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JEE Advanced 2019Paper 2
Exam
JEE Advanced
Year
2019
Paper
Paper 2
Subject
Physics
Chapter
Ray Optics

JEE Advanced 2019 · Paper 2 · Physics · Ray Optics

An optical bench has 1.5m long scale having four equal divisions in...

An optical bench has 1.5m1.5m long scale having four equal divisions in each cm.cm . While measuring the focal length of a convex lens, the lens is kept at 75cm75cm mark of the scale and the object pin is kept at 45cm45cm mark. The image of the object pin on the other side of the lens overlaps with image pin that is kept at 135cm135cm mark. In this experiment, the percentage error in the measurement of the focal length of the lens is_____An optical bench has 1.5m long scale having four equal divisions in each cm . While measuring the focal length of a

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An optical bench has 1.5m long scale having four equal divisions in each cm . While measuring the focal length of a
Let, \left(x1=distanceofobjectfromOx2=distanceoflensfromOx3=distanceofimagefromO\begin{array}{ll}{x}_{1} = distanceofobjectfromO \\ {x}_{2} = distanceoflensfromO \\ {x}_{3} = distanceofimagefromO\end{array}\rightopen=[\rightopen=[
∴u=(x2−x1)=75−45=30cm∴ u = \left({x}_{2} - {x}_{1}\right) = 75 - 45 = 30cm
⇒∆u=∆x2+∆x1=14+14=12cm⇒⇒ ∆ u = ∆ {x}_{2} + ∆ {x}_{1} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}cm ⇒ error in measurement of “u”“u”
And, v=(x3−x2)=135−75=60cmv = \left({x}_{3} - {x}_{2}\right) = 135 - 75 = 60cm
⇒∆v=∆x3+∆x2=14+14=12cm⇒⇒ ∆ v = ∆ {x}_{3} + ∆ {x}_{2} = \dfrac{1}{4} + \dfrac{1}{4} = \dfrac{1}{2}cm ⇒ error in measurement of “v”“v”
Using Lens formula ∴1v−1u=1f⇒160+130=1f⇒1f=120∴ \dfrac{1}{v} - \dfrac{1}{u} = \dfrac{1}{f} ⇒ \dfrac{1}{60} + \dfrac{1}{30} = \dfrac{1}{f} ⇒ \dfrac{1}{f} = \dfrac{1}{20}
∴f=20cm,Also,−dvv2+−duu2=−dff2→∴ f = 20cm, Also, \dfrac{- dv}{{v}^{2}} + \dfrac{- du}{{u}^{2}} = \dfrac{- df}{{f}^{2}} \to [differentiating lens formula]
⇒dff=f[dvv2+duu2]=20[1602+1302]12⇒ \dfrac{df}{f} = f\left[\dfrac{dv}{{v}^{2}} + \dfrac{du}{{u}^{2}}\right] = 20\left[\dfrac{1}{{60}^{2}} + \dfrac{1}{{30}^{2}}\right]\dfrac{1}{2}
∴ \dfrac{df}{f} \times 100 = 10\left[\dfrac{1}{36} + \dfrac{1}{9}\right] = \dfrac{50}{36} = 1.38and1.39\left(both\right)
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