A fish is a little away below the surface of a lake. If the critical angle is 49°, then the fish could see things above the water surface within an angular range of θ0 where:

Question

A fish is a little away below the surface of a lake. If the critical angle is 49°, then the fish could see things above the water surface within an angular range of θ0 where: (A) θ=49° (B) θ=90° (C) θ=98° (D) θ=24 (1°/2)

Tardigrade Admin · Accepted Answer

According to the question <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/f8589af32f90cb696f863dadf8cf5337-.png" /> θ=c+c (as by the figure) θ=2 c θ=2 × 49 θ=98°

Q. A fish is a little away below the surface of a lake. If the critical angle is $4 9^{\circ},$ then the fish could see things above the water surface within an angular range of $θ^{0}$ where:

$θ = 4 9^{\circ}$

$θ = 9 0^{\circ}$

$θ = 9 8^{\circ}$

$θ = 24 \frac{1 ^{\circ}}{2}$

According to the question

$θ = c + c$ (as by the figure)
$θ = 2 c$
$θ = 2 \times 49$
$θ = 9 8^{\circ}$

Q. A fish is a little away below the surface of a lake. If the critical angle is 49∘, then the fish could see things above the water surface within an angular range of θ0 where:

θ=49∘

θ=90∘

θ=98∘

θ=2421∘​