Appeared in: 2025 (2 sets)
OR
If an object in case (i) above is $20\text{ cm}$ from the lens and the screen is $50\text{ cm}$ away from the object, the focal length of the lens used is
(A) $10\text{ cm}$
(B) $12\text{ cm}$
(C) $16\text{ cm}$
(D) $20\text{ cm}$
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Official Evaluation Steps & EL10 Insight
Step 1 (Formula / Governing Law & Given Data): [0.25 Mark]
The lens formula is given by:
$$\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$$Given data based on sign convention:
- Object distance, $u = -20\text{ cm}$
- Total distance between object and screen = $50\text{ cm}$
- Image distance, $v = 50 - 20 = +30\text{ cm}$ (since the image is formed on a screen, it is real)
Step 2 (Step-by-Step Mathematical Working): [0.5 Mark]
Substituting the values into the lens formula:
$$\frac{1}{f} = \frac{1}{30} - \frac{1}{-20}$$ $$\frac{1}{f} = \frac{1}{30} + \frac{1}{20}$$Taking the Least Common Multiple (LCM) of 30 and 20, which is 60:
$$\frac{1}{f} = \frac{2 + 3}{60}$$ $$\frac{1}{f} = \frac{5}{60}$$ $$\frac{1}{f} = \frac{1}{12}$$Step 3 (Final Concluded Answer): [0.25 Mark]
Solving for $f$:
$$f = 12\text{ cm}$$Correct Option: (B) $12\text{ cm}$
💡 EL10 Concept Insight: Think of the lens as a bridge that takes light rays from an object and bends them to meet at a specific point on the screen; because the light has to travel a fixed distance to reach that screen, the lens must have a specific "strength" (focal length) to focus the image exactly where the screen is placed.