In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towardes A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of

Question

In a long glass tube, mixture of two liquids A and B with refractive indices 1.3 and 1.4 respectively, forms a convex refractive meniscus towards A. If an object placed at 13 cm from the vertex of the meniscus in A forms an image with a magnification of '–2' then the radius of curvature of meniscus is: [2025]

Smart Achievers · Accepted Answer

(3)

$μ_{A} = 1.3, μ_{B} = 1.4 and u = - 13 cm$

$m = \frac{\frac{v}{μ_{B}}}{\frac{u}{μ_{A}}} = - 2$

$\Rightarrow v = - 2 u (\frac{μ_{B}}{μ_{A}})$

$= - 2 (- 13 cm) (\frac{1.4}{1.3}) = + 28 cm$

Also, $\frac{μ_{B}}{v} - \frac{μ_{A}}{u} = \frac{μ_{B} - μ_{A}}{R}$

$\frac{1.4}{28} - \frac{1.3}{- 13} = \frac{0.1}{R} \Rightarrow R = \frac{2}{3} cm$

Solution

1 1 cm

2 $\frac{1}{3} cm$

3 $\frac{2}{3} cm$

4 $\frac{4}{3} cm$

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Solution

1 1 cm

2 13 cm

3 23 cm

4 43 cm

Ans. (3) μA=1.3, μB=1.4 and u=–13 cm m=vμBuμA=–2 ⇒ v=–2u(μBμA) =–2(–13 cm)(1.41.3)=+28 cm Also, μBv–μAu=μB–μAR 1.428–1.3–13=0.1R ⇒ R=23 cm

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2 $\frac{1}{3} cm$

3 $\frac{2}{3} cm$

4 $\frac{4}{3} cm$