Question

Identify the nature of the image formed when the object is placed at a distance of 20 cm from a concave mirror of focal length 10 cm.​

Answer 1

Answer:

Given:

• Object Distance = 20 cm
• Focal Length = 10 cm

Applying Sign conventions, we get:

• u = -20 cm
• f = -10 cm

According to Mirror Formula,

$\boxed{ \dfrac{1}{f} = \dfrac{1}{v} + \dfrac{1}{u}}$

Substituting the values in the formula we get:

$\implies \dfrac{1}{-10} = \dfrac{1}{v} - \dfrac{1}{20}\\\\\\\implies \dfrac{1}{v} = \dfrac{1}{20} - \dfrac{1}{10}\\\\\\\implies \dfrac{1}{v} = \dfrac{ 1 - 2}{20}\\\\\\\implies \dfrac{1}{v} = \dfrac{-1}{20}\\\\\\\implies \boxed{ \bf{ v = -20\:cm}}$

Since the image distance is equal to object distance, the size of the image would be same as that of the object. This is because the object is placed at 2F.

Since the image distance has a negative sign, it has to be real and an inverted image.

Hence the nature of the image is:

• Real, Inverted and Same size as that of the object.

Answer 2

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Question:-

• Identify the nature of the image formed when the object is placed at a distance of 20 cm from a concave mirror of focal length 10 cm.

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Given:-

• Object Distance = 20cm
• Focal Distance = 10cm

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formula Used:-

$\star{\boxed{\sf{formula \: of \: mirror \: \implies \: \frac{1}{f} = \frac{1}{v} + \frac{1}{u} }}}$

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Putting the formula:-

$\to \: \frac{1}{ - 10} = \frac{1}{v} - \frac{1}{u}$

$\to \: \frac{1}{v} = \frac{1}{20} - \frac{1}{10}$

$\to \: \frac{1}{v} = \frac{1 - 2}{20}$

$\to \: \frac{1}{v} = \frac{ - 1}{20}$

$\star{\boxed{\sf\green{v = - 20}}}$

So, Image will be real and inverted.

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