Question

An object is located on a wall, its image of equal size is to be obtained on a parallel wall with the help of a convex lens. The lens is placed at a distance '$d$' in front of the second wall. The required focal length of the lens is

• A. less than $\frac{d}{4}$
• B. only $\frac{d}{4}$
• C. more than $\frac{d}{4}$ but less than $\frac{d}{2}$
• D. only $\frac{d}{2}$

Hint:

To get a real image of the exact same size using a convex lens, the object must be placed at $2F$ and its image will form on the other side at $2F$. Therefore, the image distance $v$ is exactly equal to $2f$. Since we are given $v = d$, we can write $2f = d \implies f = \frac{d}{2}$ instantly!

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
An object is fixed on one wall, and a convex lens projects a real image of the exact same size onto an opposite parallel wall. The distance from the lens to this second wall (image distance) is given as $d$. We need to determine the exact focal length $f$ of the lens.

Step 2: Key Formula or Approach:
For a real image formed by a convex lens, the linear magnification $m$ is given by: $$m = -\frac{v}{u}$$ Since the image is real and inverted but equal in size to the object, the magnification value must be $m = -1$. This means the object distance $u$ and image distance $v$ must be numerically equal: $|u| = v$. We then use the standard lens formula: $$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$

Step 3: Detailed Explanation:
The problem states that the lens is placed at a distance $d$ in front of the image wall, so the image distance is: $$v = d$$ Since the image size equals the object size ($m = -1$): $$-1 = -\frac{v}{u} \implies u = -v = -d$$ Now, substitute these coordinate parameters into the lens maker's formula: $$\frac{1}{f} = \frac{1}{d} - \left(\frac{1}{-d}\right)$$ $$\frac{1}{f} = \frac{1}{d} + \frac{1}{d} = \frac{2}{d} \implies f = \frac{d}{2}$$ Let's re-verify the total distance setup. In standard displacement textbook variants, if the total distance between the two walls is defined as $D$, then for equal size, the lens sits exactly at the midpoint, so $v = \frac{D}{2} = 2f \implies f = \frac{D}{4}$. If the text states the lens is at a distance $d$ from the second wall, then $v = d = 2f \implies f = \frac{d}{2}$. Let's ensure alignment with the options selection architecture where option (D) specifies exactly $\frac{d}{2}$ or option (B) represents the classic $\frac{D}{4}$ setup depending on the variable definition. Here, with $v = d$, the calculation gives $f = \frac{d}{2}$.

Step 4: Final Answer:
The required focal length of the lens is exactly $\frac{d}{2}$, which corresponds to option (D).