Question

A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15cm then the location of image will be​

Answer 1

Answer:

2.5 cm

Explanation:

Here,

object size , h1 = 4.5 cm

object distance, u = –12 cm

focal length, f = + 15 cm

image distance , v = ?

magnification, m = ?

As. 1/u + 1/v = 1/f

1/v = 1/f – 1/u

Now,

putting u = –12 cm and f = +15cm , we get

1/v = 1/15 - 1/-12 = 4+5/60 = 9/60

v = 60/9 = 6.7 cm

✴image is formed 6.7 CM behind the convex mirror. it must be virtual and erect.

if h2 is size of image , then m = h2/h1 = -v/u

or,

m = h2/h1 = –(6.7)/–12 = 0.558

h2 = 0.558

h1 = 0.558 * 4.5 = 2.5 cm

⚡As the needle is moved further from the mirror, image moves away from the mirror till it is at focus F of the mirror the size of the image goes on decreasing.

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Answer 2

GIVEN:-

$• \sf Object \: size,h_1=4.5cm \: \: \: \: \: \: \: </p><p> \\ \sf •Object \: distance,u=-12cm \\ \sf •Focal \: length,f=+15cm \: \: \: \: \:$

TO CALCULATE:-

$\sf •Image \: distance,v=? \\ \sf• Magnification ,m=?$

$\longrightarrow \sf Formula \: used \begin {cases}• \sf \frac{1}{u} + \frac{1}{v}= \frac{1}{f} \\ \\ • \sf m= \frac{h_2}{h_1}=- \frac{v}{u} \\ \end{cases}$

SOLUTION :-

$\sf We \: have,$

$\longrightarrow \sf \frac{1}{u} + \frac{1}{v} = \frac{1}{f} , \\ \implies \sf\frac{1}{v} = \frac{1}{f}- \frac{1}{u}$

•Putting u=-12cm and f=+15cm ,we get

$\sf \frac{1}{v} = \frac{1}{15} - \frac{1}{-12 \: }= \frac{4 + 5}{60}$

$\sf v= \frac{60}{9} = 6.7cm$

I.e ,image is formed 6.7 cm behind the convex mirror, It must be virtual and erect

If $\sf h_2$ is size of the image ,then m=$\sf\frac{h_2}{h_1}=\frac{-v}{u}$

$\sf Or \: m= \frac{h_2}{h_1} = \frac{ - (6.7)}{ - 12} = 0.558$

$\sf \therefore h_2=0.558×h_1 \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \\ \sf h_2=0.558×4.5=2.5cm$

As the needle is moved farther from the mirror, image moves away from the mirror till it is at focus