Question

The shortest distance between an object and its real image formed by a thin convex lens of focal length 20 cm is _____ cm. (in integer)

Hint:

For a convex lens, the minimum distance between an object and its real image is \(4f\). This occurs when the object is placed at a distance of \(2f\) from the lens, which results in an image also at \(2f\) on the other side (magnification of -1). Remembering this result can be a significant shortcut in exams.

Answer 1

Correct Answer: 80

Step 1: Understanding the Concept:
For a thin convex lens, a real image is formed when the object is placed at a distance greater than the focal length. The distance between the object and its real image depends on the object's position. We need to find the minimum possible value for this distance. 
Step 2: Key Formula or Approach: 
The thin lens formula relates the object distance \(u\), image distance \(v\), and focal length \(f\): \[ \frac{1}{v} - \frac{1}{u} = \frac{1}{f} \] Let the distance of the object from the lens be \(x\). Using the standard sign convention, \(u = -x\). Let the distance of the image from the lens be \(v\). The total distance between the object and the image is \(D = x + v\). We need to find the minimum value of \(D\). 
Step 3: Detailed Explanation: 
Given \(f = 20\) cm. Let the object distance be \(u = -x\). For a real image formed by a convex lens, \(v\) is positive. Using the lens formula: \[ \frac{1}{v} - \frac{1}{-x} = \frac{1}{f} \implies \frac{1}{v} = \frac{1}{f} - \frac{1}{x} = \frac{x-f}{xf} \] \[ v = \frac{xf}{x-f} \] For a real image, \(v > 0\), which requires \(x - f > 0\), so the object distance \(x\) must be greater than the focal length \(f\). 
The distance between the object and the image is \(D = x + v\): \[ D(x) = x + \frac{xf}{x-f} = \frac{x(x-f) + xf}{x-f} = \frac{x^2 - xf + xf}{x-f} = \frac{x^2}{x-f} \] To find the minimum distance, we differentiate \(D(x)\) with respect to \(x\) and set the derivative to zero: \[ \frac{dD}{dx} = \frac{d}{dx} \left( \frac{x^2}{x-f} \right) = \frac{(x-f)(2x) - x^2(1)}{(x-f)^2} = \frac{2x^2 - 2xf - x^2}{(x-f)^2} = \frac{x^2 - 2xf}{(x-f)^2} \] Setting \(\frac{dD}{dx} = 0\): \[ x^2 - 2xf = 0 \implies x(x - 2f) = 0 \] Since \(x > f\), the only valid solution is \(x = 2f\). This is the object distance at which the separation is minimum. 
Now we find the minimum distance \(D_{min}\) by substituting \(x = 2f\) back into the expression for \(D(x)\): \[ D_{min} = \frac{(2f)^2}{2f - f} = \frac{4f^2}{f} = 4f \] Given the focal length \(f = 20\) cm: \[ D_{min} = 4 \times 20 \text{ cm} = 80 \text{ cm} \] Step 4: Final Answer: 
The shortest distance between the object and its real image is 80 cm.