Question

An object is placed 20 cm away from a concave lens with a focal length of 10 cm.

• [(a)] Calculate the distance to the image from the lens.
• [(b)] Write down the characteristics of the image.

OR A convex lens forms a virtual image at a distance of 7.5 cm away from it. Focal length of the lens is 15 cm.

• [(a)] Calculate the distance between lens and the object.
• [(b)] Write any two instances where a convex lens is used in this manner.

Hint:

Concave lens → always virtual, erect
Convex lens (object inside focus) → virtual, magnified

Answer 1

(A)
(a) Image distance
Concept: Lens formula
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] Step 1: Sign convention}
For concave lens: \[ f = -10\ \text{cm}, \quad u = -20\ \text{cm} \] Step 2: Substitute values}
\[ \frac{1}{-10} = \frac{1}{v} + \frac{1}{-20} \] \[ \frac{1}{-10} + \frac{1}{20} = \frac{1}{v} \] \[ \frac{-2 + 1}{20} = \frac{1}{v} = \frac{-1}{20} \] \[ v = -20\ \text{cm} \] Conclusion (a): \[ \text{Image distance} = -20\ \text{cm} \] (b) Characteristics of image

• Virtual
• Erect
• Diminished
• Formed on the same side as object

(B)
(a) Object distance
Step 1: Sign convention}}
Convex lens: \[ f = +15\ \text{cm}, \quad v = -7.5\ \text{cm} \] Step 2: Lens formula}}
\[ \frac{1}{f} = \frac{1}{v} + \frac{1}{u} \] \[ \frac{1}{15} = \frac{1}{-7.5} + \frac{1}{u} \] Step 3: Calculation}}
\[ \frac{1}{15} + \frac{1}{7.5} = \frac{1}{u} \] \[ \frac{1 + 2}{15} = \frac{3}{15} = \frac{1}{5} \] \[ u = 5\ \text{cm} \] Conclusion (a): \[ \text{Object distance} = 5\ \text{cm} \] (b) Uses of convex lens in this manner

• Magnifying glass
• Simple microscope