given magnification \( m \) and screen distance \( x \).
• Next, substituting back into the lens formula helped derive an expression for the focal length \( f \).
• By simplifying \( \frac{1+ m}{mx} - \frac{1+ m}{x} \), we achieved the formula for \( \frac{1}{f} \): \[ \frac{(1 + m)(1 - m)}{mx} \]
• Finally, inverting the result yielded the sought expression: \[ f = \frac{m x}{(m-1)^2} \]

This step-by-step approach is invaluable in optics problem solving, allowing us to break down complex scenarios into manageable parts. By doing so, we develop a stronger comprehension of how different variables in lenses influence the type and position of images formed.