A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is (4/3), and the diver's eyes are 15 cm below the surface of water. Then the radius of the circle is :

Question

A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is (4/3), and the diver's eyes are 15 cm below the surface of water. Then the radius of the circle is : (A) 15×3×√5 cm (B) 15×3×√7 cm (C) (15×√7/3) cm (D) (15×3/√7) cm

Tardigrade Admin · Accepted Answer

sin θC=(1/4/ 3)=(3/4) tan θ C=(3/√7)=(r/h) r=(3/√7)×15

Q. A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $\frac{4}{3}$ , and the diver's eyes are $15 c m$ below the surface of water. Then the radius of the circle is :

$15 \times 3 \times 5 c m$

$15 \times 3 \times 7 c m$

$\frac{15 \times 7}{3} c m$

$\frac{15 \times 3}{7} c m$

$sin θ_{C} = \frac{1}{4/3} = \frac{3}{4}$
$tan θ_{C} = \frac{3}{7} = \frac{r}{h}$
$r = \frac{3}{7} \times 15$

Q. A diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is 34​, and the diver's eyes are 15cm below the surface of water. Then the radius of the circle is :

15×3×5​cm

15×3×7​cm

315×7​​cm

7​15×3​cm