Question

we are given a convex lens of focal length 20cm draw Ray diagrams to show the nature and position of the image from when the object is kept at the distance of 55 CM, 35 cm, 15 cm from the lens

Answer 1

I post three photos see and take your answer

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Answer 2

Given:

f = 20cm

1. u = 55cm
2. u = 35cm
3. u = 15cm

To find:

The nature and position of the image

Solution:

1) Object at a distance of 55cm

1/f = 1/v - 1/u

1/20 = 1/v - 1/(-55)

1/20 = 1/v + 1/55

1/v = 1/20 - 1/55

1/v = (55 - 20)/(20 × 55)

1/v = 35/1100

1/v = 7/220

v = 220/7

v = 31.43cm

m = v/u

m = 31.43/(-55)

m = -0.57 (- sign indicates that the image formed is real and inverted)

1. Image formed at a distance of 31.43cm
2. Diminished
3. Real and inverted

2) Object at a distance of 35cm

1/f = 1/v - 1/u

1/20 = 1/v - 1/(-35)

1/20 = 1/v + 1/35

1/v = 1/20 - 1/35

1/v = (35 - 20)/(20 × 35)

1/v = 15/700

1/v = 3/140

v = 140/3

v = 46.67cm

m = v/u

m = 46.67/(-35)

m = -1.33

1. Image formed at a distance of 46.67cm
2. Enlarged
3. Real and inverted

3) Object at a distance of 15cm

1/f = 1/v - 1/u

1/20 = 1/v - 1/(-15)

1/20 = 1/v + 1/15

1/v = 1/20 - 1/15

1/v = (15 - 20)/(20 × 15)

1/v = -5/300

1/v = -1/60

v = -60cm (- sign indicates that the image formed is on the same side as the object)

m = v/u

m = (-60)/(-15)

m = 4

1. Image formed is at a distance of 60cm, on the same side as the object
2. Enlarged
3. Virtual and erect