Question

A small illumination bulb is at the buttom of a tank containing a liquid of refractive index u upto a hight h. Find the expression for the distance of an opaque disc, floating symmetically on the liquid surface in order to cut off the light from the bulb

Answer 1

R = $\frac{H}{\sqrt{u^{2} - 1} }$ is the required expression for the distance of an opaque disc, floating symmetrically on the liquid surface to cut off the light from the bulb.

Given that;

A small illumination bulb is at the bottom of a tank containing a liquid of refractive index μ up to a height of h

To find;

The expression for the distance of an opaque disc, floating symmetrically on the liquid surface to cut off the light from the bulb

Solution;

According to Snell's law,

"The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant".

Here in the given question, we have,

Sin C = 1/ μ...(1)

1 - $Sin^{2} C$ = 1 - 1/$u^{2}$ = $Cos^{2}$C

$Cos^{2}$C = $u^{2} - 1/u^{2}$

CosC = $\sqrt{u^{2} } - 1/u$...(2)

We know that,

TanC = SinC/CosC

Dividing equation (1) by equation (2) we get,

TanC = $\frac{1}{\sqrt{u^{2} - 1 } }$...(3).   Also,

TanC = R/H...(4)

From equations (3) and (4) we get,

R = $\frac{H}{\sqrt{u^{2} - 1} }$.

Hence, R = $\frac{H}{\sqrt{u^{2} - 1} }$ is the required expression for the distance of an opaque disc, floating symmetrically on the liquid surface to cut off the light from the bulb.

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Answer 2

Answer:

Explanation:

From the above question,

They have given :

A small illuminated bulb is at the bottom of a tank, containing a liquid of refractive index μ μ upto a height H.

Liquid surface in order to cut off the light from the bulb, but this is a temporary solution and the water needs to be changed frequently.

If it is a regular incandescent light bulb, it should be submerged in the liquid. However, depending on the type of liquid, the bulb may not be able to stay on for very long due to the heat generated by the bulb. If the liquid is flammable or corrosive, it is especially important to ensure that the bulb is not submerged.

Another option is to use a light

Here sin C = 1 μ

Area visible = π $R^{2}$

The distance of an opaque object depends on the size of the object and the focal length of the lens. The formula is given by

Distance = focal length × object size/image size

Here we have to find the expression for the distance of an opaque disc, floating symmetically on the liquid surface in order to cut off the light from the bulb

For more such related questions : https://brainly.in/question/18722533

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