Question

an object is placed at a distance of 15 cm from a concave mirror of focal length 30 cm. find the position amd nature of image formed?
give me correct answer only and which one of u give me correct ans i will mark u as a brainlist.​

Answer 1

Object distance = 15 cm

Focal length of Concave Mirror = 30 cm

Image distance = ?

Nature of image = ?

Calculation:

Applying mirror formula and keeping the sign convention in mind , we can say that :

$\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}$

$= > \dfrac{1}{ (- 30)} = \dfrac{1}{ (- 15)} + \dfrac{1}{v}$

$= > \dfrac{1}{v} = \dfrac{1}{15} - \dfrac{1}{30}$

$= > \dfrac{1}{v} = \dfrac{2 - 1}{30}$

$= > \dfrac{1}{v} = \dfrac{1}{30}$

$= > v = 30 \: cm$

So image will be produced 30 cm behind the Mirror.

Magnification $=-\dfrac{v}{u}=+2$

Since magnification is positive and greater than 1 , we can say that image is erect , virtual and magnified.

Answer 2

Answer:

30 or -30

Explanation:

image distance=2*object distance

2*15=30

And by the mirror formula

1/f=1/v+1/u

1/30=1/v+1/15

1/v=1/30-1/15

1/v=15-30/30*15

1/v= -15/450

1/v=-1/30

v=-30

I hope it will help you...