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 $2$.

1.159 Work done in uniform circular motion

1.159 Work done in uniform circular motion

Problem 1.159:.A body of mass $‘m’$ is moving in a circle of radius with a constant speed $‘v’$. Calculate the work done by the resultant force in moving the body over half circumference and full circumference respectively.

Solution:

This is motion along a circular path. Resultant force on the particle produces the necessary centripetal force.

problem-1-159

Centripetal force is directed toward the center of the path and velocity is tangential to the path.

Therefore  resultant force is always perpendicular to the velocity.

This makes work done zero as there is no change in speed and kinetic energy of the body.

Then the work done by the resultant force in moving the body over half circumference and full circumference respectively are zero.

6.99 Effective focal length

6.99 Effective focal length

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.

Problem 6.99A

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,

Problem 6.99C

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

Therefore

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

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

For second lens,

Problem 6.99D

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

Therefore

$\frac{1}{f_1}\:=\:(\frac{1.2}{1}\:-\:1)(\frac{1}{}\:-\:\frac{1}{-14}$

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

The two lenses are in contact.

Problem 6.99B

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 $F\:=\:20\:cm$.

5.154 Charge of elementary particles

5.154 Charge of elementary particles

Problem 5.154:What is the charge of $(i)$ electron $(ii)$ proton and $(iii)$ neutron?

Solution:

Problem 5.154

Charge of $(i)$ electron$ = - \:1.6 \times 10^{-19}\:C$

                           $(ii)$ Proton$ = + \:1.6 \times 10^{-19}\:C$

                           $(iii)$ Neutron= zero.

 

6.98 Time taken by light

6.98 Time taken by light

Problem 6.98:A light ray is incident from air into a glass slab as shown in the diagram. Calculate the time spend by light ray inside glass. Refractive index of glass is $\sqrt{3}$.

Solution:

Problem 6.98

Here the ray is making angle of $30^o$ with the plane of the slab. Angle of incidence is always measured with respect to the normal at the point of incidence.

So the angle of incidence, $i\:=\:60^o$.

The light ray suffers refraction and bends as shown in the next diagram. The ray travels inside glass along $PR$ and then emerges out.

Problem 6.98 B

Let us apply Snell’s law at the point of incidence, $P$.

We get $\frac{sin\: i}{sin\: r}\:=\:n$.

$\frac{sin \:60}{sin \:r}\:=\:\sqrt{3}$.

From this, $sin\:r\:=\:\frac {1}{2}$ .

Therefore angle of refraction $r\:=\:30^o$.

In Triangle $PQR$, $cos\:r\:=\:\frac{PQ}{PR}$

Or, $PR\:=\:\frac{PQ}{cos\:r}$

But $PQ\:=\:5\:cm$.

Then, $PR\:=\:\frac{5}{cos\:30}\:=\::\frac{5}{\sqrt{3}/2}$.

Or, $PR\:=\:\frac{10}{\sqrt{3}}\:cm\:=\:\frac{10}{\sqrt{3}}\:\times 10^{-2}\:m$.

Speed of light in glass, $v \:=\:\frac{c}{n}$ where $c$ is the speed of light in vacuum which is $3\times 10^8\:ms^{-1}$ and $n$ is the refractive index.

Then  $v \:=\:\frac{3\times 10^8}{\sqrt{3}}\:=\:\sqrt{3}\times 10^8\:ms^{-1}$.

Time  spent by light in glass slab, $t\:=\:\frac{PR}{v}\:=\:\frac{\frac{10}{\sqrt{3}}\:\times 10^{-2}}{\sqrt{3}\times 10^8}$,.

This will be approximately $3.33\times 10^{-10}$ second.