A point source S is placed at the bottom of a transparent block of height 10 mm and refractive index 2.72. It is immersed in a lower refractive index liquid as shown in the figure. It is found that the light emerging from the block to the liquid forms a circular bright spot of diameter 11.54 mm on the top of the block. The refractive index of the liquid is [2014]

Question

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(3)

Applying Snell's law at point P

$n_{1} \sin θ = n_{2} \sin 90 °$

$\sin θ = \frac{11.54 / 2}{\sqrt{10^{2} + {(\frac{11.54}{2})}^{2}}}$ and $n_{2} =$ refractive index of liquid.

$∴$ $n_{2} = 2.72 \times \frac{11.54 / 2}{\sqrt{10^{2} + {(\frac{11.54}{2})}^{2}}}$

$∴$ $n_{2} = 1.36$

Solution

1 1.21

2 1.30

3 1.36

4 1.42

Ans.
(3)

Applying Snell's law at point P

$n_{1} \sin θ = n_{2} \sin 90 °$

$\sin θ = \frac{11.54 / 2}{\sqrt{10^{2} + {(\frac{11.54}{2})}^{2}}}$ and $n_{2} =$ refractive index of liquid.

$∴$ $n_{2} = 2.72 \times \frac{11.54 / 2}{\sqrt{10^{2} + {(\frac{11.54}{2})}^{2}}}$

$∴$ $n_{2} = 1.36$

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