Question

An object is placed at 30 cm in front of a convex lens of focal length 20 cm. Find the position, and nature of the image.

Answer 1

Answer:

Given -

• Object distance, u = -30 cm
• Focal length, f = 20 cm
• Image distance, v = ?

From the lens formula we'll calculate the value of image distance.

$\\ \sf \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u} \\ \\ \\ \implies \sf \: \dfrac{1}{20} = \dfrac{1}{v} - \dfrac{1}{ - 30} \\ \\ \\ \implies \sf \: \dfrac{1}{20} = \dfrac{1}{v} + \dfrac{1}{30} \\ \\ \\ \implies \sf \: \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{30} \\ \\ \\ \implies \sf \: \dfrac{1}{v} = \dfrac{30 - 20}{30 \times 20} = \dfrac{10}{600} \\ \\ \\ \implies \sf \: v = \cancel \frac{600}{10} \\ \\ \\ \implies \sf \blue{v = 60 \: cm} \\ \\ \\ \bf \green{Therefore, \: image \: \: distance \: (v) = 60 \: cm.}$

Magnification, m -

$\\ \\ \sf \: \frac{height \: \: of \: \: the \: \: image}{height \: \: of \: \: the \: \: object} = \dfrac{v}{u} \\ \\ \\ \implies \sf \: m = \dfrac{60}{ - 30} = \blue{ - 2}$

Negative size indicates that the image is inverted.

• m is greater than 1, image is magnified.
• v is positive, image is real.