A point is a depth of 1 m from the water surface- The diameter of the circle of light rays on water surface refracted into the air will be -refractive index of water is dfrac - 4 - 3 - - dfrac - 3 - sqrt - 7 - - m dfrac - 6 - sqrt - 7 - - m dfrac - 9 - 7 - m dfrac - 18 - 7 - m

Question

A point is a depth of 1 m from the water surface- The diameter of the circle of light rays on water surface refracted into the air will be -refractive index of water is  dfrac - 4 - 3 - - dfrac - 3 - sqrt - 7 - - m dfrac - 6 - sqrt - 7 - - m dfrac - 9 - 7 - m dfrac - 18 - 7 - m

Toppr Admin · Accepted Answer

A circle of radius R appears on the water surface since no rays go out to the air from water surface outside of the circle, since all of them reflect back due to total internal reflection.
In the total internal reflection phenomenon, $μ s i n r = 1$

Therefore $s i n r = \frac{3}{4}$

From trigonometry, $R = 1$ $\times t a n r =$ $\frac{3}{\sqrt{7}}$

Hence the diameter of the circle is $\frac{6}{\sqrt{7}}$

A point source is at a depth of 1 m from the water surface. The diameter of the circle of light rays on water surface refracted into the air will be (refractive index of water is 43)3√7m6√7m97m187m

A point source is at a depth of 1 m from the water surface. The diameter of the circle of light rays on water surface refracted into the air will be (refractive index of water is $\frac{4}{3}$ )
$\frac{3}{\sqrt{7}} m$
$\frac{6}{\sqrt{7}} m$
$\frac{9}{7} m$
$\frac{18}{7} m$