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JEE Main 202523 JanuaryMorning Shift
Exam
JEE Main
Year
2025
Date
23 January
Shift
Morning Shift
Subject
Physics
Chapter
Ray Optics

JEE Main 2025 · 23 January · Morning Shift · Physics · Ray Optics

Given a thin convex lens (refractive index μ_2 ), kept in a liquid...

Given a thin convex lens (refractive index μ2\mu_2 ), kept in a liquid (refractive index μ1,μ1<μ2\mu_1, \mu_1 \lt \mu_2 ) having radii of curvatures \left|R_1\right| and \left|R_2\right|$. Its second surface is silver polished. Where should an object be placed on the optic axis so that a real and inverted image is formed at the same place?Given a thin convex lens (refractive index μ_2 ), kept in a liquid (refractive index μ_1, μ_1 μ_2 ) having radii of

Options

  1. Aμ1∣R1∣⋅∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R2∣\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1\left|\mathrm{R}_2\right|}
  2. Bμ1∣R1∣⋅∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R1∣\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1\left|\mathrm{R}_1\right|}
  3. C(μ2+μ1)∣R1∣(μ2−μ1)\frac{\left(\mu_2+\mu_1\right)\left|\mathrm{R}_1\right|}{\left(\mu_2-\mu_1\right)}
  4. Dμ1∣R1∣⋅∣R2∣μ2(2∣R1∣+∣R2∣)−μ1∣R1∣⋅∣R2∣\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(2\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1 \sqrt{\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}}

Correct answer

A. μ1∣R1∣⋅∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R2∣\frac{\mu_1\left|\mathrm{R}_1\right| \cdot\left|\mathrm{R}_2\right|}{\mu_2\left(\left|\mathrm{R}_1\right|+\left|\mathrm{R}_2\right|\right)-\mu_1\left|\mathrm{R}_2\right|}

Step-by-step solution

Given a thin convex lens (refractive index μ_2 ), kept in a liquid (refractive index μ_1, μ_1 μ_2 ) having radii of
1feq=2fL−1fmfm=−∣R2∣21fL=(μ2μ1−1)(1R1+1R2)\begin{aligned} & \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2}{\mathrm{f}_{\mathrm{L}}}-\frac{1}{\mathrm{f}_{\mathrm{m}}} \\ & \mathrm{f}_{\mathrm{m}}=-\frac{\left|\mathrm{R}_2\right|}{2} \\ & \frac{1}{\mathrm{f}_{\mathrm{L}}}=\left(\frac{\mu_2}{\mu_1}-1\right)\left(\frac{1}{\mathrm{R}_1}+\frac{1}{\mathrm{R}_2}\right)\end{aligned}1feq=2(μ2−μ1μ1)(R1+R2R1R2)+2R2=2R2[(μ2−μ1)(R1+R2)+μ1R1μ1R1]=2R2[μ2R1+μ2R2−μ1R1−μ1R2+μ1R1μ1R1]1feq=2[μ2R1+μ2R2−μ1R2]μ1R1R2\begin{aligned} & \frac{1}{\mathrm{f}_{\mathrm{eq}}}=2\left(\frac{\mu_2-\mu_1}{\mu_1}\right)\left(\frac{\mathrm{R}_1+\mathrm{R}_2}{\mathrm{R}_1 \mathrm{R}_2}\right)+\frac{2}{\mathrm{R}_2} \\ & =\frac{2}{\mathrm{R}_2}\left[\frac{\left(\mu_2-\mu_1\right)\left(\mathrm{R}_1+\mathrm{R}_2\right)+\mu_1 \mathrm{R}_1}{\mu_1 \mathrm{R}_1}\right] \\ & =\frac{2}{\mathrm{R}_2}\left[\frac{\mu_2 \mathrm{R}_1+\mu_2 \mathrm{R}_2-\mu_1 \mathrm{R}_1-\mu_1 \mathrm{R}_2+\mu_1 \mathrm{R}_1}{\mu_1 \mathrm{R}_1}\right] \\ & \frac{1}{\mathrm{f}_{\mathrm{eq}}}=\frac{2\left[\mu_2 \mathrm{R}_1+\mu_2 \mathrm{R}_2-\mu_1 \mathrm{R}_2\right]}{\mu_1 \mathrm{R}_1 \mathrm{R}_2} \end{aligned}ForsamesizeofimageFor same size of imageu=2fu=μ1R1R2μ2R1+μ2R2−μ1R2\begin{aligned} & \mathrm{u}=2 \mathrm{f} \\ & \mathrm{u}=\frac{\mu_1 \mathrm{R}_1 \mathrm{R}_2}{\mu_2 \mathrm{R}_1+\mu_2 \mathrm{R}_2-\mu_1 \mathrm{R}_2} \end{aligned}$
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