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Standard XII

Physics

Power and Thin Lenses in Contact

The observer ...

Question

The observer 'O' sees the distance AB as infinitely large. If refractive index of liquid is $μ_{1}$ and that of glass is $μ_{2}$ 2, then $\frac{μ_{1}}{μ_{2}}$ is

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Solution

Point 'B' should be at the focus of curved surface $f = A B = 2 R$

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{μ_{2}}{\infty} + \frac{μ_{1}}{2 R} = \frac{μ_{2} - μ_{1}}{R} - μ_{1} = 2 μ_{2} - 2 μ_{1}; μ_{1} = 2 μ_{2} \Rightarrow \frac{μ_{1}}{μ_{2}} = 2$

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The observer 'O' sees the distance AB as infinitely large. If refractive index of liquid is μ1 and that of glass is μ22, then μ1μ2 is

Point 'B' should be at the focus of curved surface f=AB=2Rμ2v−μ1u=μ2−μ1Rμ2∞+μ12R=μ2−μ1R−μ1=2μ2−2μ1;μ1=2μ2⇒μ1μ2=2

The observer 'O' sees the distance AB as infinitely large. If refractive index of liquid is $μ_{1}$ and that of glass is $μ_{2}$ 2, then $\frac{μ_{1}}{μ_{2}}$ is

Point 'B' should be at the focus of curved surface $f = A B = 2 R$

$\frac{μ_{2}}{v} - \frac{μ_{1}}{u} = \frac{μ_{2} - μ_{1}}{R}$

$\frac{μ_{2}}{\infty} + \frac{μ_{1}}{2 R} = \frac{μ_{2} - μ_{1}}{R} - μ_{1} = 2 μ_{2} - 2 μ_{1}; μ_{1} = 2 μ_{2} \Rightarrow \frac{μ_{1}}{μ_{2}} = 2$