In an experiment to measure focal length (f) of convex lens, the least counts of the measuring scales for the position of object ( u ) and for the position of image ( v ) are $\Delta \mathrm{u}$ and $\Delta \mathrm{v}$, respectively. The error in the measurement of the focal length of the convex lens will be :

(B) $\mathrm{f}^{2}\left[\frac{\Delta \mathrm{u}}{\mathrm{u}^{2}}+\frac{\Delta \mathrm{v}}{\mathrm{v}^{2}}\right]$

(A) $\frac{\Delta u}{u}+\frac{\Delta v}{v}$

(C) $2 f\left[\frac{\Delta u}{u}+\frac{\Delta v}{v}\right]$

(D) $\mathrm{f}\left[\frac{\Delta \mathrm{u}}{\mathrm{u}}+\frac{\Delta \mathrm{v}}{\mathrm{v}}\right]$

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Units & Measurement

In an experiment to measure focal length (f) of convex lens, the least counts of the measuring scales for the position of object ( u ) and for the position of image ( v ) are

Δ u

and

Δ v

, respectively. The error in the measurement of the focal length of the convex lens will be :

(A) $\frac{Δ u}{u} + \frac{Δ v}{v}$
(B) $f^{2} [\frac{Δ u}{u ^{2}} + \frac{Δ v}{v ^{2}}]$ ✓ Correct
(C) $2 f [\frac{Δ u}{u} + \frac{Δ v}{v}]$
(D) $f [\frac{Δ u}{u} + \frac{Δ v}{v}]$

Answer: (B)

Solution

The lens formula is

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

. Differentiating,

- \frac{df}{f ^{2}} = - \frac{d v}{v ^{2}} + \frac{d u}{u ^{2}} .

Taking magnitudes of the least-count errors (worst case, errors add):

Δ f = f^{2} [\frac{Δ v}{v ^{2}} + \frac{Δ u}{u ^{2}}] .

Option B.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option

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JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Units & Measurement

Solution

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