Question

an object of height 5cm is placed 10cm in front of a convex lens of focal length 15cm find nature position and height of image​

Answer 1

Position = -30 cm

Nature= virtual and erect

height of the image = 10 cm

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Answer 2

Explanation:

Object distance = u = 10 cm

Focal Length = f = 15 cm

According to Sign Covention,

u = -ve

f (convex) = +ve

By Lens Formula,

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

$\frac{1}{v} - \frac{1}{ - 10} = \frac{1}{15}$

$\frac{1}{v} = \frac{2 - 3}{30}$

$v = - 30 cm$

Magnification,

$m = \frac{v}{u}$

$m = \frac{ - 30}{ - 10}$

$m = 3$

Therefore,

$\frac{hi}{ho} = 3$

$hi = 3 \times 5$

$hi = 15 cm$

Nature of Image = Virtual, Erect, Enlarged

Hope this Helps!