Question

object placed at a distance of 30 cm from a concave lens of focal length 15cm find the position magnification and nature of image​

Answer 1

Answer:

You should already know that a concave lens produces a virtual, erect diminished image. The only question is whereabouts? and how diminished?

Using Real Is Positive RIP Convention, u=30 and f=-15

1/u + 1/v = 1/f

1/30 + 1/v = -1/15

1/v = -1/15 - 1/30 = -3/30 = -1/10

v = -10 M=v/u = -1/3

So image is erect, virtual, 1/3 size of object, and 10 cm from the lens, (between lens and object)

please mark brainliest

Answer 2

$\red{ANSWER}$

u = -30 cm , f = -15 cm , v = ? , m = ?

1/v - 1/u = 1/f

1/v = 1/f + 1/u

1/v = 1/-15 + 1/-30

1/v = 1/-15 - 1/30

= -2 - 1/30

= -3/30

v = 30/-3

v = -10 cm

m = v/u

m = -10/-30

m = 1/3

m = 0.33

$\green{virtual and errect}$

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