Question

an object is placed at a distance of 3 cm from a concave lens of focal length 12 cm find the position and nature of the image formed

Answer 1

$\huge{\star}{\underline{\boxed{\red{\sf{Answer :}}}}}{\star}$

# Refers to attachment for Explanation

V (image distance) = -2.4 cm

Nature = Virtual and Erect

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Answer 2

$\huge{\bold{\underline{\underline{....Answer....}}}}$

$\huge{\bold{\underline{Given:-}}}$

• Object distance (u) = 3 cm.
• Focal length (f) = 12 cm.
• Given lens is concave,

$\huge{\bold{\underline{Explanation:-}}}$

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Applying sign - convention,

• $\large{\tt u = - 3 \: cm}$
• $\large{\tt f = - 12 \: cm}$

From, the lens formula,

$\large{\boxed{\tt \dfrac{1}{f} = \dfrac{1}{v} - \dfrac{1}{u}}}$

Substituting the values,

$\large{\tt \dfrac{1}{- 12} = \dfrac{1}{v} - \left[\dfrac{1}{- 3}\right]}$

Now,

$\large{\tt \dfrac{- 1}{12} = \dfrac{1}{v} + \dfrac{1}{3}}$

Now,

$\large{\tt \dfrac{1}{v} = \dfrac{-1}{12} - \dfrac{1}{3}}$

$\large{\tt \dfrac{1}{v} = \dfrac{- 1 - 4}{12}}$

$\large{\tt \dfrac{1}{v} = \dfrac{- 5}{12}}$

Reciprocating it,

$\large{\tt v = \dfrac{- 12}{5}}$

$\huge{\boxed{\boxed{\tt v = - 2.4 \: cm}}}$

So, the position of the image is - 2.4 cm.

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Finding the magnification,

$\large{\boxed{\tt m = \dfrac{v}{u}}}$

Substituting the values,

$\large{\tt m = \dfrac{ - 2.4}{ - 3}}$

$\large{\tt m = \dfrac{ \cancel{-} 2.4}{ \cancel{-} 3}}$

$\large{\tt m = \dfrac{2.4}{3}}$

$\huge{\boxed{\boxed{\tt m = 0.8}}}$

As the magnification is positive,

So, the image is Virtual erect, and diminished.

[Diminished because the magnification is smaller than 1, I.e. m < 1]

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