Question

: An object of height 10 cm is placed 15 cm away from concave mirror of
focal length 12 cm find position, Nature and size of the image form.​

Answer 1

Given :

Height of object = 10cm

Distance of object = 15cm

Focal length = 12 cm

Type of mirror : concave

To Find :

Position, nature and size of the image.

Solution :

❖ Focal length of concave mirror is taken negative and that of convex mirror is taken positive.

Position of image can be calculated by using mirror formula which is given by

$:\implies\:\underline{\boxed{\bf{\orange{\dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}}}}}$

• u denotes distance of object
• v denotes distance of image
• f denotes focal length

By substituting the given values;

➙ 1/(-15) + 1/v = 1/(-12)

➙ 1/v = -1/12 + 1/15

➙ 1/v = (-5 + 4) / 60

➙ v = -60/1

➙ v = -60 cm

❖ Size of image :

➙ m = -v/u = h'/h

• h' denotes height of image
• h denotes height of object

➙ -(-60)/(-15) = h'/10

➙ -4 = h'/10

➙ h' = -4 × 10

➙ h' = -40 cm

❖ Nature of image :

• Real
• Inverted
• Enlarged

Answer 2

GIVEN :

• Height of an object = 10cm
• Object placed (from concave mirror) = 15cm
• focal length of an object = 12cm

TO FIND :

• position of the image, nature of the image and size of the image = ?

STEP-BY-STEP EXPLANATION :

Here, as per the provide question first of all we will find out position of the image –

Formula used :

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

[ substituting the values as per the formula ] :

$\frac{ \: \: \: \: 1}{ - 15} + \frac{1}{ \rm {v}} = \frac{ \: \: \: \: 1}{ - 12}$

$\frac{1}{ \rm {v}} = \frac{ \: \: \: \: 1}{ - 12} + \frac{1}{15}$

$\frac{1}{ \rm {v}} = \frac{( - 5 \: + \: 4)}{60}$

$\rm {v} = \frac{ - 60}{ \: \: \: 1} = - 60$

$\rm {v} = - 60 \rm {cm}$

Therefore, position of the image = -60cm.

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Then, we'll have to find size of the image and after all we can find the nature of the image –

→ m = -v/u = h'/h

[ substituting the values as per the formula ] :

→ m = -(-60)/(-15) = h'/10

→ m = -4 = h'/10

→ h' = -4 × 10

→ h' = -40cm

Hence, size of the image formed = -40cm.

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Now, nature of the image formed here is –

1. Inverted image
2. Enlarged image
3. Real image

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