First slide

Accuracy; precision of instruments and errors in measurements

Question

Object distance, $u = (50.1 \pm 0.5) c m$ and image distance $v = (20.1 \pm 0.2) c m,$ then focal length is

Moderate

A

$(12.4 \pm 0.4) c m$
B

$(12.4 \pm 0.1) c m$
C

$(14.3 \pm 0.4) c m$
D

$(14.3 \pm 0.1) c m$

Solution

Focal length is given by $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ or
$f = \frac{u v}{u + v} = \frac{(50.1) (20.1)}{(50.1) + (20.1)} = 14.3 c m$
Also, $\frac{Δ f}{f} = \pm (\frac{Δ u}{u} + \frac{Δ v}{v} + \frac{Δ u + Δ v}{u + v})$
$= \pm (\frac{0.5}{50.1} + \frac{0.2}{20.1} + \frac{0.5 + 0.2}{50.1 + 20.1})$
$= \pm [0.00998 + 0.00995 + 0.00997]$
$= \pm [0.0299]$
$∴ Δ f = 0.0299 \times 14.3$
$= 0.428 = 0.4 c m$
$∴ f = (14.3 \pm 0.4) c m$
In power If $Z = A^{n}, t h e n \frac{Δ Z}{Z} = n \frac{Δ A}{A}$
In more general form if $Z = \frac{A^{x} B^{y}}{C^{q}},$ then the maximum fractional error in Z is
$\frac{Δ Z}{Z} = x \frac{Δ A}{A} + y \frac{Δ B}{B} + q \frac{Δ C}{C}$
(Note that there is no negative sign)

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