Q: Differentiate between regular and diffused reflection. Does diffused reflection mean the failure of the laws of reflection ?

\textbf{Explanation :} \textbf{Regular reflection:} When the reflected rays are parallel after reflection, it is called regular reflection. It takes place in a smooth surface. \textbf{Diffused reflection:} When the reflected rays are not parallel after reflection, it is called irregular or diffused reflection. It takes place through irregular surfaces. Diffused reflection is not the failure of laws of reflection, it takes place because of irregularities in the reflecting surface.

Q: A ray of light falls on the surface of a spherical glass paperweight making an angle a with the normal and is refracted in the medium at angle b . The angle of deviation of the emergent ray from the direction of incident ray is :

From the above figure we can see that \theta =2(a-b) Exterior angle=sum of interior opposite angles. So answer is option B.

Question 1

The image of our face in a plane mirror is

Accepted Answer

the image formed by a plane mirror is a virtual , laterally inverted image at same distance from the mirror that of object and of same size of object.

Question 2

Differentiate between regular and diffused reflection. Does diffused reflection mean the failure of the laws of reflection ?

Accepted Answer

extbf{Explanation :} 	extbf{Regular&#10;reflection:}  When the reflected rays are parallel after&#10;reflection, it is called regular reflection. It takes place in a smooth&#10;surface. 	extbf{Diffused&#10;reflection:}  When the reflected rays are not parallel after&#10;reflection, it is called irregular or diffused reflection. It takes place&#10;through irregular surfaces.&#10;&#10;&#10;&#10;&#10;&#10;&#10;&#10;Diffused reflection is not the failure of laws of&#10;reflection, it takes place because of irregularities in the reflecting surface.

Question 3

Which are the two convenient rays that are chosen to construct the image by a spherical mirror for a given object ? Explain with the help of suitable ray diagrams.

Accepted Answer

Two convenient rays chosen to construct the image by a spherical mirror are : (i) A ray parallel to the principal axis which after reflection passes or appears to pass through focus.(ii) A ray passing through the center of curvature or appears to pass through center of curvature which retraces its path after reflection.

Question 4

A ray of light strikes a flat block of glass   (n=1.50)   of thickness   2.00 \mathrm{cm}   at an angle of   30.0^{\circ}   with the normal. Trace the light beam through the glass and find the angles of incidence and refraction at each surface.

Accepted Answer

At entry, the wave under refraction model, expressed as   n_{1} \sin 	heta_{1}=n_{2} \sin 	heta_{2},   gives 	heta_{2}=\sin ^{-1}\left(\dfrac{n_{1} \sin 	heta_{1}}{n_{2}}ight)=\sin ^{-1}\left(\dfrac{1.000 \sin 30.0^{\circ}}{1.50}ight)=19.5^{\circ} To do ray optics, you must remember some geometry. The surfaces of entry and exit are parallel so their normals are parallel. Then angle   	heta_{2}   of refraction at entry and the angle   	heta_{3}   of incidence at exit are alternate interior angles formed by the ray as a transversal cutting parallel lines. Therefore,   	heta_{3}=	heta_{2}=\left[19.5^{\circ}ight]  At the exit point,   n_{2} \sin 	heta_{3}=n_{1} \sin 	heta_{4}   gives 	heta_{4}=\sin ^{-1}\left(\dfrac{n_{2} \sin 	heta_{3}}{n_{1}}ight)=\sin ^{-1}\left(\dfrac{1.50 \sin 19.5^{\circ}}{1.000}ight)=30.0^{\circ} Because   	heta_{1}   and   	heta_{4}   are equal, the departing ray in air is parallel to the original ray.

Question 5

Question 6

The relation between  u,\,v\;and\;f  for a mirror is given by

Accepted Answer

The relation between  u,\,v\;and\;f  for a mirror is given by  f=\displaystyle\frac{u	imes v}{u+v}

Question 7

A ray of light traveling in the air is incident on the plane of a transparent medium. The angle of the incident is 45 ^o  and that of refraction is 30 ^o . Find the refractive index of the medium.

Accepted Answer

Snell's Law of refraction gives the relation between the incident and refracted light at the interface of two media, which is given by  \displaystyle \frac{\mu_2}{\mu_1} = \frac{\sin i}{\sin r} The initial medium is air. Thus  \mu_1 = 1 Given incidence angle  i = 45^o  and refraction angle is  30^o By Snell's law,  \mu_2 = \mu_1	imes \displaystyle \frac{\sin i}{\sin r} = 1	imes \frac{\sin 45^o}{\sin 30^o} = \frac{1/\sqrt{2}}{1/2} = \sqrt{2}

Question 8

A ray of light falls on the surface of a spherical glass paperweight making an angle  a  with the normal and is refracted in the medium at angle  b . The angle of deviation of the emergent ray from the direction of incident ray is :

Accepted Answer

From the above figure we can see that  	heta =2(a-b) Exterior angle=sum of interior opposite angles. So answer is option B.

Question 9

a. Draw the given diagram and label the following in the diagram.
i) Incident ray
ii) Refracted ray
iii) Emergent ray
iv) Angle of refraction
v) Angle of deviation
vi) Angle of emergence
b. The refractive index of diamond is 2.42. What is the meaning of this statement in relation to the speed of light?

Accepted Answer

Given: Refractive index of diamond is  2.42 Solution:We know ,  \mu = \dfrac{c}{v} Hence, 	ext{Refractive index of diamond} = \dfrac {	ext{Speed of light in air}}{	ext{Speed of light in diamond}}\\implies 	ext{Speed of light in air}=2.42	imes 	ext{Speed of light in diamond} This means the speed of light in air is  2.42  times the speed of light in diamond.

Question 10

In which of the following instruments is final image erect?

Accepted Answer

A simple microscope is magnifying glass, an ordinary double convex lens having a short focal length that produces virtual and erect image.rather a Compound microscope and astronomical telescope produces virtual and final inverted image. hence correct option is A.

Ray Optics And Optical Instruments