A parallel narrow beam of light travelling through water in a container incident on an air bubble of radius $8$ mm. What be the distance of image obtained due to refraction of beam on first face of air bubble from the centre of bubble? (take $μ_{w a t e r} = \frac{4}{3}$ )

17.2 mm

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32 mm

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41.5 mm

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25.75 mm

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Solution

The correct option is B 32 mm
$\begin{matrix} \frac{1}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R} \Rightarrow \frac{1}{v} \frac{(\frac{4}{3})}{- \infty} = \frac{1 - \frac{4}{3}}{8} \Rightarrow \frac{1}{v} = - \frac{1}{2 u} \end{matrix}$
$∴$ distance from center: $32$ mm

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A parallel narrow beam of light travelling through water in a container incident on an air bubble of radius 8 mm. What be the distance of image obtained due to refraction of beam on first face of air bubble from the centre of bubble? (take μwater=43)

The correct option is B 32 mm1v−n1u=n2−n1R⇒1v(43)−∞=1−438⇒1v=−12u∴ distance from center:32 mm

A parallel narrow beam of light travelling through water in a container incident on an air bubble of radius $8$ mm. What be the distance of image obtained due to refraction of beam on first face of air bubble from the centre of bubble? (take $μ_{w a t e r} = \frac{4}{3}$ )

The correct option is B 32 mm
$\begin{matrix} \frac{1}{v} - \frac{n_{1}}{u} = \frac{n_{2} - n_{1}}{R} \Rightarrow \frac{1}{v} \frac{(\frac{4}{3})}{- \infty} = \frac{1 - \frac{4}{3}}{8} \Rightarrow \frac{1}{v} = - \frac{1}{2 u} \end{matrix}$
$∴$ distance from center: $32$ mm