Question

An object is placed at a distance of 30 cm in front of a convex lens of focal length 15 cm. Use lens formula to determine the position of the image. What will be the size of the image in this case?

Hint:

Object at 2f of convex lens → Image at 2f, same size, inverted.

Answer 1

Given:

• Object distance, $u = -30$ cm (in front of lens)
• Focal length, $f = +15$ cm (convex lens)

Lens Formula: \[ \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \] Substitute values: \[ \frac{1}{15} = \frac{1}{v} - \left(-\frac{1}{30}\right) \] \[ \frac{1}{15} = \frac{1}{v} + \frac{1}{30} \] \[ \frac{1}{v} = \frac{1}{15} - \frac{1}{30} \] \[ \frac{1}{v} = \frac{2 - 1}{30} = \frac{1}{30} \] \[ v = +30 \text{ cm} \] Position of Image: Image is formed 30 cm on the other side of the lens. Size of Image: Magnification, \[ m = \frac{v}{u} = \frac{30}{-30} = -1 \]

• Magnitude of magnification = 1 → Image size = Object size
• Negative sign → Image is inverted

Conclusion:

• Image is formed at 30 cm on the other side of the lens.
• The image is real, inverted, and equal in size to the object.