Question

║⊕QUESTION⊕║
“However difficult life may seem, there is always something you can do and succeed at.”
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CLASS 11
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The focal length f of a mirror is given by
$\frac{1}{f} = \frac{1}{u}+ \frac{1}{v}$
Where u and v represent object and image distances respectively. Find the relative error in f, if u is measured as [u±Δu] and v is measured as [v±Δv]

Answer 1

◆【Solution】◆

1/f = 1/u + 1/v

Differentiate the equation

-(∆f/f²) = -(∆u/u²) - (∆v/v²)

∆f/f² = ∆u/u² + ∆v/v²

(∆f/f) = f (∆u/u² + ∆v/v²)

Relative error in f = (1/u + 1/v)(∆u/u² + ∆v/v²)

Differential of 1/f = -df/f²

Differential of 1/u = -du/u²

Differential of 1/v = -dv/v²

Sorry in rush I commit a mistake in your mathematics answer in which we need to find length of semi Latus rectum so now I correct it so now you can check and tell whether it helps or not.

Answer 2

Explanation:

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“However difficult life may seem, there is always something you can do and succeed at.”

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CLASS 11

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The focal length f of a mirror is given by

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

Where u and v represent object and image distances respectively. Find the relative error in f, if u is measured as [u±Δu] and v is measured as [v±Δv]