The focal length of a mirror is given by 1v−1u=2f. If equal errors α are made in measuring u and v. Then relative error in f is

δu=δv=α ⇒−1v2δv+1u2δu=-2f2δf 1u2α-1v2α=-2f2δf 1u-1v1u+1vα=-2f2δf-2f1u+1vα=-2f2δf 1u+1vα=1fδf

The focal length of a mirror is given by 1v−1u=2f. If equal errors α are made in measuring u and v. Then relative error in f is

δu=δv=α ⇒−1v2δv+1u2δu=-2f2δf 1u2α-1v2α=-2f2δf 1u-1v1u+1vα=-2f2δf-2f1u+1vα=-2f2δf 1u+1vα=1fδf

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$The focal length of a mirror is given by \frac{1}{v} - \frac{1}{u} = \frac{2}{f} . If equal errors α are made in$ $measuring u and v . Then relative error in f is$

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$\frac{2}{α}$

$α (\frac{1}{u} + \frac{1}{v})$

$α (\frac{1}{u} - \frac{1}{v})$

$\frac{3}{α}$

answer is B.

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Detailed Solution

$δ u = δ v = α \Rightarrow \frac{- 1}{v^{2}} δ v + \frac{1}{u^{2}} δ u = - \frac{2}{f^{2}} δ f \frac{1}{u^{2}} α - \frac{1}{v^{2}} α = - \frac{2}{f^{2}} δ f$

$(\frac{1}{u} - \frac{1}{v}) (\frac{1}{u} + \frac{1}{v}) α = - \frac{2}{f^{2}} δ f$

$\frac{- 2}{f} (\frac{1}{u} + \frac{1}{v}) α = - \frac{2}{f^{2}} δ f$

$(\frac{1}{u} + \frac{1}{v}) α = \frac{1}{f} δ f$

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The focal length of a mirror is given by 1v−1u=2f. If equal errors α are made in measuring u and v. Then relative error in f is

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$The focal length of a mirror is given by \frac{1}{v} - \frac{1}{u} = \frac{2}{f} . If equal errors α are made in$ $measuring u and v . Then relative error in f is$