A small bulb is placed at the bottom of a tank containing water to a depth of 7m. The refractive index of water is 43. The area of the surface of water through which light from the bulb can emerge out is xπm2. The value of x is_________.

by snell's law 43sinθC=μ1sin90° sinθC=34=rr2+ (7)2 34=rr2+7 Squaring both side 916=r2r2+7 A=r2=π*9=9π

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt[]{7} m .$ The refractive index of water is $\frac{4}{3} .$ The area of the surface of water through which light from the bulb can emerge out is xπm². The value of x is_________.

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Vishal Baghel | Contributor-Level 10

6 months ago

by snell's law

$\frac{4}{3} s i n θ_{C} = μ_{1} s i n 90 °$

$s i n θ_{C} = \frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + {(\sqrt[]{7})}^{2}}}$

$\frac{3}{4} = \frac{r}{\sqrt[]{r^{2} + 7}}$

Squaring both side

$\frac{9}{16} = \frac{r^{2}}{r^{2} + 7}$

$A = r^{2} = π * 9 = 9 π$

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