Verified Solution • 2025–26 Session moderate

QJEEM25Ph1e0a: JEE Main 2025 Physics - Class 12 JEEM25Ph1e0a Solution: Given a thin convex lens (refractive index ), kept in a liquid…

Given a thin convex lens (refractive index μ2\mu_2 ), kept in a liquid (refractive index \mu_1, \mu_1<\mu_2 ) having radii of curvatures ∣R1∣\left|R_1\right| and ∣R2∣\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?

Multiple Choice Options

  • A (μ2+μ1)∣R1∣(μ2−μ1)\frac{\left(\mu_2+\mu_1\right)\left|R_1\right|}{\left(\mu_2-\mu_1\right)}
  • B μ1∣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|}}
  • C μ1∣R1∣⋅∣R2∣μ2(∣R1∣+∣R2∣)−μ1∣R1∣\frac{\mu_1\left|R_1\right| \cdot\left|R_2\right|}{\mu_2\left(\left|R_1\right|+\left|R_2\right|\right)-\mu_1\left|R_1\right|}
  • D 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 Verified Solution

JEE Main 2025 (Online) 23rd January Morning Shift Physics - Geometrical Optics Question 11 English Explanation

amp;1feq=2fL−1fmamp;fm=−∣R2∣2amp;1fL=(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}

amp;amp;1feq=2(μ2−μ1μ1)(R1+R2R1R2)+2R2amp;=2R2[(μ2−μ1)(R1+R2)+μ1R1μ1R1]amp;=2R2[μ2R1+μ2R2−μ1R1−μ1R2+μ1R1μ1R1]amp;1feq=2[μ2R1+μ2R2−μ1R2]μ1R1R2amp;For same size of image amp;u=2famp;u=μ1R1R2μ2R1+μ2R2−μ1R2\begin{aligned}&\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}\\\\&\text {For same size of image }\\\\&\mathrm{u}=2 \mathrm{f}\\\\&\mathbf{u}=\frac{\mu_1 \mathbf{R}_1 \mathbf{R}_2}{\mu_2 \mathrm{R}_1+\mu_2 \mathrm{R}_2-\mu_1 \mathrm{R}_2}\end{aligned}

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