by Rajan M V | Nov 23, 2016 | Mechanics |

**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.**

**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.**

by Rajan M V | Apr 17, 2016 | Optics |

**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,

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

Therefore

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

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

For second lens,

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

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.

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

by Rajan M V | Mar 28, 2016 | Electromagnetism |

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

**Solution:**

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

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

$(iii)$ Neutron= zero.

by Rajan M V | Feb 3, 2016 | Miscellaneous |

**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:**

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.

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.

by Rajan M V | Jan 31, 2016 | Optics |

**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.**

**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 $\theta$.

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

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

But $tan\:45^o\:=\:1$.

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

or, $H\:=\:R$.

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