Question

A convex lens of focal length f is placed somewhere between object and screen. The distance between object and screen is x. If m be the magnification of lens, then prove that \(f = \frac{mx}{(m + 1)^2\).}

Hint:

This is a standard result in optics. A key step in such derivations is to express the object distance (\(u\)) and image distance (\(v\)) in terms of the given variables (\(x\) and \(m\)) before substituting them into the lens formula.

Answer 1

Step 1: Understanding the Concept and Variables:
We are considering a real image formed by a convex lens on a screen. \begin{itemize} \item Let \(u\) be the magnitude of the object distance. \item Let \(v\) be the magnitude of the image distance. \item The total distance between the object and the screen is \(x\), so \(x = u + v\). \item The magnification is \(m\). For a real, inverted image, the magnification is negative, but here \(m\) is treated as its magnitude, so \(m = \frac{v}{u}\). \end{itemize}

Step 2: Key Formula or Approach:
The thin lens formula connects the object distance, image distance, and focal length. For a real image formed by a convex lens, using sign convention (\(u\) is negative, \(v\) is positive), the formula \(\frac{1}{v} - \frac{1}{(-u)} = \frac{1}{f}\) becomes: \[ \frac{1}{v} + \frac{1}{u} = \frac{1}{f} \] where \(u\) and \(v\) are magnitudes.

Step 3: Detailed Explanation (Derivation):
From the magnification formula, we have \(v = mu\).
Substitute this into the distance equation \(x = u + v\): \[ x = u + mu = u(1+m) \] From this, we can express the object distance \(u\) in terms of \(x\) and \(m\): \[ u = \frac{x}{m+1} \] Now, we can find the image distance \(v\): \[ v = mu = m\left(\frac{x}{m+1}\right) = \frac{mx}{m+1} \] Now, substitute the expressions for \(u\) and \(v\) into the lens formula: \[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} = \frac{1}{\frac{mx}{m+1}} + \frac{1}{\frac{x}{m+1}} \] \[ \frac{1}{f} = \frac{m+1}{mx} + \frac{m+1}{x} \] Take \(\frac{m+1}{x}\) as a common factor: \[ \frac{1}{f} = \frac{m+1}{x} \left(\frac{1}{m} + 1\right) \] \[ \frac{1}{f} = \frac{m+1}{x} \left(\frac{1+m}{m}\right) = \frac{(m+1)^2}{mx} \] Finally, inverting the expression to find \(f\): \[ f = \frac{mx}{(m+1)^2} \] Hence, the relation is proved.