Q191 / 200 · MHT CET - Medical 2026 (21 April Shift 1) · Physics
In an experiment to measure the power of a convex lens, the least count of measuring scale for the position of object $u$ and the position of image $v$ are $\Delta u$ and $\Delta v$ respectively. The error $\Delta P$ in measurement of power of a convex lens will be
- Correct answerA$\dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}$
- B$\dfrac{\Delta v}{v} + \dfrac{\Delta u}{u}$
- C$\dfrac{1}{\dfrac{\Delta v}{v} + \dfrac{\Delta u}{u}}$
- D$\dfrac{1}{\dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}}$
Solution
The power of a lens is given by $P = \dfrac{1}{f}$
Using the lens formula, we can write the power in terms of object distance $u$ and image distance $v$ as:
$P = \dfrac{1}{v} - \dfrac{1}{u}$
Differentiating both sides to find the error in power, we get:
$dP = -\dfrac{dv}{v^2} + \dfrac{du}{u^2}$
To find the maximum permissible error, we take the absolute values of the individual error terms and add them:
$\Delta P = \dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}$
Answer: $\dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}$
Using the lens formula, we can write the power in terms of object distance $u$ and image distance $v$ as:
$P = \dfrac{1}{v} - \dfrac{1}{u}$
Differentiating both sides to find the error in power, we get:
$dP = -\dfrac{dv}{v^2} + \dfrac{du}{u^2}$
To find the maximum permissible error, we take the absolute values of the individual error terms and add them:
$\Delta P = \dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}$
Answer: $\dfrac{\Delta v}{v^2} + \dfrac{\Delta u}{u^2}$
Hence, the correct answer is option: 'A'.