Question

© An object of 1cm high is placed at 10cm in front of a concave
mirror of focal length 15cm. what is the magnification produced
by the mirror.
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Answer 1

$\large{\underline{\underline{\mathfrak{Answer :}}}}$

• Image distance (v) = 30 cm

• Magnification (m) = 3

• Nature of Image is Virtual and erect

$\rule{200}{0.5}$

$\underline{\underline{\mathfrak{Step-By-Step-Explanation :}}}$

Given :

• Height of object (Ho) = 1 cm
• Object Distance (u) = -10 cm
• Focal length (f) = -15 cm
• Concave mirror

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To Find :

• Image distance
• Magnification

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Solution :

We have mirror formula :

$\large{\boxed{\sf{\dfrac{1}{f} \: = \: \dfrac{1}{v} \: + \: \dfrac{1}{u}}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{f} \: - \: \dfrac{1}{u}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{-15} \: - \: \bigg( \dfrac{1}{-10} \bigg)}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{-1}{15} \: + \: \dfrac{1}{10}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{-2 \: + \: 3}{30}}} \\ \\ \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{30}}} \\ \\ \implies {\sf{v \: = \: 30 \: cm}} \\ \\ \underline{\sf{\therefore \: Image \: distance \: is \: 30 \: cm}}$

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And we have formula for magnification :

$\large{\boxed{\sf{m \: = \: \dfrac{-v}{u}}}} \\ \\ \implies {\sf{m \: = \: \dfrac{-30}{-10}}} \\ \\ \implies {\sf{m \: = \: 3}} \\ \\ \underline{\sf{\therefore \: Magnification \: is \: 3}}$

• Nature of Image is Virtual and Erect.

Answer 2

$\Large{\underline{\underline{\bf{Solution :}}}}$

★ Given :

Height of object (H) = 1 cm

Object Distance (u) = -10 cm

Focal length (f) = -15 cm

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★ To Find :

We have to find the magnification (m)

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★ Solution :

We know that,

$\Large{\implies{\boxed{\boxed{\sf{\frac{1}{f} = \frac{1}{u} + \frac{1}{v}}}}}} \\ \\ \sf{→ \frac{-1}{15} = \frac{-1}{10} = \frac{1}{v}} \\ \\ \sf{→ \frac{1}{v} = \frac{-1}{15} + \frac{1}{10}} \\ \\ \sf{→ \frac{1}{f} = \frac{-2 + 3}{30}} \\ \\ \sf{→v = \frac{1}{30}} \\ \\ \sf{→ v = 30 \: cm} \\ \\ \Large{\implies{\boxed{\boxed{\sf{ v = 30 \: cm}}}}}$

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Now,

$\Large{\implies{\boxed{\boxed{\sf{m = \frac{-v}{u}}}}}} \\ \\ \sf{→ m = \frac{\cancel{-}30}{\cancel{-}15}} \\ \\ \sf{→ m = \frac{\cancel{30}}{\cancel{10}}} \\ \\ \sf{→ m = 3} \\ \\ \Large{\implies{\boxed{\boxed{\sf{m = 3}}}}}$

$\therefore$ Nature of image is Virtual and erect.