Question

If an object is placed at a distance of $12,cm$ from a thin lens of power $5D$, then the distance of the image from the lens with proper sign convention is

• A. $-30,cm$
• B. $20,cm$
• C. $15,cm$
• D. $24,cm$
• E. $-12,cm$

Hint:

Physics Tip: If object is inside focal length of convex lens, image is virtual, erect and appears on same side, so $v$ becomes negative.

Answer 1

The Correct Option is A

Concept:
Power of lens: $$P=\frac{1}{f(\text{in meter})}$$ Lens formula: $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$ (using Cartesian sign convention)
Step 1: Find focal length.
Given: $$P=5D$$ $$f=\frac{1}{5}=0.2m=20cm$$ Since power is positive, lens is convex.
Step 2: Use sign convention.
Object placed on left side: $$u=-12cm$$ Now, $$\frac{1}{20}=\frac{1}{v}-\frac{1}{(-12)}$$ $$\frac{1}{20}=\frac{1}{v}+\frac{1}{12}$$
Step 3: Solve for image distance.
$$\frac{1}{v}=\frac{1}{20}-\frac{1}{12}$$ $$=\frac{3-5}{60}=-\frac{1}{30}$$ So, $$v=-30cm$$
Negative sign means virtual image on object side. Hence correct option is (A). :contentReference[oaicite:1]{index=1}