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Q33(b):
Draw ray diagrams showing the image formation by a concave lens when an object is placed (b) between focus and twice the focal length of the lens

Solution :

Given Variables & Initial Setup

In a concave lens (a diverging lens), the image formation is dictated by the refractive properties of the lens surface. Given parameters:

  • Lens Type: Concave (Diverging)
  • Object Position ($u$): Between the principal focus ($F_1$) and twice the focal length ($2F_1$).
  • Conventions: Using the Cartesian Sign Convention, light travels from left to right. The optical center ($O$) is the origin $(0,0)$.

Theoretical Principles of Ray Tracing

To locate the image, two primary rays are traced [Per the Rules for Image Formation by Concave Lenses]:

  1. Ray 1: A ray parallel to the principal axis appears to diverge from the principal focus ($F_1$) after refraction. [Law of Refraction for Parallel Rays].
  2. Ray 2: A ray passing through the optical center ($O$) of the lens proceeds without any deviation. [Property of Optical Center].

Visual Representation of Ray Diagram

F1 2F1 F2 2F2 Object Image

Analytical Characteristics of the Image

Based on the intersection of the refracted ray and the virtual extension of the parallel ray, the image formed exhibits the following properties:

Attribute Resulting Property
Position Between the optical center ($O$) and the principal focus ($F_1$).
Nature Virtual and Erect [Property of Concave Lenses].
Size Diminished (smaller than the object).

Mathematical Verification

Per the Gaussian Lens Formula: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.

Given the object is placed at $u$ such that $f < |u| < 2f$, and knowing that for a concave lens $f$ is negative ($-f$), we derive:

$\frac{1}{v} = \frac{1}{u} - \frac{1}{f}$

Since both $u$ and $f$ are negative, the resulting image distance $v$ must also be negative, confirming the image is formed on the same side of the lens as the object (virtual).

Final Solution: The image formed by a concave lens when the object is between $F_1$ and $2F_1$ is virtual, erect, diminished, and located between the optical center and the principal focus on the same side of the lens.


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