A converging mirror with a radius of 20cm creates a real image at 30cm

1.	A converging mirror with a radius of 20cm creates a real image at 30cm from the mirror what is the objects distance
Answer» $\large \underline{ \red{\underline{ \sf Question}}} : - \\ \sf A \: converging \: mirror \: with \: a \: radius \: of \: 20cm \\ \sf creates \: a \: real \: image \: at \: 30cm \: from \: the \: \\ \sf mirror \: what \: is \: the \: <klux>OBJECT</klux> \: distance \: .$ $\large \underline{ \green{\underline{ \sf \: Answer \: }}} : - \\ \sf object \: distance \: is \: 1 5 <klux>CM</klux> \:$ $\large \underline{ \red{\underline{ \sf To \: Find }}} \: - \\ \to \sf object \: distance \:$ $\large \underline{ \pink{\underline{ \sf Explanation }}} \: : - \\$ Given that : Image distance (v) = 30 cm Radius of CURVATURE (R) = 20 cm Object distance (u) = ? Note :- → Focal length of a converging mirror is positive . $\large \underline{ \pink{\underline{ \sf Solution }}} \: : - \\$ we know that , $\to \sf \frac{2}{R} = \frac{1}{v} + \frac{1}{u} \\ \\ \rm putting \: the \: given \: values \: \\ \\ \to \sf \frac{2}{20} = \frac{1}{30} + \frac{1}{u} \\ \\ \to \sf \frac{1}{u} = \frac{1}{10} - \frac{1}{30} \\ \\ \to \sf \frac{1}{u} = \frac{3- 1}{30} \\ \\ \to \sf \frac{1}{u} = \frac{2}{30} \\ \\ \to \sf \frac{1}{u} = \frac{1}{15} \\ \\ \to \boxed{ \sf u = 15cm \: }$ •°• object distance is 15 cm

A converging mirror with a radius of 20cm creates a real image at 30cm from the mirror what is the objects distance

Answer»

$\large \underline{ \red{\underline{ \sf Question}}} : - \\ \sf A \: converging \: mirror \: with \: a \: radius \: of \: 20cm \\ \sf creates \: a \: real \: image \: at \: 30cm \: from \: the \: \\ \sf mirror \: what \: is \: the \: <klux>OBJECT</klux> \: distance \: .$

$\large \underline{ \green{\underline{ \sf \: Answer \: }}} : - \\ \sf object \: distance \: is \: 1 5 <klux>CM</klux> \:$

$\large \underline{ \red{\underline{ \sf To \: Find }}} \: - \\ \to \sf object \: distance \:$

$\large \underline{ \pink{\underline{ \sf Explanation }}} \: : - \\$

Given that :

Image distance (v) = 30 cm

Radius of CURVATURE (R) = 20 cm

Object distance (u) = ?

Note :-

→ Focal length of a converging mirror is positive .

$\large \underline{ \pink{\underline{ \sf Solution }}} \: : - \\$

we know that ,

$\to \sf \frac{2}{R} = \frac{1}{v} + \frac{1}{u} \\ \\ \rm putting \: the \: given \: values \: \\ \\ \to \sf \frac{2}{20} = \frac{1}{30} + \frac{1}{u} \\ \\ \to \sf \frac{1}{u} = \frac{1}{10} - \frac{1}{30} \\ \\ \to \sf \frac{1}{u} = \frac{3- 1}{30} \\ \\ \to \sf \frac{1}{u} = \frac{2}{30} \\ \\ \to \sf \frac{1}{u} = \frac{1}{15} \\ \\ \to \boxed{ \sf u = 15cm \: }$

•°• object distance is 15 cm

A converging mirror with a radius of 20cm creates a real image at 30cm from the mirror what is the objects distance

Note :-

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