For a thin convex lens of focal length f, the image of an object at O is formed at I, as shown in the figure below. The distances of object and image from the two focal points (\(F_O\) and \(F_I\)) are \(x_O\) and \(x_I\), respectively. Which of the following graphs correctly represent(s) the variation of the quantities shown in the figure?
Show Hint
Newton's lens equation, \(x_O x_I = f^2\), is a powerful alternative to the standard thin lens formula, especially when distances are measured from the focal points. Remembering this formula can save you the derivation time in an exam. Always check the axes of the graph carefully and rearrange the formula to match the \(y=mx+c\) form to verify linear relationships and their slopes.
Step 1: Understanding the Concept:
This problem relates the object and image distances measured from the focal points of a thin convex lens. This is described by Newton's lens equation. We need to derive this equation and then check which of the given graphs correctly represents the relationships derived from it. Step 2: Key Formula or Approach:
The standard thin lens formula is \(\frac{1}{v} - \frac{1}{u} = \frac{1}{f}\).
From the figure, we can relate the object distance \(u\) and image distance \(v\) to the distances \(x_O\) and \(x_I\).
Using the sign convention (light travels from left to right, optic center is the origin):
Object distance \(u = -(f + x_O)\). (Note: \(x_O\) is given as a distance, so it's a positive value).
Image distance \(v = +(f + x_I)\).
Substituting these into the thin lens formula will give us a relationship between \(x_O\) and \(x_I\), known as Newton's formula. Step 3: Detailed Explanation:
Let's substitute the expressions for \(u\) and \(v\) into the thin lens formula:
\[ \frac{1}{f + x_I} - \frac{1}{-(f + x_O)} = \frac{1}{f} \]
\[ \frac{1}{f + x_I} + \frac{1}{f + x_O} = \frac{1}{f} \]
Now, let's find a common denominator for the left side:
\[ \frac{(f + x_O) + (f + x_I)}{(f + x_I)(f + x_O)} = \frac{1}{f} \]
\[ \frac{2f + x_O + x_I}{f^2 + f x_O + f x_I + x_O x_I} = \frac{1}{f} \]
Cross-multiply:
\[ f(2f + x_O + x_I) = f^2 + f x_O + f x_I + x_O x_I \]
\[ 2f^2 + f x_O + f x_I = f^2 + f x_O + f x_I + x_O x_I \]
Cancel the terms \(f x_O\) and \(f x_I\) from both sides:
\[ 2f^2 = f^2 + x_O x_I \]
\[ x_O x_I = f^2 \]
This is Newton's lens formula. The problem uses magnitudes \(|x_O|\) and \(|x_I|\), but since \(x_O\) and \(x_I\) are defined as distances in the diagram, they are positive. So, \(|x_O| |x_I| = f^2\).
Now let's check the graphs: Graph (A):
This graph plots \(|x_O x_I|\) versus \(|x_I|\). Our derived formula is \(|x_O| |x_I| = f^2\). Since \(f\) is a constant focal length, the product \(|x_O| |x_I|\) is a constant, equal to \(f^2\). The graph shows exactly this: a constant value for the product, independent of \(|x_I|\). Thus, graph (A) is correct. Graph (B):
This graph plots \(\frac{1}{|x_I|}\) versus \(|x_O|\). From \(|x_O| |x_I| = f^2\), we can write \(\frac{1}{|x_I|} = \frac{|x_O|}{f^2}\).
This is a linear relationship of the form \(y = mx\), where \(y = \frac{1}{|x_I|}\), \(x = |x_O|\), and the slope is \(m = \frac{1}{f^2}\).
The graph shows a straight line passing through the origin, which is correct. However, it states the slope is \(f\). This is incorrect. The slope should be \(\frac{1}{f^2}\). So, graph (B) is incorrect. Graph (C):
This graph plots \(|x_I|\) versus \(\frac{1}{|x_O|}\). From \(|x_O| |x_I| = f^2\), we can write \(|x_I| = f^2 \left(\frac{1}{|x_O|}\right)\).
This is a linear relationship of the form \(y = mx\), where \(y = |x_I|\), \(x = \frac{1}{|x_O|}\), and the slope is \(m = f^2\).
The graph shows a straight line passing through the origin with a slope of \(f^2\). This matches our derivation. Thus, graph (C) is correct. Graph (D):
This graph plots \(|x_I|\) versus \(|x_O|\). From \(|x_O| |x_I| = f^2\), we have \(|x_I| = \frac{f^2}{|x_O|}\). This is an inverse relationship, representing a hyperbola, not a straight line. The graph incorrectly shows a linear relationship with slope \(f\). So, graph (D) is incorrect. Step 4: Final Answer:
The correct graphs are (A) and (C).
