Given a thin convex lens (refractive index ), kept in a liquid (refractive indes ) having radii of curvature || and ||. 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 a

Question

Given a thin convex lens (refractive index $μ_{2}$ ), kept in a liquid (refractive index $μ_{1}, μ_{1} < μ_{2}$ ) having radii of curvature | $R_{1}$ | and | $R_{2}$ |. 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? [2025]

Smart Achievers · Accepted Answer

(2)

$μ_{1}, μ_{1} < μ_{2}$

$f_{m} = - \frac{R_{2}}{2}$

$P_{eq} = 2 P_{l} + P_{m}$

$\frac{1}{| f |} = \frac{2 (μ_{2} - μ_{1})}{μ_{1}} (\frac{1}{R_{1}} + \frac{1}{R_{2}}) + \frac{2}{R_{2}}$

$= \frac{2 (μ_{2} - μ_{1}) (R_{1} + R_{2})}{μ_{1} (R_{1} R_{2})} + \frac{2}{R_{2}}$

$\frac{1}{f} = \frac{2 μ_{2} R_{1} + 2 μ_{2} R_{2} - 2 μ_{1} R_{1} - 2 μ_{1} R_{2} + 2 μ_{1} R_{1}}{μ_{1} R_{1} R_{2}}$

$\Rightarrow f = \frac{μ_{1} R_{1} R_{2}}{2 μ_{2} R_{1} + 2 μ_{2} R_{2} - 2 μ_{1} R_{2}}$

Required distance = $2 f$

Solution

1 $\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (| R_{1} | + | R_{2} |) - μ_{1} | R_{1} |}$

2 $\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (| R_{1} | + | R_{2} |) - μ_{1} | R_{2} |}$

3 $\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (2 | R_{1} | + | R_{2} |) - μ_{1} \sqrt{| R_{1} | \cdot | R_{2} |}}$

4 $\frac{(μ_{2} + μ_{1}) | R_{1} |}{(μ_{2} - μ_{1})}$

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Solution

Q. Given a thin convex lens (refractive index μ2), kept in a liquid (refractive index μ1,μ1<μ2) having radii of curvature |R1| and |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 [2025]

1 μ1|R1|·|R2|μ2(|R1|+|R2|)–μ1|R1|

2 μ1|R1|·|R2|μ2(|R1|+|R2|)–μ1|R2|

3 μ1|R1|·|R2|μ2(2|R1|+|R2|)–μ1|R1|·|R2|

4 (μ2+μ1)|R1|(μ2–μ1)

Ans. (2) 1fl=(μ2μ1–1)(1R1+1R2) fm=–R22 Peq=2Pl+Pm 1|f|=2(μ2–μ1)μ1(1R1+1R2)+2R2 =2(μ2–μ1)(R1+R2)μ1(R1R2)+2R2 1f=2μ2R1+2μ2R2–2μ1R1–2μ1R2+2μ1R1μ1R1R2 ⇒ f=μ1R1R22μ2R1+2μ2R2–2μ1R2 Required distance = 2f

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1 $\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (| R_{1} | + | R_{2} |) - μ_{1} | R_{1} |}$

2 $\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (| R_{1} | + | R_{2} |) - μ_{1} | R_{2} |}$

3 $\frac{μ_{1} | R_{1} | \cdot | R_{2} |}{μ_{2} (2 | R_{1} | + | R_{2} |) - μ_{1} \sqrt{| R_{1} | \cdot | R_{2} |}}$

4 $\frac{(μ_{2} + μ_{1}) | R_{1} |}{(μ_{2} - μ_{1})}$