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An equi-concave lens of radius of curvature 15 cm and μ = 1.5 is placed in water (μ = 1.33). If one surface is silvered, then image distance from lens when an object is placed at distance of 14 cm from the lens is

Option 1 -

8.5 cm

Option 2 -

4.2 cm

Option 3 -

10.5 cm

Option 4 -

6.4 cm

0 38 Views | Posted a month ago
Asked by Shiksha User

  • 1 Answer

  • V

    Answered by

    Vishal Baghel | Contributor-Level 10

    a month ago
    Correct Option - 2


    Detailed Solution:

    1/f_lens = (1.5/ (4/3) - 1) ( - 2/15) = f_lens = -60cm
    ∴ 1/fe = 2/ (-60) + 2/ (-15) => fe = -6 cm
    ∴ V = (14 × (-6)/ (14 + 6) = 4.2 cm from lens (virtual)

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