6.100 refractive index of this material

Problem 6.100:The speed of light in a certain material is 50% of its speed in vacuum. What is the refractive index of this material?

Solution:

Refractive index of a material is defined as the ratio of speed of light in vacuum to the speed of light in that material.

Or, Refractive index = $n\:=\:\frac{speed\:of\:light\:in\:vacuum(c)}{speed\:of\:light\:in\:material(v)}$.

Or $n\:=\:\frac{c}{v}$.

Given that The speed of light in the material $($v$)$  is 50% of its speed in vacuum$($c$)$.

Or, $v\:=\:\frac{50}{100}\times c$.

Or, $v\:=\:\frac{c}{2}$.

Then Refractive index,  $n\:=\:\frac{c}{\frac{c}{2}}\:=\:2$.

Therefore refractive index of the  material is $2$.

6.99 Effective focal length of thin lenses in contact

6.99 Effective focal length of thin lenses in contact

6.99: A bi-convex lens is formed with two thin plano-convex lenses as shown in the figure. Refractive index $n$ of the first lens is $1.5$ and that of the second lens is $1.2$. Both the curved surface are of the same radius of curvature $R \:=\: 14\:cm$. Find the effective focal length of the combination.

Solution:

To find the position of image, effective focal length of the lens must be calculated. 

We will assume that the lens is placed in air.

It is the combination of two plano convex lenses.

Focal length of a lens made out of a material of refractive index $n_1$ in a medium of refractive index $n_2$ is given by the equation $\frac{1}{f}\:=\:(\frac{n_2}{n_1}\:-\:1)(\frac{1}{R_1}\:-\:\frac{1}{R_2})$ where $R_1$ and $R_2$ are the radius of curvature of the two surfaces.

Problem in ray optics needs sign convention. In this problem Cartesian sign convention is followed.

For first lens,

The first surface is spherical with radius of curvature of 14 cm and the second surface is plane with infinite $(\:=\:\alpha)$ radius.

$n_1\:=\:1$, $n_2\:=\:1.5$,$R_1\:=\:+14$,$R_2\:=\:\alpha$.

Therefore

$\frac{1}{f_1}\:=\:(\frac{1.5}{1}\:-\:1)(\frac{1}{14}\:-\:\frac{1}{\alpha})$

Or, $f_1\:=\:28\:cm$.

For second lens,

The first surface is plane with infinite $(\:=\:\alpha)$ radius and the second surface is spherical with radius of curvature of 14 cm.

$n_1\:=\:1$, $n_2\:=\:1.2$,$R_1\:=\:\alpha$,  $R_2\:=\:-14$.

Therefore

$\frac{1}{f_2}\:=\:(\frac{1.2}{1}\:-\:1)(\frac{1}{\alpha}\:-\:\frac{1}{-14})$

Or, $f_1\:=\:70\:cm$.

The two lenses are in contact.

Therefore effective focal length is given by,

$\frac{1}{F}\:=\:\frac{1}{f_1}\:+\:\frac{1}{f_2}$

Or, $\frac{1}{F}\:=\:\frac{1}{28}\:+\:\frac{1}{70}$

Or effective focal length is $F\:=\:20\:cm$.

6.97 Law of reflection.

6.97 Law of reflection.

Problem 6.97:A ray of light initially travelling horizontally,  is incident on the reflecting surface of a hemispherical object and gets deflected vertically upward as shown in the diagram. Find the value of height, $H$ at which the ray hit the object.


Problem 6.97

Solution:

The problem can be solved using laws of reflection which says that the angle of incidence is equal to angle of reflection.

Angle between incident and reflected rays will be always equal to twice the angle of incidence.

Here incident ray is horizontal and reflected ray is vertical.

Hence angle between incident and reflected rays will be $90^o$.

So angle of incidence will be $\frac{90^o}{2}\:=\:45^o$.

From the second diagram, angle of incidence is $\theta$.

Problem 6.97B

Also, $sin\theta\:=\:\frac{H}{R}$

Therefore, $sin\:45^o\:=\:\frac{H}{R}$.

But $sin\:45^o\:=\:\frac{1}{\sqrt{2}}$.

Then, $\frac{H}{R}\:=\:\frac{1}{\sqrt{2}}$.

or, $H\:=\:\frac{R}{\sqrt{2}}$.

So the horizontal ray can get reflected vertically upward if it hits the hemispherical surface at a height $H\:=\:\frac{R}{\sqrt{2}}$. from the base.

6.96 Angle of incidence

6.96 Angle of incidence

Problem 6.96: A ray is incident on a transparent  slab of refractive index $\sqrt{3}$.   The reflected and the refracted rays are mutually perpendicular.  Find the  angle of incidence.

Solution:

Problem 6.96

With reference to the diagram, $AO$ is the incident ray that makes angle $i$ with the normal. $OB$ is the reflected ray which according to law of reflection, will also make the same angle $i$ with the normal.

$OC$ is the refracted ray that makes angle $r$ with the normal.

From the geometry of the diagram, we get the sum of angles $i$ , $r$ and $90^o$ must be equal to $180^o$.

That is , $i\:+\:r\:+\:90^o\:=\:180^o$.

Or, $i\:+\:r\:=\:90^o$.

Or $r\:=\:(90-i)$.

Using Snell's law for refraction at point of incidence, O, we get

$\frac{sin\:i}{sin\:r}\:=\:\frac{n_1}{n_2}$.

Substituting the values we get,

$\frac{sin\:i}{sin\:(90-i)}\:=\:\frac{n_1}{n_2}$.

Or, $\frac{sin\:i}{cos\:i}\:=\:\frac{n_1}{n_2}$.

Or, $tan\:i\:=\:\frac{n_1}{n_2}$.

Here $n_1$ is the refractive index of air which is equal to 1 and $n_2$ is the refractive of the slab which is $\sqrt{3}$.

Then $tan\:i\:=\:\frac{1}{\sqrt{3}}$.

From this angle of incidence, $i\:=\:30^o$.

The angle of refraction $r\:=\:(90-i)\:=\:(90-30)\:=\:60^o$.

6.95 Refractive index of a medium

6.95 Refractive index of a medium

Problem 6.95:Speed of light in a medium is one-third of the speed of light in vacuum. Determine the refractive index of the medium

Solution:

10-01-2016 Problem 6.95

By definition refractive index $(n)$ of a medium is the ratio of speed of light in vacuum $(c)$to speed of light in that medium $(v)$.

Therefore, refractive index $n\:=\:\frac{c}{v}$.

Given that speed of light in the medium $(v)$ = $\frac{1}{3}$ X speed of light in vacuum $(c)$

Therefore , $n\:=\:\frac{c}{v}\:=\:\frac{c}{\frac{1}{3}}\:=\:3$.

Hence refractive index of the medium$n\:=\:3$.

6.94 Angle of deviation in refraction

6.94 Angle of deviation in refraction

Problem 6.94: A plane separates a transparent media of refractive index of $\sqrt{3}$ from air. A ray of light coming from air is incident at an angle of $60^o$. Determine the angle of deviation of the ray.

Problem 6.94Solution:

The light ray gets refracted as it enters the medium from air. It deviates from the original direction and follows a new path. Angle between the initial and new path is called angle of deviation.


Problem 6.94-1

In the diagram $AO$ represents the incident ray,.

$OC$ represents the refracted ray.

$OB$ is the initial path of the ray.

$NN$ is the normal at the point of incidence.

Angle between $NN$ and $AO$ is the angle of incidence which is $60^o$.

Angle between $NN$ and $OC$ is the angle of refraction which is $r$.

Angle between $OB$ and $OC$ is the angle of deviation which is $\theta$.

From the geometry, Angle between $OB$ and $NN$ equal to angle of incidence which is $60^o$.

Also, angle of deviation $\theta\:=\:60^o\:-\:r$....$(1)$

Angle of deviation can be calculated by applying Snell’s law at the point of incidence $O$.

$\frac{sin\:i}{sin\:r}\:=\:\frac{n_2}{n_1}$

Or, $\frac{sin\:60^o}{sin\:r}\:=\:\frac{\sqrt{3}}{1}$

Or $sin\:i\:=\:\frac{1}{2}$

So angle of refraction $r\:=\:30^o$

Now from equation $(1)$, we get angle of deviation $\theta\:=\:60^o\:-\:30^o\:=\:30^o$.

So the calculation shows that the ray will suffer a deviation of $30^o$ from its initial path.