Question

An object is placed at a distance of 0.4m from a lens having focal length 0.3 m. The object is moving towards the lens at a speed of 0.01 m/s. What is the rates of change of position of image and lateral magnification of image ?

• A. 3, 0.03 m/s
• B. 9, 0.09 m/s
• C. 3, 0.09 m/s
• D. 9, 0.03 m/s

Answer 1

Answer: (C)

3, 0.09 m/s

Answer 2

u = – 0.4 m = – 40 cm, v = ?

f = 0.3 m = 30 cm

$\frac { 1 } { \mathrm { f } } = \frac { 1 } { \mathrm { v } } - \frac { 1 } { \mathrm { u } }$

$\frac { 1 } { 30 } = \frac { 1 } { v } - \frac { 1 } { - 40 } \Rightarrow v = 120 \mathrm {~cm}$ Lateral

magnification $\mathrm { m } = \frac { \mathrm { v } } { \mathrm { u } } = \frac { 120 \mathrm {~cm} } { - 40 \mathrm {~cm} } = - 3$

$\mathrm { m } = - 3$

Velocity of image = m² × velocity of object

= 3² × 0.01 m/s

$\mathrm { v } _ { \mathrm { I } } = 0.09 \mathrm {~m} / \mathrm { s }$

Option (3) is correct