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Geometrical Optics question

2025 · 23 Jan · Shift 1 · Q70
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  5. /2025 · 23 Jan · Shift 1 · Q70

Geometrical Optics question

2025 · 23 Jan · Shift 1 · Q70

JEE MainPhysicsGeometrical OpticsMCQ+4 / −1
Given a thin convex lens (refractive index μ2\mu_2μ2​), kept in a liquid (refractive index μ1,μ1<μ2\mu_1, \mu_1\lt \mu_2μ1​,μ1​<μ2​) having radii of curvatures ∣R1∣\left|R_1\right|∣R1​∣ and ∣R2∣\left|R_2\right|∣R2​∣. 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?
  1. A
    (μ2+μ1)∣R1∣(μ2−μ1)\frac{\left(\mu_2+\mu_1\right)\left|R_1\right|}{\left(\mu_2-\mu_1\right)}(μ2​−μ1​)(μ2​+μ1​)∣R1​∣​
  2. 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|}}μ2​(2∣R1​∣+∣R2​∣)−μ1​∣R1​∣⋅∣R2​∣​μ1​∣R1​∣⋅∣R2​∣​
  3. 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|}μ2​(∣R1​∣+∣R2​∣)−μ1​∣R1​∣μ1​∣R1​∣⋅∣R2​∣​
  4. 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|}μ2​(∣R1​∣+∣R2​∣)−μ1​∣R2​∣μ1​∣R1​∣⋅∣R2​∣​
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