JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Geometrical Optics

The power of a lens (biconvex) is

1.25 m^{- 1}

in particular medium. Refractive index of the lens is 1.5 and radii of curvature are 20 cm and 40 cm respectively. The refractive index of surrounding medium :

(A)1.0
(B) $\frac{9}{7}$ ✓ Correct
(C) $\frac{3}{2}$
(D) $\frac{4}{3}$

Answer: (B)

Solution

In a surrounding medium the lens-maker relation uses the index of the lens relative to the medium,

P = (\frac{μ _{lens}}{μ _{med}} - 1) (\frac{1}{R _{1}} - \frac{1}{R _{2}})

. For the biconvex lens

\frac{1}{R _{1}} - \frac{1}{R _{2}} = \frac{1}{0.20} - \frac{1}{- 0.40} = 7.5 m^{- 1}

, so

1.25 = (\frac{1.5}{μ _{med}} - 1) (7.5)

. This gives

\frac{1.5}{μ _{med}} - 1 = \frac{1}{6}

, hence

\frac{1.5}{μ _{med}} = \frac{7}{6}

and

μ_{med} = 1.5 \times \frac{6}{7} = \frac{9}{7}

. Using the difference

μ_{lens} - μ_{med}

in place of the ratio is what wrongly produces

\frac{4}{3}

JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Single correct option

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JEE Main · 2022 · 28 Jul 2022 · Shift 2 · Geometrical Optics

Solution

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