The focal length of mirror is given by 2-f-1-v-1-u if du- dv- df are the differentials in u-v-f and du-dv then df-f-

The correct option is B (1/u+1/v) 2f^2df= #10; 1v^2dv+ duu^2u #10; #10; 2f^2df= du ( #160;1v 1u ) ( #10;1v+ 1u ) #10; #10; ...

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Mirror Formula Examples

The focal len...

Question

The focal length of mirror is given by $\frac{2}{f} = \frac{1}{v} - \frac{1}{u}$ if du, dv, df are the differentials in u,v,f and $d u = d v$ then $\frac{d f}{f} =$

(\frac{1}{u} - \frac{1}{v})

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(\frac{1}{u} + \frac{1}{v})

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1

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(\frac{1}{u^{2}} + \frac{1}{v^{2}})

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Similar questions

u = f (cot x)

v = g (cosec x)

f^{'} (1) = \sqrt{2}

g^{'} (\sqrt{2}) = 2

{(\frac{d u}{d v})}_{x = \frac{π}{4}} = 1

u = f (x), v = g (x)

f

g

\frac{d u}{d v} = \frac{d u / d x}{d v / d x}

f

\frac{1}{f} = \frac{1}{u} + \frac{1}{v}

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\frac{1}{v} - \frac{1}{u} = \frac{2}{f}

α

u

v

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u = {tan}^{- 1} (\frac{\sqrt{1 - x^{2}} - 1}{x})

v = {sin}^{- 1} x

\frac{d u}{d v}

(u)

(v)

(f)

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