Question

an object of height 4 cm is kept at a distance of 25cm in front of a concave mirror of focal length of the mirror is 15 CM at what distance from the mirror should a screen be kept so as to get a clear image what will be the size and nature of the image? This is numerical I want easy step.

Answer 1

$\boxed{AhoyMatey!!!}$

Given that f = -15 cm, u = -25 cm, ho = 4cm, v=?, hi = ?

By mirror formula,

1/f = 1/u + 1/v

1/-15 = 1/-25 + 1/v

1/v = 1/25 - 1/15

1/v = 3-5 / 75 = -2 / 75

v = -37.5 cm

The screen should be placed 37.5 cm from the pole of mirror and the image is real...

magnification (m) = hi / ho = -v / u

⇒ hi / 4 = -(-37.5) / (-25)

⇒ hi = - 37.5 / 25 × 4

= -6 cm

∴ the image is enlarged and inverted ...

$\small{Wishithelpsyouu!!!}$

$\small{Guddday!!!}$

Answer 2

$\color{red}\huge\boxed{Numerical :-}$

{We are provided the following values}

$h = 4cm$
$u = - 25cm$
$f = - 15cm$

They all are taken as per sign convention}

$<u>{Using the mirror Formula} </u>$

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

$\binom{1}{v} + \binom{1}{ - 25} = \binom{1}{ - 15}$

$\binom{1}{v} = \binom{1}{ - 15} - \binom{1}{ - 25}$

$\binom{1}{v} = \binom{ - 5 + 3}{75}$

$\binom{1}{v} = \binom{ - 2}{75}$

$\color{blue}\large{Therefore}$

$v = - 37.5$

{Therefore the screen must be placed 37.5 cm in front of the concave mirror}

{The (- ve) sign of ' v' shows that the image is real and inverted}

Now, For finding the size of the image, using mirror Formula

$m = \binom{h1}{h2} = - \binom{v}{u}$

Where( h1) is the object distance and( h2)
Is the image distance...

$h1 = \binom{h2 \times - v}{u}$

$h2 = \binom{ - ( - 37.5) \times 4}{ - 25}$

$h2 = \binom{150}{ - 25}$

$\color{blue}\large{Therefore}$

$h2 = - 6 \: cm$
$\color{blue}\large {Hence The Image Size Is - 6 cm}$