Question

Three mediums of refractive indices${\mu _0}$, ${\mu _1}$ and ${\mu _2}$ are shown in fig. $\left( {{\mu _1} > {\mu _0}{\text{ }}and{\text{ }}{\mu _2} > {\mu _0}} \right)$. The lamps $A$ and $B$ are placed at the bottom and top of first and third mediums of the same thickness. If the bottom layer of the middle medium is illuminated for a circle of half of the radius for which the upper layer of this medium is illuminated, the relationship between ${\mu _1}$ and ${\mu _2}$ ​is (given${\mu _0} = 1$):

$\left( a \right)$ $2{\mu _2} = \sqrt {\mu _1^2 + 3}$
$\left( b \right)$ ${\mu _2} = \sqrt {\mu _1^2 + 4}$
$\left( c \right)$ ${\mu _2} = \sqrt {\mu _1^2 + 2}$
$\left( d \right)$ ${\mu _2} = \sqrt {\mu _1^2 + 1}$

Answer 1

Hint So to find the relationship between them, we will use some of the trigonometric formula and the geometry properties. So as we know from the figure $\sin c = \dfrac{1}{\mu }$and$\tan c = \dfrac{1}{{\sqrt {{\mu ^2} - 1} }}$. So by using these two, we will find the relation between R.I.

Complete Step By Step Solution
As we have seen in the figure, let’s first make the ray diagram and see the angle and the distance made.
As we know,
$\sin c = \dfrac{1}{\mu }$, and
$\tan c = \dfrac{1}{{\sqrt {{\mu ^2} - 1} }}$
So from the figure, we can write it as
$\tan {c_1} = \dfrac{r}{h}$
Here,
$r$, is the distance
And $h$, will be the height
So,
$\Rightarrow r = h\tan {c_1}$
Now for ${c_{_2}}$
$\tan {c_2} = \dfrac{{2r}}{h}$
And also it can be written as
$\Rightarrow 2r = h\tan {c_2}$
So from the formula, we can write the $r$as
$\Rightarrow r = h \times \dfrac{1}{{\sqrt {u_1^2 - 1} }}$
And for ${\mu _2}$, it can be written as
$\Rightarrow 2r = h \times \dfrac{1}{{\sqrt {u_2^2 - 1} }}$
Now on dividing both the above equation, we get
$\Rightarrow \dfrac{1}{2} = \dfrac{{\sqrt {\mu _2^2 - 1} }}{{\sqrt {\mu _1^2 - 1} }}$
Now on removing the under root, we get
$\Rightarrow \dfrac{1}{4} = \dfrac{{\mu _2^2 - 1}}{{\mu _1^2 - 1}}$
So by doing the cross-multiplication, we get
$\Rightarrow \mu _1^2 - 1 = 4\mu _2^2 - 4$
And on solving the above line again, we get
$2{\mu _2} = \sqrt {\mu _1^2 + 3}$

Therefore the option $A$ is correct.

Additional information Well, an electromagnetic wave is only photons with an Energy. That kind of wave only interacts with the electrons of the material. Only on very high frequencies (gamma-ray) can interact with the nucleus (protons).
Then the refractive index only depends on the electrons on the material and how to interact with the wave. In a quantum theory, photons interact with the electrons, but there isn't a solution to that problem. Even a photon interacting with an electron is a hard problem, imagine a lot of photons interacting with a lot of electrons.
But the refractive index is not unique. Depends on wavelength and this is well known about optics. On the lens, red light has a different focal point than blue light, that's called "chromatic aberration".
If different wavelengths have a different refractive index, means that different wavelengths have different speed of light.

Note The refractive index is measured as the change of velocity of light through water and the sample, the electrical permittivity of the material and the magnetically permeability of the material define these differences If the light is propagating through a medium then the higher the refractive index of the medium lower is the velocity of light and vice versa. The most important physical significance is the fact that we can see through many transparent or partially objects like glass.