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Q31(b):
Draw ray diagrams showing the image formation by a convex lens when an object is placed (b) between focus and twice the focal length of the lens

Solution :

Step 1: Theoretical Framework and Given Parameters

To determine the image formation of an object placed between the principal focus ($F_1$) and twice the focal length ($2F_1$) of a convex lens, we apply the rules of refraction through a thin spherical lens:

  • Object Position: Between $F_1$ and $2F_1$.
  • Lens Type: Convex (Converging lens).
  • Rule 1: A ray passing parallel to the principal axis will refract through the lens and pass through the focus ($F_2$) on the other side [Per the Laws of Refraction].
  • Rule 2: A ray passing through the optical center ($O$) of the lens will emerge without any deviation [By the property of thin lenses].

Step 2: Visual Representation (Ray Diagram)

F1 2F1 F2 2F2 Obj Img

Step 3: Analytical Derivation of Image Properties

Based on the geometric construction, the characteristics of the image formed are as follows:

Characteristic Description
Position Beyond $2F_2$ on the opposite side of the lens.
Nature Real and Inverted [Since rays actually converge].
Size Magnified ($h' > h$) [Geometric similarity $ \triangle OAB \sim \triangle OA'B' $].

Step 4: Mathematical Justification

Using the Lens Formula: $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$

Given $f < |u| < 2f$, let $u = -1.5f$.

$\frac{1}{v} = \frac{1}{f} + \frac{1}{-1.5f} = \frac{1.5 - 1}{1.5f} = \frac{0.5}{1.5f} = \frac{1}{3f}$

$v = 3f$

Since $v > 2f$, the image is mathematically confirmed to lie beyond $2F_2$. The magnification $m = \frac{v}{u} = \frac{3f}{-1.5f} = -2$, confirming the image is inverted (negative sign) and magnified (magnitude > 1).

Final Solution: When an object is placed between $F_1$ and $2F_1$, a real, inverted, and magnified image is formed beyond $2F_2$ on the other side of the lens.


More Questions from Class 10 Science Light – Reflection and Refraction CHAPTER_10


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