Chapter 23: Problem 35
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) \(\frac{(m-1)^{2}}{m} x\)
Short Answer
Step by step solution
Understand the Problem
Recall Lens Formula and Magnification Equation
Express Image and Object Distance in Terms of Given Variables
Substitute into the Equation for v
Determine Image Distance v
Use Lens Formula to Find Focal Length
Simplify to Find Expression for f
Match with Given Options
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Magnification and Lens Formula
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:
- \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \]
- \[ m = \frac{v}{u} \]
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:
- \[ u + v = x \]
- \[ v = m \cdot u \]
- \[ u (1 + m) = x \]
- \[ u = \frac{x}{1 + m} \]
- \[ v = \frac{mx}{1 + m} \]
Optics Problem Solving
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: \[ f = \frac{m x}{(m-1)^2} \]