Question

An object 4cm high is placed at 25cm in front of a concave mirror of focal length 15cm. At what distance from the mirror should a screen be placed in order to obtained a sharp image ? find the nature and size of the object

Answer 1

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հίί ʍαtε_______✯◡✯
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.
.
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_________________________ ✶
✰һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________
.
✯
♦ Given that ,

●object "h○ = 4cm

● focal length of concave mirror is 15cm,
we know that focal length of concave mirror always negative ,

so, "f "= -15

●distance " u " = -25

●v = ?

♦Now , using the mirror formula ,
=》
$\frac{1}{f} = \frac{1}{u} + \frac{1}{v}$

=》
$\frac{1}{ - 15} = \frac{1}{ - 25 } + \frac{1}{v}$

=》

$\frac{1}{v} = \frac{1}{25} - \frac{1}{15}$
=》

$\frac{1}{v} = \frac{3 - 5}{75} = \frac{ - 2}{75}$

♦--▶v = -37.5 cm

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

♦Magnification =   h¡ / h○ = -v / u

h¡ / 4 = 75 /2 * (-25)

h¡ / 4 = -3 / 2 *4

h¡ = -3 * 2 = -6 cm

▶So image is real and inverted.

______________★

$hope \: it \: helps$


★BRAINLY★

Answer 2

★ ✪

հίί ʍαtε_______✯◡✯

.

.

.

.

_________________________ ✶

✰һєяє' ʏȏȗя ѧṅśwєя ʟȏȏҡıṅɢ ғȏя________

.

✯

♦ Given that ,

●object "h○ = 4cm

● focal length of concave mirror is 15cm,

we know that focal length of concave mirror always negative ,

so, "f "= -15

●distance " u " = -25

●v = ?

♦Now , using the mirror formula ,

=》

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

=》

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

=》

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

=》

$\frac{1}{v} = \frac{3 - 5}{75} = \frac{ - 2}{75}$

v

1

♦--▶v = -37.5 cm

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

♦Magnification = h¡ / h○ = -v / u

h¡ / 4 = 75 /2 * (-25)

h¡ / 4 = -3 / 2 *4

h¡ = -3 * 2 = -6 cm

▶So image is real and inverted.