Question

Find the size, nature and position of the image formed when an object of size 1cm is placed at a distance of 15cm from a concave mirror of focal length 10cm.

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Answer 1

Answer:

Here is your answer sis

Explanation:

ANSWER

u=−15cm

f=−10cm

h=1cm

From mirror formula

f1=u1+v1

v1=f1−u1

v1=−101−−11

v=−30cm

The image is formed at a distance 30cm on the left side of the mirror fromthe pole.

m=hh′=u−v

h′=−1×−15−30

h=−2cm

The image size is 2cm and it is real and inverted.

Hope it's help you

Answer 2

» Given Question:

Find the size, nature and position of the image formed when an object of size 1 cm is placed at a distance of 15cm from a concave mirror of focal length 10 cm.

» Required Solution:

Position of the image:

To find the position of the image, we'll use here mirror formula-

Given:

• » u ( Object distance ) = -15cm

( displacement taken left side of the mirror is taken always negatively)

• » f ( focal length ) = -10cm

( As it is concave mirror )

To find:

• v ( image distance i.e position of the image)

Calculation:

We know,

• $\large \tt { \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}$

putting the values -

$\tt { : \implies \dfrac{1}{v} + \dfrac{1}{u} = \dfrac{1}{f}}$

$\tt { : \implies \dfrac{1}{v} + \dfrac{1}{-15} = \dfrac{1}{-10}}$

$\tt { : \implies \dfrac{1}{v} - \dfrac{1}{15} = - \dfrac{1}{10}}$

$\tt { : \implies \dfrac{1}{v} = - \dfrac{1}{10}+ \dfrac{1}{15} }$

$\tt { : \implies \dfrac{1}{v} = \dfrac{-3+2}{30} }$

$\tt { : \implies \dfrac{1}{v} = - \dfrac{1}{30} }$

$\tt { : \implies v = - 30 \: cm}$

Therefore, the position of the image is -30cm i.e 30cm to the left side of the mirror. $\purple \bigstar$

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Nature of the image:

$\purple \bigstar$ As the image is formed to the left side of the mirror, so it will be real and inverted image. $\purple \bigstar$

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Size of the image:

To find the nature of the image we need to calculate its magnification first.

• $\large \tt {m = - \dfrac{v}{u}}$

Where,

• m = magnification = ?
• v = image distance = -30cm
• u = object distance = - 15cm

Put the values-

$\tt {: \implies m = - \dfrac{\cancel {-} 30}{\cancel {-} 15}}$

$\tt {: \implies m = \cancel { - \dfrac{30}{ 15}} }$

$\tt {: \implies m = - 2}$

Therefore the size of the image is 2cm.( – ) sign indicates that the image is formed below the principal axis which is real and inverted. $\purple \bigstar$

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