Question

Derive the relationship between speed of object and image formed by spherical mirror.

Give Derivation (use Differentiation for it). Give Proper Explanation. Don't skip any step. ​

Answer 1

$\bigstar\huge\boxed{\mathtt{\red{Spherical\:Mirrors:}}}$

When the object moves parallel to principal axis:-

• Let f ,u ,v be the focal length, object distance and image distance respectively

• When an object moves parallel to the principal axis u , v changes but f remains constant

As per the mirror formula ,

$\implies\mathtt{\dfrac{1}{f} =\dfrac{1}{u} +\dfrac{1}{v}}$

$\implies\mathtt{f^{-1} = u^{-1} +v^{-1} }$

Differentiating wrt time we get ,

$\implies\mathtt{0 = -u^{-2} \dfrac{du}{dt} -v^{-2} \dfrac{dv}{dt} }$

$\implies\mathtt{ v^{-2} \dfrac{dv}{dt} = - u^{-2} \dfrac{du}{dt} }$

$\implies\mathtt{ \dfrac{1}{v^{2} }\dfrac{dv}{dt} = - \dfrac{1}{u^{2} } \dfrac{du}{dt} }$

$\implies\mathtt{ \dfrac{dv}{dt} = - \dfrac{v^{2} }{u^{2} } \dfrac{du}{dt} }$

The below equation is only when the mirror is at rest

$\implies\mathtt{\pink{ \overrightarrow V_I = - ( \dfrac{v^{2} }{u^{2} }) \overrightarrow V_O }}$

Mirror in motion

$\implies\boxed{\mathtt{\pink{ \overrightarrow V_I- \overrightarrow V_M = - ( \dfrac{v^{2} }{u^{2} }) \overrightarrow V_O - \overrightarrow V_M}}}$

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When the object moves perpendicular to principal axis :-

• When an object moves perpendicular to the principal axis u , v ,f  remains constant but the hi (height of image)and ho (height of object )will vary with time

we know that ,

$\implies\mathtt{m= \dfrac{h_i}{h_o} =\dfrac{-v}{u} }$

$\implies\mathtt{\dfrac{h_i}{h_o} =\dfrac{-v}{u} }$

$\implies\mathtt{h_i =(\dfrac{-v}{u}) h_o}$

Differentiating wrt time ,

$\implies\mathtt{\dfrac{dh_i }{dt}=(\dfrac{-v}{u})\dfrac{dh_o}{dt} }$

$\implies\boxed{\mathtt{\pink{\overrightarrow V_I=(\dfrac{-v}{u}) \overrightarrow V_o}}}$