Question Details

A bi-convex lens forms a real, inverted image twice as large as the object. Where is the object located relative to the lens?

Options

Between one focal length and twice the focal length from the lens

Beyond twice the focal length from the lens

At twice the focal length from the lens

At the focal length from the lens

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Correct Answer :

Option A

Between one focal length and twice the focal length from the lens

Solution :

The correct option is: Between one focal length and twice the focal length from the lens.

Let us understand this step-by-step using the lens formula and the definition of magnification.

Step 1: Understand the given information
1. We are using a bi-convex lens, which is a converging lens. Therefore, its focal length _f is positive (_{f > 0}).
2. The lens forms a real and inverted image.
3. The image is twice as large as the object. This means the magnitude of magnification _|m| is 2.
Since the image is inverted, the magnification _m is negative:
$m = - 2$

Step 2: Relate magnification to object and image distances
For a thin lens, the magnification _m is given by the ratio of the image distance (_v) to the object distance (_u):
$m = \frac{v}{u}$
Substituting _{m = -2} into the formula:
$- 2 = \frac{v}{u} \Rightarrow v = - 2 u$

Step 3: Apply the Lens Formula
The lens formula relates the focal length (_f), object distance (_u), and image distance (_v):
$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Substitute _{v = -2u} into the lens formula:
$\frac{1}{f} = \frac{1}{- 2 u} - \frac{1}{u}$
Simplify the right-hand side by finding a common denominator:
$\frac{1}{f} = \frac{- 1 - 2}{2 u}$
$\frac{1}{f} = - \frac{3}{2 u}$
Solving for _u:
$u = - \frac{3 f}{2} = - 1.5 f$

Step 4: Interpret the result
The negative sign indicates that the object is placed on the real side (in front of the lens) according to standard sign conventions.
The magnitude of the object distance is:
$| u | = 1.5 f$
Since _1.5f lies between _f (one focal length) and _2f (twice the focal length), we conclude that the object must be located between one focal length and twice the focal length from the lens.