Bihar Board - Class 12 Physics - Chapter 9: Ray Optics And Optical Instruments Long Answer Question
Long Answer Type Questions
1. Draw a ray diagram of a reflecting type telescope. State two advantages of this telescope over a refracting telescope.
Answer:
(a)
(b) Advantages of reflecting telescope over a refracting telescope:
Due to the large aperture of the mirror used, the reflecting telescopes have high resolving power.
This type of telescope is free from chromatic aberration (formation of coloured image of a white object).
The use of a paraboloidal mirror reduces the spherical aberration (formation of non-point, blurred image of a point object).
Image formed by a reflecting telescope is brighter than a refracting telescope.
A lens of large aperture tends to be very heavy and therefore difficult to make and support by its edges. On the other hand, a mirror of equivalent optical quality weighs less and can be supported over its entire back surface.
2. Draw a ray diagram of an astronomical telescope in the normal adjustment position. State two drawbacks of this type of telescope.
Answer:
(i) Magnifying power m = - f0fe . It does not change with increase of aperture of objective lens, because focal length of a lens has no concern with the aperture of lens.
(ii) Drawbacks :
Images formed by these telescopes have chromatic aberrations.
Lesser resolving power.
The image formed is inverted and fainted.
3. Explain the following, giving reasons :
(i) When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency.
(ii) When light travels from a rarer to a denser medium, the speed decreases. Does this decrease in speed imply a reduction in the energy carried by the wave?
(iii) In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity in the photon picture of light?
Answer:
(i) Reflection and refraction arise through interaction of incident light with atomic constituents of matter which vibrate with the same frequency as that of the incident light. Hence frequency of reflected and refracted light both have the same frequency as the incident frequency.
(ii) No, energy carried by a wave depends on the amplitude of the wave and not on the speed of wave propagation.
(iii) For a given frequency, intensity of light in the photon picture is determined by the number of photons incident normally on crossing a unit area per unit time.
4. Draw a labeled ray diagram of a reflecting telescope. Mention its advantages over the refracting telescope.
Answer:
Advantages over the refracting telescope :
There is no chromatic aberration as the objective is a mirror.
Spherical aberration is reduced using mirror lenses in the form of a paraboloid.
Image is brighter compared to that in a refracting type telescope.
Higher resolving power.
5. Write the conditions for observing a rainbow. Show, by drawing suitable diagrams, how one understands the formation of a rainbow.
Answer:
The conditions for observing a rainbow are:
The Sun comes out after a rainfall.
The observer stands with the Sun towards his/her back.
Formation of a rainbow :
The rays of light reach the observer through a refraction, followed by a reflection followed by a refraction.
Figure shows red light from drop 1 and violet light from drop 2, reaching the observer’s eye.
6. (i) What do you mean by refraction of light?
(ii) Define absolute refractive index of a medium.
Answer:
(i) The phenomenon of bending of light rays as they pass from one medium to another is called refraction of light. The bending of light rays is due to the fact that the speed of light changes as it passes from one medium to another. If the light goes from rarer medium to denser medium, it bends towards the normal and if it goes from denser to rarer medium, it bends away from the normal. The refraction of light obey two laws
(a) the incident ray, r refracted ray and the normal at the point of incidence all lie on the same plane and
(b) The ratio of sine of angle of incidence and sine of angle of refraction is constant for a given pair of media.
(ii) The absolute refractive index of a medium (µ) is defined as the ratio of velocity of light in vacuum (c) to the velocity of light in the medium (v)i.e.
µ = cv
As c > v ∴ µ > 1.
7. (a) A mobile phone lies along the principal axis of a concave mirror. Show, with the help of a suitable diagram, the formation of its image. Explain why magnification is not uniform.
(b) Suppose the lower half of the concave mirror reflecting surface is covered with an opaque material. What effect will this have on the image of the object? Explain
Answer:
(a) The formation of image is shown in the
The magnification is not uniform and the image is distorted because the parts of the mobile are situated at different distances from the mirror.
(b) When the lower half of the concave mirror reflecting surface is covered with an opaque material, the image will be of the whole object, i.e. mobile. However, as the area of the reflecting surface has been reduced, the intensity of image (brightness) will be reduced to half.
8. Answer the following questions:
(a) Plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances? Explain
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a vitrual image, we are obviously bringing it on to the ‘screen’ (i.e. the retina) of our eye. Is there a contradiction?
(c) A diver underwater, looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter to the diver than what he actually is?
(d) Does the apparatus depth of a tank of water change if viewed obliquely? If so, does the apparatus depth increase or decrease?
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter?
Answer:
(a) Rays converging to a point ‘behind’ a plane 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. Convince yourself by drawing an appropriate ray diagram.
(b) When the reflected or refracted rays are divergent, the image is virtual. The divergent rays can be converged onto a screen by means of an appropriate converging lens. The convex lens of the eye does just that. The virtual image here serves as an object for the lens to produce a real image. Note, the screen here is not located at the position of the virtual image. There is no contradiction.
(c) Taller
(d) The apparent depth for oblique viewing decreases from its value for near-normal viewing. Convince yourself of this fact by drawing ray diagrams for different positions of the observer.
(e) Refractive index of a diamond is about 2.42, much larger than that of ordinary glass (about 1.5). The critical angle of diamond is about 24°, much less than that of glass. A skilled diamond-cutter exploits the larger range of angles of incidence (in the diamond), 24° to 90°, to ensure that light entering the diamond is totally reflected from many faces before getting out, thus producing a sparkling effect.
9. How does the surface of the earth appear to a fish or a person sitting inside clear water?
Answer:
Whole of the earth will appear to him within 97° instead of actual 180°. A ray of light from an incident at an angle of 90° at N' will go towards NE at an angle of 48.5°. [Refractive index of water is 1.33 and critical angle is 48.5°]. It shall appear to fish to come from A instead of A'. Similarly another ray from B will appear to come from B'. Whole of the earth will appear within a cone of apex angle 48.5° + 48.5° = 97°.
10. How is the working of a telescope different from that of a microscope?
Answer:
| Telescope | Microscope |
1. | Resolving power should be higher for certain magnification. | The Resolution power is not so large but the magnification should be higher. |
2. | Focal length of objective should be kept larger while eyepiece focal length should be small for better magnification. | Both objective and eyepiece should have less focal length for better magnification. |
3. | The Object should be of large aperture. | Eyepiece should be of large aperture. |
4. | Distance between objective and eyepiece is adjusted to focus the object at infinity. | Distance between objective and eyepiece is fixed, for focusing an object the distance of the objective is changed. |
11. Describe a reflecting telescope. Give its merits and demerits.
Answer:
It is a telescope which uses a converging mirror to view distant objects like stars etc.
This telescope was initially constructed by Newton for observing distant stars, and was modified from time to time. In Fig. Cassegrain type of telescope is shown.
Construction:
It consists of an objective and eyepiece. Objective is a concave mirror of large aperture and large focal length fixed at one end of a long tube. A plane mirror M is placed before the formation of image by the concave mirror at 45° to the axis of the tube. This mirror deviates the rays towards the eyepiece which consists of two plano-convex lenses capable of moving in and out with the help of rack and pinion arrangement T as shown in the figure.
Working:
Parallel rays from distant objects like stars are made to fall on the concave mirror. These rays are reflected towards the focus F, but the plane mirror M deviates the rays towards the eye piece and real image I is formed in front of the eye piece. The rays after passing through the eyepiece become parallel. The eye piece acts as a magnifier and the final image is virtual, erect and magnified.
Merits:
- The image is free from chromatic aberration.
- The image is free from spherical aberration.
- Concave mirror of large aperture can be easily made.
- The objective i.e. concave mirror does not absorb light.
- The image is stable.
Demerits:
- It cannot be used for general purpose.
- It is less convenient to use.
- Large areas of sky cannot be photographed.
12. What is total internal reflection ? Explain its applications.
Answer:
Total internal reflection : Consider the incident ray I1 which retracts as R1. When the angle of incidence is increased to ie the incident ray I2 refracts as R and glazes the surface. When angle of incidence is increased beyond the value of ie the complete light comes back to the same medium and phenomenon is called total internal reflection. Ie, total internal reflection i, the angle of incidence for which angle of refraction becomes 90° is called critical angle.
Applications of total internal reflection :
(i) Mirage : The atmosphere can be considered as made up of layers of air. The atmosphere gets heat from the sun directly through radiation. Therefore the air in the contact of the surface of earth remains warmer as the upper layers of air in the atmosphere. Due to this variation of temperature, the density of upper layers of air remains more than the density of lower layers of air. When light passes through the atmosphere the angle of incidence at each layer of air increases and for certain layer becomes greater than the critical angle and total internal reflection takes place due to which an inverted image of the object is formed just below the object as through reflected from pool of water and gives an illusion of presence of water.
Thus, mirage is the phenomenon of optical illusion based upon the principle of total internal reflection. It is very common in hot countries and in deserts.
13. Show the position, nature and size of the image when an object is placed between pole and focus in front of a concave mirror.
Answer: Let a ray parallel to the principal axis incident from point A on the mirror and reflected through F. Another ray is incident normally and reflects back tracing the same path. These rays after reflection appear to come from point A’. Hence A ‘B’ be the virtual image of object AB.
Thus the image is virtual, erect and larger in size than the object.
14. What are the factors upon which the refractive index of the medium depends?
Answer:
On the pair of mediums : Refractive index of the denser medium with respect to the rarer medium is always greater than 1 but the refractive index of the rarer medium with respect to denser medium is less than 1.
Density of the medium: If the optical density of any medium is more than its refractive index will also be more.
Wavelength of light: If the wavelength is less, more is the refractive index and more the wavelength less is the refractive index.
Temperature of medium: On increasing temperature, it becomes optically rarer hence refractive index decreases.
15. What are the factors on which angle of minimum deviation depends?
Answer:
1. Angle of prism:
It is clear from eqn. (6) that δm ∝A , so if the value of angle of prism is greater than the value of minimum angle of deviation will also be greater.
2. Refractive index of material of prism:
δm ∝ (µ – 1), so greater the value of refractive index of material of prism greater will be the value of angle of minimum deviation.
3. Color of light:
The value of angle of minimum deviation also depends upon the color of light. Refractive index of the prism is greater for violet color and is less for red color. So, violet color is deviated most and minimum deviation will be for red.
16. Establish the relation between focal length and radius of curvature for spherical mirrors.
Or
Prove for spherical mirror: f = R2
Answer:
Let MPN be a concave mirror and its pole is P, focus F, center of curvature is C and PC is principal axis.
Now, a ray AB is incident on the mirror parallel to the principal axis. Join CB, which will be the normal to the mirror.
Now, by laws of reflection,
Incident ∠ABC = Reflected ∠CBF = θ (let)
AB ∥ PC and BC is a transversal line
∠ABC = ∠PCB = θ, (alternate angles)
∠CBF = ∠PCB = θ
∴ In ∆BCF, the opposite sides will be equal.
∴ FC = BF
As the aperture of the mirror is small.
∴ BF = PF (approx)
∴ FC = PF
or PC = PF
or R = 2f
or f = R2
17. What are optical fibers? Write its construction and working method with a diagram. Write any two uses of optical fibers.
Or
What are optical fibers? Write its working principle. How does it transmit light signals? Write its main uses.
Answer:
An optical fiber is a thin strand of highly pure glass or quartz core with outside coating of glass or slightly different chemical composition having different refractive index, as shown in fig.
The main parts of optical fiber are as follows :
1.Central core:
It is the innermost core of the fiber made of thin and fine quality transparent glass. The refractive index of the material of the core is about 1.7.
2.Cladding:
It is a layer made of glass or plastic which surrounds the central core of optical fiber. Its refractive index is less than that of the central core about µ = 1.5.
3.Plastic jacket:
The central core and cladding are enclosed in a protective jacket made of plastic for providing safety and strength to optical fiber.
Working:
When light falls on the interface separating the core and coating at an angle of incidence greater than the critical angle, then the phenomena of total internal reflection takes place because the refractive index of coating is less than the core. This phenomena takes place continuously and light travels the entire length of the fiber even if it is curved or twisted.
Uses:
- For transmission of light without any loss in its intensity.
- In endoscopy, to examine the internal parts of the body such as the stomach and intestines.
18. What are the differences between the primary and secondary rainbow?
Answer:
Differences between primary and secondary rainbow :
Primary rainbow:
- In this rainbow, red color is outside and violet is inside.
- Its brightness is more.
- Its angular width is 2°.
- It lies below the secondary rainbow.
- It is formed by two refractions and one internal reflection.
Secondary rainbow:
- In this rainbow, violet color is outside and red is inside.
- It is less bright.
- Its angular width is 3°.
- It lies above the primary rainbow.
- It is formed by two refractions and two internal reflections.
19. A person looking at a person wearing a shirt with a pattern comprising vertical and horizontal lines is able to see the vertical lines more distinctly than the horizontal ones. What is this defect due to? How is such a defect of vision corrected?
Answer:
This defect is called astigmatism. It arises due to non spherical cornea. The eye lens is ideally spherical and has the same curvature in different planes, but in an astigmatic eye due to non spherical cornea the curvature may be insufficient in different planes.
In the given situation the curvature in the vertical plane is enough, so vertical lines are visible distinctly. But the curvature is insufficient in the horizontal plane, hence horizontal lines appear blurred. The defect can be corrected by using a cylindrical lens with its axis along the vertical. The parallel rays in the vertical plane will suffer no extra refraction but the parallel rays in the horizontal plane will be refracted largely and converge at the retina, according to the requirement to form a clear image of horizontal lines.
20. Answer the following question:
(a) The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virtual image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) When viewing through a compound microscope, our eyes should be positioned not on the eyepiece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer:
(a) In magnifying glass the object is placed closer than 25 cm, which produces an image at 25 cm. This closer object has larger angular size than the same object at 25 cm. In this way although the angle subtended by virtual image and object is the same at eye but angular magnification is achieved.
(b) On moving the eye backward away from the lens the angular magnification decreases slightly, as both the angle subtended by the image at eye ‘a’ and by the object at eye ‘α’ decreases. Although the decrease in angle subtended by object a is relatively smaller.
(c) If we decrease focal length, the lens has to be thick with a smaller radius of curvature. In a thick lens both the spherical aberrations and chromatic aberrations become pronounced. Further, grinding for small focal lengths is not easy. Practically we can not get magnifying power more than 3 with a simple convex lens.
(d) If we place our eye too close to the eyepiece, we shall not collect much of the light and also reduce our field of view. When we position our eye slightly away and the area of the pupil of our eye is greater, our eye will collect all the light refracted by the objective, and a clear image is observed by the eye.
21. Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm?
Answer:
1f = (n – 1) [ 1R1-1R2 ]
f = 20 cm.
n = 1.55.
R1 = R,
R2= – R
∴ 120 = (1.55 – 1) [ 1R+1R ]
= 0.55 x 2R
R = 0.55 x 2 x 20 = 22 cm
22. You are given prisms made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will
(a) deviate a pencil of white light without much dispersion
(b) disperse (and displace) a pencil of white light without much deviation.
Answer:
Two identical prisms made of the same glass placed with their bases on opposite sides (of the incident white light) and faces touching (or parallel) will neither deviate nor disperse, but will merely produce a parallel displacement of the beam.
(a) To deviate without dispersion, choose, say, the first prism to be of crown glass, and take for the second prism a flint glass prism of suitably chosen refracting angle (smaller than that of crown glass prism because the flint glass prism disperses more) so that dispersion due to the first is nullified by the second.
(b) To disperse without deviation, increase the angle of flint glass prisms (i.e., try flint glass prisms of greater and greater angle) so that deviations due to the two prisms are equal and opposite. (The flint glass prism angle will still be smaller than that of crown glass because flint glass has a higher refractive index than that of crown glass). Because of the adjustments involved for so many colors, these are not meant to be precise arrangements for the purpose required.
23. A myopic person has been using spectacles of power -1.0 dioptre for distant vision. During old age, he also needs to use a separate reading glass of power + 2.0 dioptres. Explain what may have happened.
Answer:
The far point of the person is 100 cm, while his near point may have been normal (about 25 cm). Objects at infinity produce virtual images at 100 cm (using spectacles). To view closer objects i.e., those which are (or whose images using the spectacles are) between 100 cm and 25 cm, the person uses the ability of accommodation of his eye-lens. This ability usually gets partially lost in old age (presbyopia). The near point of the person recedes to 50 cm. To view objects at 25 cm clearly, the person needs a converging lens of power +2 dioptres.
24. (a) What is the scattering of light?
(b) Explain Rayleigh scattering.
(c)Why does the sky appear blue?
(d) Sky blue extends over a wide region. Why?
Answer:
(a) The irregular reflection of light in all directions at random from extremely small particles such as dust, molecules of air etc is called scattering of light.
(b) According to Rayleigh’s scattering law, the intensity of scattered light is inversely proportional to the fourth power of the wavelength.
i.e.. Intensity,
I ∝ 14
It is found that shorter wavelengths are scattered more than longer wavelengths.
(c) The blue color of the sky is due to the scattering of light by small particles of the atmosphere. According to Rayleigh’s scattering law, the scattering intensity is maximum for smaller wavelengths. The size of the particles in the upper atmosphere is smaller so that the radiations of shorter wavelengths are scattered with greater intensity. Sunlight is rich with blue and therefore the most intense scattered color is blue. Thus the sky appears blue.
(d) Scattering of upper atmospheric particles produces blue of the sky. Since the region of observation is extended the blue color thus extends over a wide region.
25. A small telescope has an objective lens of focal length 144 cm eyepiece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?
Answer:
(a) For normal adjustment.
M.P. of telescope = f0fe
= 1446 = 24
(b) The length of the telescope in normal adjustment
L = f0 + fe = 144 + 6
= 150 cm.
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