# 2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments

Students can Download 2nd PUC Physics Chapter 9 Ray Optics and Optical Instruments Questions and Answers, Notes Pdf, 2nd PUC Physics Question Bank with Answers helps you to revise the complete Karnataka State Board Syllabus and to clear all their doubts, score well in final exams.

## Karnataka 2nd PUC Physics Question Bank Chapter 9 Ray Optics and Optical Instruments

### 2nd PUC Physics Ray Optics and Optical Instruments One Marks Questions and Answers

Question 1.
Which, among frequency, wave length and velocity of light, will remain constant as light enters from air into glass?
Frequency. This is because f = $$\frac{c}{\lambda}=\frac{v}{\lambda^{\prime}}$$ As velocity of light decreases in a denser medium, so does the wavelength. The ratio still remains a constant.

Question 2.
As light is transmitted through a parallel sided glass slab, what relation holds good between the angle of incidence and angle of emergence?
The angle of incidence is equal to the angle of emergence.

Question 3.
How does lateral shift depend on thickness of the medium?
Lateral shift in a parallel sided glass slab increases with increase of thickness of the medium. LS α t

Question 4.
What is lateral refraction of light?
As light undergoes refraction through parallel sided glass slab the emergent ray remains parallel and shifts sidewards. This phenomenon is called as lateral refraction of light.

Question 5.
Can lateral shift be zero?
Yes, for normal incidence the angle of incidence ‘V and angle of refraction ‘r’ both will be zero.

Question 6.
When will lateral shift be maximum or equal to the thickness of the medium?
When the angle of incidence i = 90° (grazing incidence), the emergent ray grazes along the opposite parallel face and the lateral shift becomes maximum.

Question 7.
What is lateral shift?
It is the perpendicular distance between the incident ray and the emergent ray as light undergoes refraction through a parallel sided glass slab.

Question 8.
What is normal refraction of light?
When an object in the denser medium is viewed from a rarer medium, it appears to be raised along the normal towards the surface. This phenomenon is known as the normal refraction.

Question 9.
What is normal shift?
The difference between the real depth and apparent depth of an object along the normal field of view is known as the normal shift.

Question 10.
Does normal shift produced by a medium depend on the position of an object below the surface?
No. It depends on the thickness and R.I of the medium.

Question 11.
Define critical angle for a pair of media.
It may be defined as the angle of incidence in a denser medium for which the refracted ray grazes the interface and is a constant for a pair of media and the given wavelength of light.

Question 12.
What is light?
Light is that form of energy, belonging to the spectrum of electromagnetic waves, which produces visibility of objects around us after interaction with the retinal cells of the eyes.

Question 13.
What does optics deal with?
Optics is a branch of physics that deals with the nature of light, sources and properties of light and construction of optical instruments based on the effect of light through optical media.

Question 14.
What is an optical medium?
A material medium which is transparent to light is known as the optical medium.
e.g., glass, ice crystal, quartz, calcite, inorganic colourless liquids, organic liquids, air, free space.

Question 15.
Define a ray of light?
It is a geometrical representation of direction of propagation of light. A straight line with an arrow head indicates a ray of light pointing the direction of flow of luminous (light) energy.

Question 16.
Is glass an isotropic medium?
Yes, it is.

Question 17.
What is an isotropic medium?
An optical medium in which the speed of light remains the same irrespective of the direction of incident beam of light on it, is known as an isotropic medium.

Question 18.
Is quartz crystal an isotropic medium?
No. Quartz crystal is an anisotropic medium.

Question 19.
Define anisotropic medium.
An optical medium in which certain optical properties such a speed of light and refractive index depend on the direction of the incident beam of light on it, is known as an anisotropic medium.

Question 20.
A ray of light in a medium (1) bends towards the normal as it passes through a medium (2). Is medium (1) a rarer or denser than (2)?
Medium (1) is rarer than medium (2).

Question 21.
A ray of light in a medium (1) bends away from the normal as it passes out through a medium (2). Which of the two is a denser medium?
Medium (1) is denser than medium (2).

Question 22.
A ray of light in the denser medium grazes the interface after refraction through it. What is the angle of incidence in the denser medium called?
Critical angle.

Question 23.
Tiie velocity of light decreases while traveling from medium (1) into medium (2). Which of the two is a rarer medium?
Medium (1).

Question 24.
Define refractive index between a pair of media in terms of velocities of light in them.
Refractive index of medium (2) w.r.t. medium (1) may be defined as the ratio of the velocity of light in medium (1) to the velocity of light in medium (2).

Question 25.
What is real depth?
The actual distance from an object below the refracting surface to the surface, without considering the effect of refraction of the object is known as real depth.

Question 26.
What is apparent depth?
The distance from the surface to the apparent position of the object along the normal view, is known as apparent depth.

Question 27.
An object in air is viewed through a denser medium. Does the object appear closer to the surface?
No. The apparent position of the object will be farther away from the surface.

Question 28.
Why is that the outside world appears limited for any marine creatures?
This is because of the phenomenon of total reflection of light. The outside world will be limited to an angular separation of twice the critical angle for a given pair of media and wavelength of light.

Question 29.
A ray of light grazes the top surface of a glass slab. Can it be traced along the opposite bottom surface of the glass slab?
Yes, it can be traced along the bottom surface because the lateral shift produced will be equal to the thickness of the slab.

Question 30.
What will be the lateral shift for a normal incidence of light on a glass slab?
Zero. Because as i = 0, r = 0, sin (i – r) = 0.
Hence lateral shift = 0

Question 31.
Explain why there is an early sunrise or a late sunset.
It is due to the atmospheric refraction of light. Light from the sun bends towards the normal at the point of incidence on the free space – air interface. When the refracted beam is projected backwards, it gives the position of the sun w.r.t an observer.

Question 32.
Will there be an early sunrise or sunset on the surface of the moon?
No. There is no atmosphere around the moon to cause refraction of light.

Question 33.
Out of two cables, one of copper and the other of optical fiber of same thickness, one is to be selected for transmission, of larger bandwidth of signals. Which one of the two do you prefer?
Optical fiber. Because for the given thickness of the two, optical fiber cable can handle more signals than copper cables.

Question 34.
What is a prism?
A prism is an optical medium, bounded by three rectangular faces and two parallel triangular faces.

Question 35.
What is meant by base of a prism?
The non refracting face of a prism is known as the base of the prism.

Question 36.
Under symmetric refraction of light through a prism, how does the refracted ray appear with respect to the base of the prism?
The refracted ray will be parallel to the base of the prism.

Question 37.
Define angle of minimum deviation.
Angle of minimum deviation is the minimum angle at which the angle of incidence will be equal to the angle of emergence and the refracted ray will be parallel to the base of the prism.

Question 38.
How is angle of incidence related to the angle of minimum deviation?
Angle of incidence i = $$\frac{A+D}{2}$$ where
‘A’ is angle of the prism.
‘D’ is angle of minimum deviation.

Question 39.
What is dispersion of light?
The phenomenon of splitting up of composite white light into its constituent colours, as the white light passes through a dispersive medium such as a refracting prism, is known as dispersion of light.

Question 40.
Define the term angle of deviation.
It is the angle between the emergent ray and the incident ray with regard to refraction of light through an optical medium.

Question 41.
Given the limiting angle of a prism in which grazing incidence results in grazing emergence of a ray of light then how are ‘A’ and ‘C’ related?

A = 2C

Question 42.
What is a spectrum?
The seven constituent colours of white light deviated by a dispersive medium are known as spectrum. A spectrum, formed due to the dispersion of light by water particles in the atmosphere, is known as a rainbow.

Question 43.
What is the function of the achromatic lens used in the spectrometer?
Achromatic lens removes chromatic or colour defect. All the colours will be brought to the same focal plane after refraction through the achromatic lens.

Question 44.
What is a thin prism?
A thin prism is that prism whose angle of refraction is less than 10°.

Question 45.
In what way a refracting prism is different from a reflecting prism?
A refracting prism has two refracting rectangular faces adjacent to each other and the rest are non-refracting, whereas a reflecting prism has all the sides capable of refraction.

Question 46.
What kind of prisms are used in the binoculars for turning light at right angles?
Total reflecting prisms are used to turn light at right angles.

Question 47.
Is dispersion in the prism due to a property of light?
Yes it is. This is because of velocity of different colours are different and they travel in different paths in a dispersive medium such as a refracting prism.

Question 48.
Can two prisms be arranged to get dispersion without deviation?
Yes: With two prisms of the same kind with their refracting angles facing opposite to each other.

Question 49.
Why refraction of light through lenses causes convergence or divergence?
This is because a lens can be assumed to consist of a large number of prisms. Depending upon the position of the base of such prisms, rays of light either converge or diverge.

Question 50.
Why do we see violet light ray bending the most?
The velocity of violet light is the least among the constituent colours. Therefore its refractive index is maximum and the angle of deviation is maximum.

Question 51.
Name the type of prisms that can be used in direct vision spectroscopes.
Crown and flint glass prisms.

Question 52.
Out of crown and flint glass prisms which has the larger refrangibility?
Flint glass prism.

Question 53.
For what purpose is a refracting prism used along with a spectrometer?
A refracting prism is used to measure the refractive indices of lights of different colours and hence their wavelengths.

Question 54.
What is the deviation produced by a thin prism of angle 8° and of R.I. 1.5?
d = (n – 1)A = (1.5 – 1)8° = 4°

Question 55.
If the expression n = $$\frac{\sin \left(\frac{A+D}{2}\right)}{\sin (A / 2)}$$ true for all the position of the prism?
No. This expression is true for symmetric refraction only.

Question 56.
What is a spherical surface?
A surface that forms a part of a sphere is called a spherical surface.

Question 57.
What is an aperture?
The area of a spherical surface available for refraction of light is called an aperture.

Question 58.
Define pole of spherical refracting surface.
The centre of aperture of a spherical surface is called the pole

Question 59.
Define object space.
The refracting medium that contains the incident ray is called the object space.

Question 60.
Define image space.
The refracting medium that contains the refracted ray is called the image space.

Question 61.
Define radios of curvature of a spherical surface.
The radius of the sphere of which the given spherical surface forms a part, is called the radius of curvature.

Question 62.
In which of the following spherical surfaces, the radius of curvature is taken as positive.

In the case of fig (i), because the parallel beam of light bends towards the principal axis.

Question 63.
In which of the following spherical surfaces the radius of curvature is taken as negative.

In the fig (ii), because parallel beam of light bends away from the principal axis.

Question 64.
On a spherical surface, a line is drawn normal to it. Where will it meet when extended to the principal axis?
It will meet the principal axis at the centre of curvature of the spherical surface.

Question 65.
Define power of a refracting surface. (Define the power of a lens).
Refracting power of a refracting surface or lens may be defined as its ability to bend the incident beam of light towards or away from the principal axis.

Question 66.
What is the unit of measurement of refracting power of a spherical surface?
The unit of measurement of power of a refracting spherical surface is dioptre (D).

Question 67.
A parallel beam of li^tit after refraction through a lens, converges toward its principal focus. What kind of lens is it?
Convex lens, (also called converging lens).

Question 68.
A parallel beam of light after refraction appears to diverge from the principal focus. What kind of lens is it?
Concave lens, (also called diverging lens)

Question 69.
What is the refracting power of a plane refracting surface?
Refracting power is zero because, the radius of curvature of a plane surface is infinity and hence its focal length is also infinity.

Question 70.
How is the power of a lens measured?
The power of a lens is measured as the reciprocal of the focal length of a lens.

Question 71.
Mention the unit of the power of a lens.
The SI unit of the power of a lens is dioptre.

Question 72.
Can power of a lens be negative?
Yes, it can be. The refracting powers of a piano concave and biconcave lenses are negative.

Question 73.
What kind of lens is an air bubble in water?
Concave lens. This is because both the convex surfaces face towards denser (water) medium and air inside the bubble is rarer.

Question 74.
A liquid of higher refractive index forms a bubble inside water. What kind of lens does it act like?
It acts as a convex lens. This is because, the convex surfaces face the rarer medium.

Question 75.
What is meant by linear magnification?
The ratio of the image height, to the object height is known as linear magnification.

Question 76.
A piano concave lens is silvered at the plane surface. How does it behave?
It behaves like a convex mirror.

Question 77.
The optical medium of higher refractive index is bonded byan optical medium of lower refractive index. This process of bonding is known as cladding.

Question 78.
Does normal shift depend on the position of the object below the denser medium?
Normal shift does not depend on the position of the object below the glass slab.

### 2nd PUC Physics Ray Optics and Optical Instruments Two Marks Questions and Answers

Question 1.
State the laws of reflection of light.
I Law: The incident ray, the reflected ray and the normal drawn at the point of incidence all lie in the same plane.
II Law: The angle of incidence is equal to the angle of reflection.

Question 2.
Define the terms
(a) pole and
(b) centre of curvature of a spherical mirror.
(a) The centre of the reflecting surface of the aperture of the mirror is called the pole of the spherical mirror.
(b) The centre of curvature of a mirror is the centre of the spherical mirror of which the mirror (concave or convex) is a part of it.

Question 3.
Define the terms
(b) principal axis of a mirror.
(a) The distance between the pole and the centre of the mirror is called the radius of curvature of the concave or convex mirror.
(b) The line joining the pole of the mirror and the centre of curvature and extending on either side of the mirror, is called the principal axis of the concave/convex mirror.

Question 4.
Give the Cartesian sign convention for measuring distances in spherical mirrors and lenses.
1. The distances measured in the same direction as the incident light are taken as positive and those measured in the direction opposite to the direction of the incident light are taken as negative.

2. The heights measured upwards with respect to the X -axis and normal to the principal axis of the mirror/lens are taken as positive and heights measured downwards below the axis are taken as negative.

Question 5.
Identify the terms
(a) paraxial rays and
(b) marginal rays of light.
(a) The rays of light close to the pole and which make small angles with the principal axis, are called paraxial rays.
(b) The rays of light farther away from the principal axis and falling close to the end surface of mirrors/lenses are called marginal rays.

Question 6.
Define the terms
(a) principal focus F and
(b) focal plane of the mirror.
(a) The principal focus is a point on the principal axis of a mirror or lens at which rays of light after reflection or refraction either converge or appear to diverge from that point.
(b) A plane drawn normal to the principal axis and containing the principal focus of the mirror or lens is called the focal plane.

Question 7.
Draw ray diagrams to show rays of light converging or appear diverging from a point due to reflection in a spherical mirror.

Question 8.
Define focal length of a mirror and hence relate focal length and radius of curvature of a mirror.
(a) The distance between the pole and the principal focus of the mirror is called focal length (f).
(b) f = $$\frac{\mathrm{R}}{2}$$ where R – radius of curvature of a spherical mirror.

Question 9.
Thick lenses could be assumed as made of many prisms. Each of these prisms disperse composite white light into its constitutent colours. Therefore thick lenses produce chromatic aberration.

Question 10.
Explain why convex lenses converge incident beam of light whereas concave lenses diverge light.
Lenses can be assumed to be made up of a large number of prisms. Light after refraction has a tendency to move towards the base having a larger area. Hence light beam converges inside a convex lens and diverges inside a concave lens.

Question 11.
Draw a neat labelled diagram to show (i) primary rainbow and (ii) secondary rainbow.

Question 12.
Explain why the colour of the sky is blue (Cyan).
The particle size of the atmosphere is smaller than the wavelength of light. Rayleigh type (coherent type) of scattering of light takes place. According to Rayleigh’s law, the intensity of light scattered is directly proportional to the intensity of incident light and square of the size of the particle and is inversely proportional to the fourth power of wavelength. Among the lower wavelengths, intensity of blue light is stronger and more sensitive to the eye than the violet. Hence blue light is scattered and the sky appears blue.
I ∝ $$\frac{I_{0} V^{2}}{\lambda}$$
where, V = volume of the particle,
X – wavelength the of light scattered

Question 13.
Define angular magnification.
The ratio of the angle subtended at the eye by an image to the angle subtended at the eye by the object is known as the angular magnification.

Question 14.
Draw a neat labelled diagram of Cassegrain reflecting telescope

Note : India’s largest reflecting telescope used at Kavalur in Tamil Nadu has a concave mirror, 2.34 m in diameter and the one at Hawaii in USA has a concave mirror 10 m in diameter.

Question 15.
Can plane and convex mirrors produce real images? Give an explanation to your answer.
The rays of light converging to a point behind the plane mirror or convex mirror are reflected to a point in front of the mirror on a screen. In other words, a plane or convex mirror can produce a real image if the object is virtual.

Question 16.
A virtual image though cannot be caught on the screen, is actually caught on the retina of the eye. Explain.
When the reflected or refracted rays are divergent, the image is virtual. The convex lens converges this divergent rays. Here the virtual image serves as an object for the eye lens to produce a real image.

Question 17.
Write the formula for the lateral shift and explain the symbols used.
Lateral shift, LS = $$\frac{t}{\cos r}$$ sin(i – r)
where, ‘t’ – thickness of the medium,
i – angle of incidence,
LS – lateral shift.

Question 18.
Write the formula for normal shift and explain the symbols used.

where, NS = normal shift,
t = thickness of a medium,
n2 = R.I. of medium (2) w.r.t medium (1)

Question 19.
Give the reason for early sunrise and late sunset.
During sunrise and sunset, the sun will be at the horizon. The light from the sun travelling from vacuum into air undergoes refraction. Hence the sun appears to rise earlier and lingers a little longer after the sunset.

Question 20.
Give any two consequences of refraction of light.

1. The phenomena of lateral shift and normal shift are due to refraction of light.
2. The early sunrise and late sunset are due to atmospheric refraction of light.

Question 21.
Write any two practical applications of optical fibers.
Optical fibers are used in the field of medical endoscopy and stellar spectroscopy.

Question 22.
What is an optical fibre? Name the principle on which it works.
Optical fibre is a transparent medium used to conduct light along selected paths. It works on the principle of total internal reflection.

Question 23.
Distinguish between a pure and an impure spectrum.
A pure spectrum is that spectrum in which there is no overlapping of colours. Different colours fall on the screen at different points distinctly.

An impure spectrum is that spectrum in which there is an overlapping of colours. There will be intermixing of colours so that different colours do not fall on the screen in a regular sequence.

Question 24.
Define Power of a lens.
The power of a lens is defined as the tangent of the angle by which it converges or diverges a beam of light falling at one metre distant from the optical centre.

Question 25.
Draw a neat diagram to show dispersion of light in a prism.
The different constituent colours obtained on the screen are known as the spectrum.

Question 26.
Why vaccum is a non-dispersive medium?
In Vacuum, the speed of light is independent of wavelength. The refractive index of the medium for any colours is unity. Therefore vacuum is a non dispersive medium.

Question 27.
Can dispersive power of a prism be negative? Give reason for your answer.
No, it is always positive because R.I. for violet colour is always greater than the R.I. for red colour.

Question 28.
Is the dispersive power independent on the angle of the prism? Give reason.
Yes, it is. The dispersive power of the material of a prism depends only on the nature of the material

Question 29.
Draw a graph of angle of deviation versus the angle of incidence in a refracting prism.

For d = D called the angle of minimum deviation, the refraction in the prism becomes symmetric. For a thin prism, deviation produced by it, does not depend on the angle of incidence. Deviation is a constant quantity for a given prism.

Question 30.
Define dispersive power of the material of a prism. How can it be measured?
The dispersive power of the material is its ability to disperse the constituent colours of incident light. The dispersive power of the material of the prism can be measured by the ratio of angular dispersion between any two colours to the mean of the angle of deviation of the two colours. Dispersive power has no unit. It is a dimensionless quantity.

Question 31.
Draw a neat labeled diagram of a prism.

AD – Refracting edge of a prism
∠A – angle c f the prism
BCFE base of the prism
ACFD and ABED are refracting rectangular faces.

Question 32.
A ray of light incident at 60° on a prism undergoes a deviation of 20°. If the angle of the prism is 40, S.T. the emergent ray is normal to the second face of the
prism.
d = (i1 + i2) -A
20° = 60° + i2 – 40°
∴ i2 = 0.
Hence, emergent ray is normal to the face of the prism.

Question 33.
Draw a neat labeled diagram to obtain an inverted image in the case of a total reflecting prism.

Question 34.
Draw a neat labeled diagram to show the turning of light rays through 90° in a total reflecting prism.

Question 35.
Draw a neat labeled diagram to show the reflected rays turned through 180°.

Question 36.
Write the expression for R.I. of the material of the prism for symmetric refraction.

where, ‘A’ -angle of the prism
‘D’ – angle of minimum deviation
1n2 – R.I. of medium (2) w.r.t medium (1) .

Question 37.
Write the expression for the power of a lens in terms of powers of its refracting surfaces.

Question 38.
Write the formula for the equivalent focal length of two thin lenses separated by a small distance and explain the symbols used.

where, f1,f2 are focal lengths of two lenses in contact, f is equivalent focal length and d – distance of separation between the two lenses.

Question 39.
How can the power of the combination expressed, if two lenses are equibiconcave and separated by a distance?

Question 40.
Two lenses having focal lengths + 0.20 m and 0.30 m are separated by a distance of 0.15 m. Find the resultant power of the combination.

P = 5 – 3.33 + 2.5
P = 4.17D

Question 41.
An extended object is placed at the principal focus of a lens. Where will the final image be formed? Comment on the nature of the image.

∴ $$\frac{1}{v}$$ = 0 ⇒ v = ∞
The image is formed at infinity and the image will be real and inverted.

Question 42.
State Snell’s law of refraction of light?
When refraction of light takes place, the ratio of sine of the angle of incidence to sine of the angle of refraction is a constant for a given pair of media and wavelength of light.

Question 43.
Write Snell’s law in general mathematical terms.
The general form of Snell’s law may be written as n1 sin i1 = n2 sin i2 where i1 – angle of incidence in a medium (1) and i2 – angle of refraction in a medium (2) for a pair of media (1) and (2), n1 – refractive index of medium (1) and n2 – refractive index of medium (2).

Question 44.
Define angle of deviation.
The difference between the angle of incidence in a medium (1) and the angle of refraction in a medium (2) is known as the angle of deviation. If i > r then d = i – r.
If i < r then d = r – i.
In general, angle of deviation = d = i ~ r

Question 45.
A riding glass though formed by spherical surfaces has zero power. Give reason.
The radii of curvature of convex and concave surfaces are equal but one is positive and the other negative. These cancel with each other. Hence its power becomes zero.

### 2nd PUC Physics Ray Optics and Optical Instruments Three Marks Questions and Answers

Question 1.
Explain the phenomenon of mirage.
During a hot day, air near the earth’s surface becomes hotter and rarer and whereas air slightly above the earth’s surface will be relatively colder and denser. The light from the object on land suffers total internal reflection as it travels from a denser into a rarer media.

When the reflected rays are drawn backwards, we notice that the rays appear to diverge from the image below the object.

The optical pseudo image formation due to total internal reflection is called mirage. The images near the hot surface of the land, chimney where water like ripples are formed are called inferior images.

Question 2.
Name any three devices which work on the principle of total internal reflection. (March 2014)
The devices which work on the principle of total internal reflection are

1. total reflecting prisms (used in periscopes, binoculars, and optical lanterns),
2. optical fibres (used in optical fibre communication),
3. stellar spectroscopes.

Note: The principle of total internal reflection is used in the designing of diamond ornaments.

Question 3.
Explain myopia or short-sightedness with a neat labelled diagram.
If the light from a distant object gets focussed at a point in front of the retina, then the defect of the eye is known as short or near sightedness. This defect or myopic eye can be corrected by having a biconcave lens in front of the eye.
In the case of myopic eye, objects at short distances only can be seen.

Question 4.
Explain hypermetropia or long sightedness with a neat labelled diagram.
If the eye lens focuses the incoming light at a point behind the retina then that defect is known as hypermetropia. Hypermetropic eye is able to see far off objects clearly but not nearby objects. This defect can be overcome by having a biconvex lens in front of the eye.

Question 5.
What is a simple microscope? Obtain an expression for magnification obtained in a simple microscope for final image at near point and at infinity. (March 2015)
A simple microscope is a magnifier of a small object placed or held on one side of it and viewed from the other side.
An erect and enlarged image of the object is obtained by placing the object in between the optic centre and the focus of the convex lens.

We know that $$\frac{1}{f}=\frac{1}{v}-\frac{1}{u}$$. Magnification m = $$m=\frac{v}{u}=v\left(\frac{1}{v}-\frac{1}{u}\right)=1-\frac{v}{f}$$
For image at a nearby point (at the least distance of vision), v = -D . The magnification m = $$\left(1+\frac{D}{f}\right)$$
For image at infinity, (object placed at F), m = $$\frac{D}{f}$$

Question 6.
What advantages has optical fibre communication over cable communication?
The advantages of using optical fibres in communications are:-
(a) large bandwidth of operation,
(b) high efficiency, no noise or interference of signals,
(c) high speed operation, and
(d) multi signal transmission at reduced cost.

Question 7.
Draw a neat labelled diagram of a compound microscope and give the expression for its overall magnification.

The image due to the object piece will be formed at or in front of focus of the eye piece. The distance between the second focal length of the objective and the first focal length of the eye piece is called the tube length (L) of the compound microscope.

Question 8.
Give any three differences between a compound microscope and a telescope.
The focal length and the size of the objective will be less than the eye piece in the case of a compound microscope whereas in the case of a telescope, the focal length of the objective and its size will be larger than in the microscope.

The magnification in the case of a compound microscope is m = $$\frac{\mathrm{L}}{-f_{0}}\left(\frac{\mathrm{D}}{f_{e}}\right)$$, whereas in the case ofa telescope m = $$\frac{f_{0}}{f_{e}}$$ for image at infinity and ‘L’ is the length ofthe tube of the microscope.
The length of the telescope = L = f0 + fe and the length of compound microscope is v0 + fe where v0 is the image distahce obtained in the case of refraction through the object lens.

Question 9.
Draw a neat labelled diagram of image formed in a refracting telescope. Give the expression for magnification of an object, for an image formed at infinity.

Here, L = f0 + fe
tan α = $$-\frac{h}{f_{o}}$$ = α for small angles.
tan β = $$\frac{h}{f_{e}}$$ = β for small angles.
and m = $$\frac{\beta}{\alpha}=-\frac{f_{0}}{f_{e}}$$. The – ve sign indicates that the final image is inverted.
Note : The largest lens objective in use has a diameter of 40 inches, which is at the Yerkes observatory in Wisconsin, USA.

Question 10.
Show that n = $$\frac{1}{\sin C}$$ where symbols have their usual notation.
Applying Snell’s Law
n1 Sin i1 = n2 Sin i2
n sin c = 1. sin 90° 1
∴ n = $$\frac{1}{\sin C}$$

Question 11.
Mention the three factors affecting lateral shift.
Factors affecting lateral shift are
(a) thickness of the medium,
(b) refractive index of the object space & surrounding medium,
(c) angle of incidence and
(d) wavelength of incident light.

Question 12.
State the conditions for dispersion without deviation.
If the angles of two prisms are so adjusted that the deviation produced for the mean ray by the first prism is equal and opposite to that produced by the second prism, then the emergent mean ray will be parallel to the incident ray. In this case the combination of such prisms produces dispersion without deviation. The net deviation of the mean ray from the two prisms is zero. The refracting edges of two prisms will be placed opposite to each other. The two prisms are made up of different materials and have different reflacting angles.

Question 13.
Show that m = $$\frac{f-v}{f}$$ for a lens

Question 14.
Show that m = $$\frac{f}{u+f}$$ for a lens.

Question 15.
On what factors does the focal length of a lens depend?
The focal length of a lens depends on,
(a) refractive index of the given medium,
(b) refractive index of the surrounding medium,
(c) radii of curvature of the two surfaces of the lens, and
(d) wavelength of light used.

### 2nd PUC Physics Ray Optics and Optical Instruments Five Marks Questions and Answers

Question 1.
S.T. f = $$\frac{\mathbf{R}}{2}$$ in the case of a spherical mirror where symbols have their usual notations.

Question 2.
Draw a ray diagram to obtain the virtual image formation in (i) a concave mirror and (ii) a convex mirror.

Question 3.
Derive mirror equation

MPN is a concave mirror of radius of curvature 2f and focal length f. AB represents an extended object and A’B’ its image. For objects beyond C, the image is real and inverted and is formed in b/w F and C. The image will be diminished. Let u be the object distance from P and v be the image distance from P.
Sign convention:

• All distances are measured from the pole of the mirror.
• Distances measured from P and to the opposite direction of incident light are taken as negative.

Let ‘M’ be very close to P. Then the arc length MP ≈ perpendicular length PM.
Consider right angled triangles A’B’F and MPF

Question 4.
Explain the phenomenon of total internal reflection.
Consider a point object in the denser medium. There will be cone of rays striking the interface. For a ray of light OA, it suffers normal refraction. For any angle of incidence less than critical angle such as ray OB, the refracted ray bends away from the normal. However for a particular angle of incidence called critical angle, the refracted ray of light grazes the interface. Therefore critical angle for the given pair of media and given wavelength of light may be defined as the angle of incidence in the denser medium for which the refracted ray is at right angles to the normal drawn at the point of incidence.

For a ray of light such as OD, whose angle of incidence is greater than the critical angle, there will be no refraction of light, instead the light bounces back into the same denser medium. This phenomenon of light is known as the total internal reflection.
Conditions for the total internal reflection.
(a) Object should be in the denser medium and viewed through rarer medium.
(b) Angle of incidence in the denser medium should be more than the critical angle for the given pair of media and wavelength of light.

Question 5.
Write a short note on optical fibers.

Optical fiber works on the principle of TIR. It consists of a central core of transparent medium of higher refractive index surrounded by another transparent medium of lower refractive index. This is called cladding. At the outer surface, silicon coating is used as sheath. To prevent the fiber from being contaminated with water, a fine layer of silicon gel can be applied on the sheath. This prevents the optical fiber system from breaking due to mechanical stress. A single fiber is of the thickness of hair. As a single fiber may not be able to handle more light, several optical fibers are bundled to form a cable.

From the fig., we see that the incident light undergoes TIR several times before emerging at the end. A thin optical cable can handle more information and also can transmit more efficiently than the conventional electric cables.

•  In stellarscopy, optical fibers are used as a funnel in order to receive more information over a wider area.
• As endoscopes, optical fibers are used to scan the internal organs of humans for more information. The diseased portion of the internal organs can be scanned and the same can be displayed on the monitor for diagnosis and suitable treatment.

Optical fibers are used in the field of telecommunication. They can handle more information, higher frequency bandwidth with less noise and no interference by an external signal. The signal transmission is faster than with the conventional electrical cables. Unlike the present telecommunication system, less repeaters are required in using optical fibers in the communication system.

Question 6.
Derive the refraction formula (for object in air and ¡mage in the denser medium) for refraction of light at a spherical surface.

Question 7.
Derive lens maker’s formula.
Lens maker’s formula

Let r1 and r2 be the radii of curvature of the thin lens.
Let O be the point object at a distance u from the pole of the first curved surface of the lens. Real image is formed at I’, at a distance v’ from the pole P1 This image is formed in the denser medium.
We know that from the refraction formula,

where n1 is the refractive index of rarer medium and n2 that of the denser medium (n2 > n1). For the second surface, real image at I’ will serve as a virtual object and the distance is taken as -ve. The object space is the lens medium for refraction through the second curved surface. Final image is formed in air at I and at a distance of ‘y’ from P2

Question 8.
(a) Obtain an expression for the equivalent focal length of two thin lenses in contact with each other.
(b) Express the combined power of two lenses, one of focal length +f1 and of the other -f2 in contact with each other.
(a) Thin lenses in contact:

A thin lens is that lens whose thickness can be neglected when compared to its radius of curvature. By applying universal lens formula of refraction through L1,

Image at T serves as a virtual object for the thin lens L2 The virtual object distance is taken as+ve in the direction of incident ray.

where f is known as the focal length of equivalent lens.
An equivalent lens is that single lens that produces the same effect as those of combined lenses in contact.
In general

Since, P = $$\frac{1}{f}$$ = power of the lens
P = P1 +P2 + P3 + ……. + Pn
P1, P2 may be +ve or – ve depending upon the type of lenses

(b) $$\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{-f_{2}}$$
where ‘f’ is in metre.
P = (P1 – P2)
and ‘P’ is in dioptre.

Question 9.
Show that n = $$\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$ where symbols have their usual notations (March 2015)
BC – Base of the prism.
AB and AC – Refracting sides representing the planes
ABC – Principle section of prism.
PQ – incident ray.
RS – emergent ray
d1 – angle of deviation d1 = i1 – r1
d2 – angle of deviation d2 = i2 – r2
uT̂s – d – total angle of deviation or net deviation
i.e., d = d1 + d2 (exterior angle = sum of two opposite interior angles)

NM is normal to AB at Q
N’M is normal to AC at R
i1 – angle of incidence
i2 – angle of emergence.
D = angle of minimum deviation,
From figure (1)
A Q̂ M = AR̂M = 90° (NM, N’M are normal to AB and BC)
So that A + M = 180° …(1)
∴ AQMR is a cyclic quadrilateral. From the triangle QMR
r1 + r2 + M̂ = 180° (Property of a A) …… (2)
From (1) and (2)
A + M = r1 + r2 + M
i.e., A = r1 + r2 …… (3)
For non-symmetric condition i1 ≠ i
d1 ≠ d2 also net deviation d.
d = d1 + d2
d = i1 – r1 + i2 – r2
∴ d = i1 + i2 – (r1 + r2)
using (3) we write,
d = i1 + i2 – A ……(4)
from the fig., we note that for non-symmetric condition for two angles of incidence, angle of net deviation is the same. However for a symmetric condition, i1 = i2 = i and r1 = r2 = r net deviation becomes the minimum angle of deviation. In this case the refracted ray will be parallel to the base.
∴ for symmetric condition,
(4) can be written as D = 2i – A
i.e., i = $$\left(\frac{\mathrm{A}+\mathrm{D}}{2}\right)$$ …… (5)
(3) can be written as,
A = 2r
or r = $$\left(\frac{\mathrm{A}}{2}\right)$$ …… (6)
From Snell ’s law of refraction at AB,
n = $$\frac{\sin i_{1}}{\sin r_{1}}$$
but for symmetric refraction i1 = i2 = i and r1 = r2 = r
using (5), (6) and (7) we get,
n = $$\frac{\sin \left(\frac{A+D}{2}\right)}{\sin \left(\frac{A}{2}\right)}$$
This formula is applicable for symmetric refraction only.

Question 10.
Obtain an expression for the deviation produced by a thin prism. Hence mention the expression for dispersive power of the material of the prism.
We know that the angle of net deviation produced by a prism,2

For a thin prism, and for small angle of incidence, sin i1 ≈ i1 (radian), sin r01 ≈ r1 (rad) sin i2 ≈ i2 (radian), sin r02 ≈ r2 (rad)

from (1) and (2) for a thin prism and for small angles of incidence d = D = (n – 1) A. which remains a constant for the given material and the prism.

### 2nd PUC Physics Ray Optics and Optical Instruments Numericals and Solutions

Question 1.
Show that lateral shift is equal to the thickness of the slab for grazing incidence.

i = 90°
r = C = critical angle

∴ For grazing incidence, there will be grazing emergence. So L.S = t.

Question 2.
Refractive index of a prism of angIe 8° is 1.55. Find the angle of deviation of a monochromatic light passing through it.
d = (n – 1)A = 0.55 × 8 = 4.4 = 4.4°

Question 3.
A transparent cube of side 0.18 m has an air bubble in it. When viewed normally through one face the bubble appears to be at a distance of 0.08 rn from that surface. When viewed normally through the opposite face, the distance of the bubble appears to be 0.04 m. Find the actual distance of the air bubble from the first face and refractive index of the material of the cube.
Let ‘x’ be the actual distance of the air Lubble from one face.

from (1) and (2)
$$\frac{x}{0.08}=\frac{0.18-x}{0.04}$$
0.04 x = 0.0144 – 0.08 x
0.12 x = 0.0144
∴ x = 0.12m
using x in (1) we get
R.I = n = $$\frac{0.12}{0.08}$$ = 1.5
Hence refractive index of the material of glass is 1.5

Question 4.
A ray of light is incident an angle of 50° on one face of a cube of side 0.10 m. If the refractive index of the material of the glass cube is 1.55 then calculate the amount of lateral shift produced by it.
Given n = 1.55 t = 0.10m i = 50° L.S = ?

∴ r = 29°.37′ and cos r = cos 29°37′
= 0.8694
sin (50° – 29° 37′) = sin (49°60′ – 29°37′)
= sin (20° (20° 23′) = 0.3493
∴ using the values of cos r and sin (i – r) in (1) we get

Lateral shift produced is 0.040 m

Question 5.
Show that area of circular patch of light on water as seen from a water medium is A = πh2 (n2 – 1) where symbols haves their usual meaning.

Question 6.
Given that the angle of minimum deviation for a colour is 43°48′ and its RI = 1.588, Calculate refracting angle of the prism.
D = 43°48′ , n = 1.588

tan A/2 = 0.5649
A/2 = 29°28′
∴ A = 58°56′

Question 7.
Given that the angle of the prism is 60° and its RI for a certain colour 1.645. Calculate the angle of minimum deviation.
Given A = 60° n = 1.645 D = ?

Question 8.
A ray of light is incident on a prism at an angle 500 and angle of prism is 600 and RI 1.5. Calculate the angle of total deviation (for non symmetric refraction).
A = 60°, n = 1.5, i = 500
Applying Snell’s law,

r1 = sin-1 0.5 106
r1 = 30°42’
But A = r1 + r2
r2 = A – r1 = 59°60’ – 30°42’ = 29°18’

sin i2 = 1.5 × 0.4894 = 0.7341
i2 = sin-1(0.7341) = 47°14′
∴ d = i1 i2 – A = 50 + 47°14′ 60° = 37°14′

Question 9.
An air bubble is situated at a distance of 0.08 m from the centre of a sphere of radius 0.12 m. When viewed from the nearest side it appears to be at distance of 0.09 m from the centre. Find the refractive index of the material of the sphere. Where will the bubble appear to be when viewed from the farther side?
Object distance of the air bubble from P]
-u = 0.12 – 0.08 = 0.04 m for real object.
r1 = 0.12 m (P1O)
v = -(0.12 – 0.09)
i.e., v = – 0.03 m (P1B)

(i) using the formula $$\frac{n_{0}}{-u}+\frac{n_{1}}{v}=\frac{n_{0} \sim n_{1}}{r}$$
V is negative because it is measured opposite to the direction of incident light.

$$\frac{n_{0}}{0.04}+\frac{1}{-0.03}=\frac{n_{0}-1}{0.12}$$
25 n0 – 33.33 = (n0 – 1) 8.33
25 n0 – 8.33 n0 = 33.33 – 8.33
16.67 n0 – 25
∴ n0 = 1.4997
n0 ≈ 1.50

(ii) u from P2 = (0.12 + 0.08) = 0.20 m

Question 10.
An equibiconvex lens of radius of curvature 0.20 and refractive index 1.5 immersed half inside water of RI 4/3 and the rest outside in air. A parallel beam of light in air is incident on it. Find the final position of the image.
Given  r = 0.20 m
nw = $$\frac{4}{3}$$ = 1.333 3
ng = 1.50 8
u = ∞
For reflection through / surface
n0 = 1, n1 = 1.50, u = ∞ v = ?

Real image serves as virtual object for II surface. Put u’ = 0.6 m

v ≈ 0.40 m from the second surface in the dfrection of incident light.

Question 11.
A parallel beam of light strikes the first surface of a glass sphere of R.I 1.5 and radius of curvature 0.10 m. Find the position of the final image.
U = ∞
ng = 1.5 na = 1
r = 0.10m
(1) for I surface using the formula

The image will serve as a virtual object for II surface.
Distance of I’ from P2 = (0.3 – 0.2) = 0.10 m

∴ The final image will be at 0.05 from p2.

Question 12.
An angular magnification of 30 X is desired using an object lens of focal length 1.25 cm and an eyepiece of focal length 5 cm. how will you set up the compound microscope?
Angular magnification of the eye piece = 1 + $$\frac{D}{f_{e}}$$
i.e., me = 1 + $$\frac { 25 }{ 6 }$$ = 6 where D = 25 cm (least distance of vision)

∴ u = 1.5 cm. For a real object at u = – 1.5 cm
v = |5u| = 7.5 cm
The object should be placed 1.5 cm from the object lens.

Question 13.
Two lenses of focal lengths 0.20 m and 0.30 m are kept in contact with each other. Calculate the resultant focal length of the combination. Also calculate the powers of the individual lenses and that of the equivalent lens. (March 2014)

Question 14.
An equibiconvex lens has a focal length of 20 cm. A ball pin of length 5 cm is placed on one side of the lens, such that the mid point of the pin is at a distance of 30 cm from the centre of the lens. Calculate the size of the image of the pin and its magnification f = 20 cm 1 5 cm.

$$\frac{1}{f}=\frac{1}{-u}+\frac{1}{v}$$
Given : -uA = 27.5 cm

Question 15.
A convex lens has a focal length of 0.1 m in air. Calculate its power. If the lens is completely dipped in CS2 of refractive index 1.66, then what will be the change in power of the lens? GivendtJofthe convex lens = 1.50.
f = 0.1m

Lens behaves as a diverging lens of power = – 1.93 D
Change in power ∆P = Pf – P1 = -1.93 – (+10) = -11.93 D

Question 16.
A piano – convex lens has a focal length of 0.25 m and is made of glass of refractive index 1.5. Find the radius of curvature of its curved surface. If two such lenses are placed with their curved surfaces in contact then what will be the focal length of the combination? If the space between them is filled with a liquid of refractive index 1.7, what will be the focal length of the combination?

Given r = ? f = 0.25 m, n = 1.5, n1 = 1.7
For a piano convex lens,

r = 0.5 × 0.25 = 0.125 m
r = 0.125m

(ii) When thin lenses are in contact,

Question 17.
A point object is placed at a distance of 0.12 m on the axis of a convex lens of focal length 0.10 m. On the other side of the lens a convex mirror is placed at a distance of 0.10 m from the lens such that the image formed by the combination coincides with the object itself. What is the focal length of the convex mirror?

Given : -u = 0.12 m
x = 0.10 m, f = 0.10 m

From the fig., radius of curvature of the mirror is 0.60 – 0.10 = 0.50 m so that the focal length
of the convex mirror f = $$\frac{r}{2}=\frac{0.50}{2}$$ = 0.25 m.

Question 18.
Photographs of the ground are taken from an aircraft at an altitude of 2000 m by a camera with a lens of focal length 0.50 m. The size of the film in the camera is 0.18 m × 0.18 . m. What area of the ground can be photographed by this camera in a single shot?
-u = 2000 m f = 0.50 m
Size of the image 0.18 m × 0.18 m

∴ Image will be formed at the focus.

i.e., Area on the ground = 720 m × 720 m.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the . opposite wall 3m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose?

i.e., f = $$\frac{3}{4}$$ = 0.75 m J 4
or f = 75 cm.

Question 20.
A small telescope has an objective lens of focal length 144 cm and an eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the object lens and the eye piece?
Given : f0 = 144 cm, fe = 6.0 cm.
Magnifying power of the telescope

Separation between objective and eye piece
= |fo + f|
= 144 + 6
= 150 cm

Question 21.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m.
(i) If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m and the radius of the lunar orbit is 3.8 × 108m.
(i) Given f0 = 15m, fe = 1.0cm
Angular magnification

(ii) Angular diameter

Diameter of the image of the moon = 9.15 × 10-3 × 1500 = 13.725 = 13.72 cm

Question 22.
A Cassegrain telescope uses two mirrors of radii of curvature 220 mm and 140 mm. The distance b/w the two mirrors is 20 mm. Where will the final image of an object at infinity be?