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Geometrical Optics question

2022 · 28 Jul · Shift 2 · Q65
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Geometrical Optics question

2022 · 28 Jul · Shift 2 · Q65

JEE MainPhysicsGeometrical OpticsNumerical+4 / −1
In an experiment with a convex lens, The plot of the image distance (v′)\left(v^{\prime}\right)(v′) against the object distance (μ′)\left.\mu^{\prime}\right)μ′) measured from the focus gives a curve v′μ′=225v^{\prime} \mu^{\prime}=225v′μ′=225. If all the distances are measured in cm\mathrm{cm}cm. The magnitude of the focal length of the lens is ‾\underline{\hspace{2cm}}​ cm.
Numerical answer
View written solutionFree

Correct answer: 15

  1. Interpret the given quantities

    For a convex lens, let:

    • actual object distance from the lens be uuu
    • actual image distance from the lens be vvv
    • focal length be fff

    The question says that the distances are measured from the focus.

    So define:

    • object distance from focus: u′u'u′
    • image distance from focus: v′v'v′

    Then, u=f+u′u = f + u'u=f+u′ v=f+v′v = f + v'v=f+v′

  2. Use Newton’s form of the lens formula

    For a thin lens, when distances are measured from the focal points, Newton’s formula is: u′v′=f2u'v' = f^2u′v′=f2

    The graph/equation given in the problem is: u′v′=225u'v' = 225u′v′=225

    Comparing with Newton’s formula, f2=225f^2 = 225f2=225

  3. Calculate the focal length

    f=225=15 cmf = \sqrt{225} = 15\ \text{cm}f=225​=15 cm

    Since magnitude is asked, ∣f∣=15 cm|f| = 15\ \text{cm}∣f∣=15 cm

  4. Compare with stored answer

    Derived answer: 151515

    Stored correct answer: 151515

    Hence, they agree.

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