The observer -O- sees the distance AB as infinitely large- If refractive index of liquid is mu -1 and that of glass is mu -22- then displaystyle frac-mu -1-mu -2- is

Point -apos-B-apos- should be at the focus of curved surface f-AB-2R-x3BC-2v-x2212-x3BC-1u-x3BC-2-x2212-x3BC-1R-x3BC-2-x221E-x3BC-12R-x3BC-2-x2212-x3BC-1R-x2212-x3BC-1-2-x3BC-2-x2212-2-x3BC-1-x3BC-1-2-x3BC-2-x21D2-x3BC-1-x3BC-2-2

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Standard XII

Physics

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

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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

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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 $f = A B = 2 R$

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

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