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 | Nov 16, 2015 | Miscellaneous |

**Problem 6.89: Two waves of the same frequency have amplitudes $1.00$ and $2.00$. They interfere at a point where their phase difference is $60^0$.What is the resultant amplitude?**

**Solution:**

When two waves of amplitude $A$ and $B$ interfere at a point with a phase difference of $\phi$, then the resultant amplitude is given by $R\:=\:\sqrt{A^2+B^@+2AB cos\:\phi}$.

Here $A\:=\:1.00$ and $B\:=\:2.00$ with $\phi\:=\:60^0\:=\:\frac{\pi}{3}$.

Therefore resultant amplitude, $R\:=\:\sqrt{(1.00)^2+(2.00)^@+2(1.00)(2.00) cos\:(\frac{\pi}{3})}$.

Thus, $R\:=\:\sqrt{1+4+2}\:=\:\sqrt{7}$.

by Rajan M V | Nov 6, 2015 | Miscellaneous |

**Problem 6.88:A concave mirror forms an image, on a screen $2.00 \:m$ in front of the mirror, of an object $20.0 \:cm$ in front of the mirror. What is the image height if the object height is 5.00 mm?**

**Solution:**

Here the object is placed at a distance of $20.0 \:cm$ from the mirror and the image is formed on the screen at a distance of $3.00 \:m$. Both image and object are on the same side of the mirror.

Object distance, $u\:=\:-20\:cm$ and image distance $v\:=\:-2\:m\:=\:-200\:cm$.

Lateral magnification is the ra tio of the size of the image to the size of the object.

Or, $m\:=\frac{h’}{h}$ Where $h$ is the size or height of the object and $h’$ is the height of the image.

Lateral magnification is the ratio of the image distance to the object distance.

Or, $m\:=\:\frac{-image\:distance}{object\:distance}\:=\:\frac{-v}{u}\:=\:\frac{-(-200)}{-20}\:=\:-10$.

The image will be ten times larger than the object. Negative sign indicates that the image is inverted with respect to the object.

Image size $h’\:=\:m\times h\:=\:-10\times 5\:cm\:=\:-50\:cm$.

So the mirror forms an inverted image of size of $50\:cm$.

by Rajan M V | Oct 31, 2015 | Miscellaneous |

**Problem 6.87: A cylindrical rod made of glass with refractive index $1.5$ is placed in air. It has a hemispherical end surface with radius $2.0\:cm$. A small object is placed at a distance of $6.0\:cm$ on the axis of the rod. Find the location of the image.**

**Solution:**

Here light falls from the object to the lens.

So that direction from left to right is taken as positive.

The object distance is measured from the lens to the object which is in the negative direction.

So object distance $u\:=\:-6.0\:cm$.

The center of curvature of the spherical surface is on the right side of the surface, so the radius is positive.

First medium is air. So $n_1\:=\:1$.

Second medium is glass. So $n_2\:=\:1.5$.

For refraction on a convex surface, we have $\frac{n_2}{v}\:-\:\frac{n_1}{u}\:=\:\frac{n_2\:-\:n_1}{R}$.

Substituting the given values,

$\frac{1.5}{v}\:-\:\frac{ 1}{-6.0}\:=\:\frac{1.5\:-\:1}{+2.0}$.

Or, $\frac{1.5}{v}\:=\:\frac{1}{4.0}\:-\:\frac{ 1}{6.0}\:=\:\frac{1}{12}$.

So image distance $v\:=\:18\:cm$.

The image is located in the direction of the light well within the glass rod.

by Rajan M V | Oct 28, 2015 | Miscellaneous |

**Problem 6.86: A concave mirror forms an image, on a screen $2.00 \:m$ in front of the mirror, of a lamp $20.0 \:cm$ in front of the mirror. Find the radius of curvature of the mirror?**

**Solution:**

Here the object, that is the lamp is at a distance of $20.0 \:cm$ and the image on the screen is at $3.00 \:m$. Both image and object are on the same side of the mirror.

Object distance, $u\:=\:-20\:cm$ and image distance $v\:=\:-2\:m\:=\:-200\:cm$.

Let us use mirror equation $\frac{1}{f}\:=\:\frac{1}{v}\:+\:\frac{1}{u}$.

Therefore, $\frac{1}{f}\:=\:\frac{1}{-200}\:+\:\frac{1}{-20}\:=\:\frac{-11}{200}$.

Or focal length $f\:=\:\frac{-200}{11}$.

Radius of curvature is twice focal length.

by Rajan M V | Oct 19, 2015 | Miscellaneous |

**Problem 6.82: A concave spherical mirror has radius of curvature of $20.0 \:cm$. A linear object of height $3.5 \:cm$ is placed $15.0 \:cm$ from the center of the mirror along the optic axis, as shown in the figure. Calculate the location of the image.**

**Solution:**

Focal length of a spherical mirror of radius of curvature $R$ is $f\:=\:\frac{R}{2}$.

Given that radius of curvature of = $20.0 \:cm$.

Therefore focal length $f\:=\:\frac{20}{2}\:=\:10\:cm$.

**Cartesian sign convention for spherical mirror.**

Mirror equation $\frac{1}{f}\:=\:\frac{1}{v}\:+\:\frac{1}{u}$.

In the given problem, object distance, $u\:=\:-15\:cm$.

Foal length, $f\:=\:-10\:cm$.

Therefore,

$\frac{1}{-10}\:=\:\frac{1}{v}\:+\:\frac{1}{-15}$.

Or, $\frac{1}{v}\:=\:\frac{1}{-10}\:-\:\frac{1}{-15}$.

Or $\frac{1}{v}\:=\:\frac{1}{-10}\:+\:\frac{1}{15}$.

Or $\frac{1}{v}\:=\:\frac{10-15}{150}\:=\:\frac{-5}{150}$.

Thus image distance $v\:=\:-30\:cm$.

So the image is formed at a distance of $30\:cm$ from the mirror.