Question

Q33. An object of height 5 cm is held 30 cm away from a converging lens of focal length
10 cm. Find the position, size and nature of the image formed. 3
SECTION D​

Answer 1

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To find position and height of image .

1/f = 1/v - 1/u

1/10 = 1/v - 1/(-30)

1/10 = 1/v + 1/30

1/v = 1/10 - 1/30

1/v = 3-1/30

1/v = 2/30

1/v = 1/15 = 15

ho/hi = v/u

ho/5 = 15/-30

ho = 15×5/30

ho = 2.5cm

nature of image is = inverted ,large in size , real

Answer 2

$\huge\star\underline\mathfrak\red{Answer:-}$

To find position and height of image .

1/f = 1/v - 1/u

1/10 = 1/v - 1/(-30)

1/10 = 1/v + 1/30

1/v = 1/10 - 1/30

1/v = 3-1/30

1/v = 2/30

1/v = 1/15 = 15

ho/hi = v/u

ho/5 = 15/-30

ho = 15×5/30

ho = 2.5cm

nature of image is = inverted ,large in size , real

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