A fish looks outside water- It is situated a depth of 12 cm below water surface- the refractive index of water is 4-3 then radius of the circle through which it can see will be 12times dfrac - 3 - sqrt - 7 - - cm12 times 3 cm 12times 3sqrt - 5 - cm 12times dfrac - sqrt - 5 - - 3 - cm

Question

Toppr Admin · Accepted Answer

As shown in the figure it can see an conical view with cone angle -x3B8-c -critical angle-xA0-x3B8-c-sin-x2212-11-x3BC-sin-x2212-134R-12tan-x3B8-c-12-xD7-3-x221A-7-

A fish looks outside water. It is situated at a depth of 12 cm below water surface. if the refractive index of water is 4/3 then radius of the circle through which it can see will be
$12 \times \frac{3}{\sqrt{7}} c m$
$12 \times 3 c m$
$12 \times 3 \sqrt{5} c m$
$12 \times \frac{\sqrt{5}}{3} c m$

As shown in the figure it can see an conical view with cone angle $θ_{c}$ (critical angle)

$θ_{c} = s i n^{- 1} \frac{1}{μ} = s i n^{- 1} \frac{3}{4}$

$R = 12 t a n θ_{c} = 12 \times \frac{3}{\sqrt{7}}$ .

A fish looks outside water. It is situated at a depth of 12 cm below water surface. if the refractive index of water is 4/3 then radius of the circle through which it can see will be12×3√7cm12×3cm12×3√5cm12×√53cm

As shown in the figure it can see an conical view with cone angle θc (critical angle) θc=sin−11μ=sin−134R=12tanθc=12×3√7.

A fish looks outside water. It is situated at a depth of 12 cm below water surface. if the refractive index of water is 4/3 then radius of the circle through which it can see will be
$12 \times \frac{3}{\sqrt{7}} c m$
$12 \times 3 c m$
$12 \times 3 \sqrt{5} c m$
$12 \times \frac{\sqrt{5}}{3} c m$

As shown in the figure it can see an conical view with cone angle $θ_{c}$ (critical angle)

$θ_{c} = s i n^{- 1} \frac{1}{μ} = s i n^{- 1} \frac{3}{4}$

$R = 12 t a n θ_{c} = 12 \times \frac{3}{\sqrt{7}}$ .