Question

A 2.0 cm tall object is placed 40 cm from a diverging lens of focal lenght 15 cm . find the position and size of image.

Answer 1

Answer:

• Focal length = -15 cm
• Object distance = -40 cm
• Diverging lens/Concave lens

We know, 1/f = 1/v - 1/u [LENS FORMULA]

1/v = 1/f + 1/u

1/v = (-1/15) + (-1/40)

1/v = -1/15-1/40

1/v = (-40-15)/600

1/v = -55/600

v = 600/-55

Hence, Image distance = -10.90 cm

Hence, the image will be formed between the the optical centre and the focus. [POSITION]

It will be virtual and erect. [NATURE OF THE IMAGE]

Now, h'/h = v/u

h'/2 = -10.90/-40

-40 * h' = -10.90 * 2

-40 * h' = -21.8

h' = -21.8/-40

h' = 0.5 cm

Hence, the image will be diminished. [SIZE OF THE IMAGE]