Question

An object is placed at a distance of 10 cm from a lens with a focal length of 30 cm. What is the magnification of the image?

• A. $ -1.5 $
• B. $ +1.5 $
• C. $ -2.0 $
• D. $ +2.0 $

Hint:

When an object is placed between the optical center and the focus of a convex lens, the image is virtual, erect, and magnified. The magnification is positive.

Answer 1

The Correct Option is B

Given:
Object distance, \( u = -10\, \text{cm} \) (negative by convention)
Focal length, \( f = +30\, \text{cm} \) (positive for convex lens) Step 1: Lens Formula
The lens formula is: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] where \( v \) is the image distance. Substituting values: \[ \frac{1}{30} = \frac{1}{v} - \frac{1}{-10} \implies \frac{1}{30} = \frac{1}{v} + \frac{1}{10} \] Step 2: Solve for Image Distance \( v \)
\[ \frac{1}{v} = \frac{1}{30} - \frac{1}{10} = \frac{1 - 3}{30} = -\frac{2}{30} = -\frac{1}{15} \] \[ v = -15\, \text{cm} \] The negative sign indicates the image is virtual and on the same side as the object. Step 3: Calculate Magnification \( m \)
\[ m = \frac{v}{u} = \frac{-15}{-10} = +1.5 \] The positive magnification indicates the image is upright. Conclusion
The magnification of the image is \( {+1.5} \), corresponding to option \( {B} \).