In an experiment to measure focal length of convex lens, the least count of the measuring scales for the position of object () and for the position of image () are and respectively. The error in the measurement of the focal length of the convex lens wi

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Q.
In an experiment to measure focal length $(f)$ of convex lens, the least count of the measuring scales for the position of object ( $u$ ) and for the position of image ( $v$ ) are $∆ u$ and $∆ v,$ respectively. The error in the measurement of the focal length of the convex lens will be [2024]

1 $2 f [\frac{∆ u}{u} + \frac{∆ v}{v}]$

2 $f [\frac{∆ u}{u} + \frac{∆ v}{v}]$

3 $\frac{∆ u}{u} + \frac{∆ v}{v}$

4 $f^{2} [\frac{∆ u}{u^{2}} + \frac{∆ v}{v^{2}}]$

Ans.
(4)

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

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

$d f = f^{2} [\frac{d v}{v^{2}} + \frac{d u}{u^{2}}] \Rightarrow ∆ f = f^{2} (\frac{∆ v}{v^{2}} + \frac{∆ u}{u^{2}})$

Solution

Q.
In an experiment to measure focal length $(f)$ of convex lens, the least count of the measuring scales for the position of object ( $u$ ) and for the position of image ( $v$ ) are $∆ u$ and $∆ v,$ respectively. The error in the measurement of the focal length of the convex lens will be [2024]

1 $2 f [\frac{∆ u}{u} + \frac{∆ v}{v}]$

2 $f [\frac{∆ u}{u} + \frac{∆ v}{v}]$

3 $\frac{∆ u}{u} + \frac{∆ v}{v}$

4 $f^{2} [\frac{∆ u}{u^{2}} + \frac{∆ v}{v^{2}}]$

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Solution

Q. In an experiment to measure focal length (f) of convex lens, the least count of the measuring scales for the position of object (u) and for the position of image (v) are ∆u and ∆v, respectively. The error in the measurement of the focal length of the convex lens will be [2024]

1 2f[∆uu+∆vv]

2 f[∆uu+∆vv]

3 ∆uu+∆vv

4 f2[∆uu2+∆vv2]

Ans. (4) 1v-1u=1f -1v2dv+1u2du=-1f2df ⇒dff2=dvv2+duu2 df=f2[dvv2+duu2]⇒∆f=f2(∆vv2+∆uu2)

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1 $2 f [\frac{∆ u}{u} + \frac{∆ v}{v}]$

2 $f [\frac{∆ u}{u} + \frac{∆ v}{v}]$

3 $\frac{∆ u}{u} + \frac{∆ v}{v}$

4 $f^{2} [\frac{∆ u}{u^{2}} + \frac{∆ v}{v^{2}}]$