Question

An object is placed 60cm from a concave mirror of focal length 15cm.Find the position and nature of the image Pls answer me It's answer will be come 12cm(Behind the mirror) virtual image pls solve it fast

Answer 1

Explanation:

u= -60cm

f=15cm

1/f=1/v+1/u

1/15=1/v+1/-60

1/v=1/15-1/-60

1/v=1/15+1/60

1/v=5/60

v=60/5

v=12cm

virtual image

because the law of ray 6th

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Answer 2

QUESTION:

An object is placed 60cm from a concave mirror of focal length 15cm.Find the position and nature of the image.

ANSWER :

SIGN COVENTION RELATED TO CONCAVE MIRROR :

Object distance

$(always \: negative)$

Focal length

$(positive )$

Now come to main question:

GIVEN :

object distance = -60 cm

Focal length = -15 cm

TO FIND :

Nature and size of the image ?

by using mirror formula :

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

where.

$v = image \: distance \\ u = \: object \: distance \\ f = focal \: length$

so

$\frac{1}{v} + \frac{1}{ - 60} = \frac{1}{ 15} \\ \frac{1}{v} - \frac{1}{60} = \frac{ 1}{15} \\ \frac{1}{v} = \frac{ 1}{15} + \frac{1}{60} \\ \frac{1}{v} = \frac{ 4 + 1}{60} \\ \frac{1}{v} = \frac{ 5}{60} \\ v = \frac{ 60}{5} \\ v = 12cm.$

IMAGE DISTANCE = 12 CM.

nature of the image

1. virtual and erect

2. diminished and behind the mirror

TO find whether it is magnified or diminished we use magnification formula :

m= -v/u

m= -12/-60

M = 1/5

hence, it is less than 1. so image will be diminished.

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