Question

A convex lens of focal length 15 cm, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively :

• A. \(-15 \, \text{cm and} -15 \, \text{cm from lens}\)
• B. \(-15 \, \text{cm and} +15 \, \text{cm from lens}\)
• C. \(-30 \, \text{cm and} +30 \, \text{cm from lens}\)
• D. \(-30 \, \text{cm and} -30 \, \text{cm from lens}\)

Hint:

For convex lens:

• Object at \(2f\) (\(u = -2f\)) \(\Rightarrow\) Image at \(2f\) (\(v = +2f\)), same size, real and inverted
• Object at \(f\) \(\Rightarrow\) Image at infinity
• Object between \(f\) and \(2f\) \(\Rightarrow\) Image beyond \(2f\), enlarged

Answer 1

The Correct Option is C

Step 1: Recall sign conventions for convex lens.
Using Cartesian sign convention (New Cartesian Sign Convention):

• Object distance (u) is negative (object placed to the left of lens)
• For real image, image distance (v) is positive (image formed on the right side)
• Focal length (f) of convex lens is positive

Given: \(f = +15 \, \text{cm}\)
Step 2: Condition for same size image.
For a convex lens, when the image size equals the object size, the magnification \(m = -1\) (negative sign indicates real and inverted image). \[ m = \frac{v}{u} = -1 \] \[ v = -u \]
Step 3: Apply lens formula.
Lens formula: \(\frac{1}{f} = \frac{1}{v} - \frac{1}{u}\) Substitute \(v = -u\) and \(f = +15 \, \text{cm}\): \[ \frac{1}{15} = \frac{1}{-u} - \frac{1}{u} \]
Step 4: Simplify the equation.
\[ \frac{1}{15} = -\frac{1}{u} - \frac{1}{u} \] \[ \frac{1}{15} = -\frac{2}{u} \]
Step 5: Solve for u.
\[ -\frac{2}{u} = \frac{1}{15} \] \[ -2 \times 15 = u \] \[ u = -30 \, \text{cm} \]
Step 6: Find v.
Since \(v = -u\): \[ v = -(-30) = +30 \, \text{cm} \]
Step 7: Interpret the result.

• Object is placed at \(-30 \, \text{cm}\) (30 cm to the left of the lens)
• Image is formed at \(+30 \, \text{cm}\) (30 cm to the right of the lens)
• This corresponds to the object being at \(2f\) (center of curvature)

Step 8: Analysis of options.

• (A) \(-15\) and \(-15\): Incorrect. Image distance cannot be negative for real image.
• (B) \(-15\) and \(+15\): Incorrect. This is at focus, would give infinite magnification.
• (C) \(-30\) and \(+30\): Correct. Matches our calculation.
• (D) \(-30\) and \(-30\): Incorrect. Image distance cannot be negative for real image.

Final Answer: (C) \(-30 \, \text{cm and} +30 \, \text{cm from lens}\)