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QJEEM19Phf71e: JEE Main 2019 Physics - Class 12 JEEM19Phf71e Solution: The graph shows how the magnification m produced by a thin lens …

The graph shows how the magnification m produced by a thin lens varies with image distance v. What is the focal length of the lens used? JEE Main 2019 (Online) 10th April Evening Slot Physics - Geometrical Optics Question 160 English

Multiple Choice Options

  • A b2ac{{{b^2}} \over {ac}}
  • B b2ca{{{b^2}c} \over a}
  • C ac{a \over c}
  • D bc{b \over c}

Step-by-Step Verified Solution

As the graph between magnification (m) and image distance (v) varies linearly, then
m = k1v + k2

⇒vu=k1v+k2\Rightarrow {v \over u} = {k_1}v + {k_2}

⇒1u=k1+k2v\Rightarrow {1 \over u} = {k_1} + {{{k_2}} \over v}

⇒k2v−1u=k1\Rightarrow {{{k_2}} \over v} - {1 \over u} = {k_1}

Clearly, k1=1f{k_1} = {1 \over f} and k2 = 1 here

∴f=1slope of m−y graph=bc\therefore f = {1 \over {slope\,of\,m - y\,graph}} = {b \over c}

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