A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12cm below the surface, the radius of this circle (in cm) is:

Name: A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12cm below the surface, the radius of this circle (in cm) is:
Uploaded: 2020-01-06
Description: Click here👆to get an answer to your question ✍️ A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12cm below the surface, the radius of this circle (in cm) is:

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Answer 1

$sin i_{c} = \frac{1}{μ} = \frac{r}{r ^{2} + h ^{2}}$
Using $h = 12$ cm, $μ = 4 / 3$
We get $r = \frac{3 6}{7}$ cm.

Answer 2

sin i_{c} = \frac{1}{μ} = \frac{r}{r ^{2} + h ^{2}}

Using

h = 12

cm,

μ = 4 / 3

We get

r = \frac{3 6}{7}

cm.

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12cm below the surface, the radius of this circle (in cm) is:

$365$

$45$

$367$

$\frac{3 6}{7}$

$sin i_{c} = \frac{1}{μ} = \frac{r}{r ^{2} + h ^{2}}$
Using $h = 12$ cm, $μ = 4 / 3$
We get $r = \frac{3 6}{7}$ cm.

A small bulb is placed at the bottom of a tank containing water to a depth of $1 m$ . Find the critical angle for water air interface and also calculate the diameter of the circular bright patch of light formed on the surface of water [Refractive index of water $= 1.33$ ]

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

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 $\frac{4}{3}$ )

If the refractive index of water is 4/3 and that of glass is 5/3, then the critical angle of light entering from glass into water will be

A fish looks upward at an unobstructed overcast sky. What total angle does the sky appear to subtend? (Take refractive index of water as $2)$ .

Revise with Concepts

Total Internal Reflection and Critical angle

Optical Fibres

Critical Angle - Formula

Applications of Total Internal Reflection

Prism binoculars

A fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is 4/3 and the fish is 12cm below the surface, the radius of this circle (in cm) is:

365​

45​

367​

7​36​

sinic​=μ1​=r2+h2​r​ Using h=12 cm, μ=4/3 We get r=7​36​ cm.

A small bulb is placed at the bottom of a tank containing water to a depth of 1 m. Find the critical angle for water air interface and also calculate the diameter of the circular bright patch of light formed on the surface of water [Refractive index of water=1.33]

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

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 34​)

If the refractive index of water is 4/3 and that of glass is 5/3, then the critical angle of light entering from glass into water will be

A fish looks upward at an unobstructed overcast sky. What total angle does the sky appear to subtend? (Take refractive index of water as 2​).

Revise with Concepts

Total Internal Reflection and Critical angle

Optical Fibres

Critical Angle - Formula

Applications of Total Internal Reflection

Prism binoculars

$365$

$45$

$367$

$\frac{3 6}{7}$

$sin i_{c} = \frac{1}{μ} = \frac{r}{r ^{2} + h ^{2}}$
Using $h = 12$ cm, $μ = 4 / 3$
We get $r = \frac{3 6}{7}$ cm.

A small bulb is placed at the bottom of a tank containing water to a depth of $1 m$ . Find the critical angle for water air interface and also calculate the diameter of the circular bright patch of light formed on the surface of water [Refractive index of water $= 1.33$ ]

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 $\frac{4}{3}$ )

A fish looks upward at an unobstructed overcast sky. What total angle does the sky appear to subtend? (Take refractive index of water as $2)$ .