Question

♦ PHYSICS ♦
$A \: short \: linear \: object \: of \: length \\ b \: lies \: along \: the \: axis \: of \: a \: concave \\ mirror \: of \: focal \: length \: f \: at \: a \\ distance \: u \: from \: pole \: of \: mirror. \\ What \: is \: the \: size \: of \: image?$
#Explanation of each and every step needed
#Spammed and copied answers are not accepted​

Answer 1

$\huge{\underline{\underline{\red{\mathfrak{Answer :}}}}}$

$\tt {\blue{Given}} \begin{cases} \sf{\green{Height \: of \: object \: = \: b}} \\ \sf{\orange{Focal \: length \: = \: f}} \\ \sf{\pink{Object \: Distance \: = \: u}} \\ \sf{\purple{Height \: of \: Image \: = \: b'}} \end{cases}$

As we know that Formula for magnification of mirror is :

$\Large{\boxed{\boxed{\sf{m \: = \: \dfrac{b'}{b} \: = \: - \: \dfrac{v}{u}}}}}$

Where as we can also write it was very small change :

$\large : \implies {\sf{\dfrac{b'}{b} \: = \: - \bigg( \dfrac{dv}{du} \bigg) -----(1)}}$

Also the mirror formula is :

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

$\large : \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{1}{f} \: - \: \dfrac{1}{u}}}$

Take L.C.M

$\large : \implies {\sf{\dfrac{1}{v} \: = \: \dfrac{u \: - \: f}{uf}}}$

$\large : \implies {\sf{v \: = \: \dfrac{uf}{u \: - \: f}}}$

Now take

$\Large \leadsto {\sf{\dfrac{dv}{du}}}$

So,

Put value of (1) here

$\large : \implies {\sf{\dfrac{dv}{du} \: = \: \dfrac{d}{du} \: \bigg( \dfrac{uf}{u \: - \: f} \bigg)}}$

Now differentiate v with respect to d/du

We get,

$\large : \implies {\sf{\dfrac{dv}{du} \: = \: \bigg( \dfrac{-f}{u \: - \: f} \bigg)^2}}$

$\large : \implies {\sf{-\dfrac{dv}{du} \: = \: \bigg( \dfrac{f}{u \: - \: f} \bigg)^2}}$

Now put value of -dv/du from (1)

$\Large : \implies {\sf{\dfrac{b'}{b} \: = \: \bigg( \dfrac{f}{u \: - \: f} \bigg)^2}}$

$\LARGE {\underline{\boxed{\sf{b' \: = \: b \bigg( \dfrac{f}{u \: - \: f} \bigg) ^2}}}}$

Answer 2

$\large\bf\underline{Given:-}$

⠀

• a short linear object of length

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• b lies along the axis of a concave mirror of focal length

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• f at a distance u from pole of mirror.

⠀

$\large\bf\underline {To \: find:-}$

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• what is the size of image?

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$\large\bf\underline{Formula \: Used:-}$

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MIRROR FORMULA:-

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• $\huge \underline{\boxed{ \pink{ \bf \: \frac{1}{v} + \frac{1}{u} + \frac{1}{f} \longrightarrow(1) }}}$

⠀

$\huge\bf\underline{Solution:-}$

⠀

We Get,

⠀

$\sf \longrightarrow - {v}^{ - 2} dv - {u}^{ - 2} dv = 0$

⠀

⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀⠀or

⠀

$\sf \longrightarrow |dv| = | \frac{ {v}^{2} }{ {u}^{2} } | dv\longrightarrow(2)$

⠀

Here |dv| = Size of image

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|dv| = Size of object (= b)

⠀

From the Equation------------(1), We write

⠀

$\sf \longrightarrow \frac{ {v}^{2} }{ {u}^{2} } = ( \frac{f}{u - f} {)}^{2}$

⠀

Substituting in Equation-----------(2), We get

⠀

$\large{ \underline{ \sf \therefore{ Size \: of \: the \: image \: dv = b \: ( \frac{f}{u - f } }}} {)}^{2}$

⠀

$\rule{200}3$