11th Class Physics Chapter 10 Optical instruments Short Questions Answer
11th Class Physics Chapter 10 Optical instruments Short Questions Answer
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1. What do you understand by linear magnification and angular magnification? Explain how a convex lens is used as a magnifier ?
Linear Magnification:- It is defined as the ratio of the size of the image to the size of the object. It has no unit but only a simple number because this is the ratio between two similar quantities. Angular Magnification: – It is defined as the ratio of the angle subtended by the image as seen through the optical device to the angle subtended by the object at the unaided eye placed at equal distance. It has also no units but a simple number because this is a ratio of two similar quantities. Convex lens as a magnifier:- A convex lens of a shorter focal length can be used as magnifier when the object is placed very closed to it i.e. When the object lies between the lens and its focus. The image thus formed is virtual, erect and magnified. For detail, see theory art 1 0. 3.
2. Explain the difference between the angular magnification and resolving power of an optical instrument. What limits the magnification of an optical instrument?
Angular magnification simply increases the apparent size of the image of an object when seen through an optical device. It can be made as large as we wish by using lenses of suitable focal lights. But the magnification alone is of no use unless we can see the details of the object distinctly. The resolving power of an optical instrument is its ability to reveal the minor details of an object under examination. It is minimum angle ‘α min’ between two point sources that allow the images to be resolved as two distinct spots of light rather than one. The magnification of an optical instrument is limited due to defects of the lens as chromatic and spherical aberrations. The image does not remain well defined and details of the object cannot be seen clearly.
3. Why would it be advantageous to use blue light which a compound microscope?
An objective lens of large aperture and use of blue light of short focal length produces less diffraction and increases its resolving power. Thus, more details of the object can be seen by the eye.
4. One can buy a cheap microscope for use by the children. The images seen in such a microscope have colored images. Why is it so?
The images seen in the cheap microscope have colored edges due to defect of the lenses. Such a defect is known as chromatic aberration. A lens acts like two prism placed end to end. This gives rise to dispersion of lights. When rays of white length parallel to the principle axis pass through a convex lens, they are dispersed into seven colors in such a way that different colors are brought to focus at different distance from the lens as shown in fig since blue rays are lens cannot bring all the rays of white light from a point on the object to a signal point on the images. As a result, the images is not sharp and is colored at edges and not clear. This images defect is called chromatic aberration.
5. Describe with the help of diagramed, how a single biconvex lens can be used as a magnifying glass biconvex lenses can be arranged to from a microscope (compound)?
a) A biconvex lens is used as a magnifying glass is shown in the following fig (a). The object is placed between the lens and focus. Simple Microscope OR (b) Ray diagram of two biconvex lenses arranged to make a compound microscope. Compound Microscope The objects of the height h is a placed just beyond the principal focus of the objective, this produces a real, magnified image of height hl inside e the focal length of the eye- piece. It is further magnified by the eye- piece.
6.If a person were looking through a telescope at the full moon, how would the appearance of the moon be changed by covering half of the objective lens?
If half of the objective lens of a telescope is covered, the moon will appear full to the person looking at it. But the intensity of light depends upon the diameter of the object lens, therefore the intensity of the light received from the moon will decrease. Thus, its brightness is reduced by the half- covered objective lens.
7. How the light signal is transmitted through the optical fiber?
Light signal is transmitted by (i) Total internal reflection from the surface of the outer layer(cladding) (ii) Continuous refraction. Both of these totally confine the light withen fiber.
8. How the power is last in optical fiber through dispersion? Explain.
When the source of light signal is not perfectly monochromatic, then the narrow band of wave-lengths will disperse in different directions. When the light signals enter the glass fiber, the light spread into different wavelength λ1, λ2 and λ3 as shown in fig. As λ1 meets the core and cladding at the critical angle and λ2 and λ3 are at slightly greater angles. The light paths have thus different lights lengths. So the light of different wavelength reaches the other end of the fiber at different times. In the presence of power losses due to dispersion and spreading of the light signal, the information received at the other end of a fiber will be faulty and distorted.
9. Define resolving power?
The resolving power of an optical instrument is its ability to reveal (show) the minor details of an object under examination. It is the minimum angle α ‘min’ between two point source that allow the images to be resolved as two distinct spots of light rather than one.
10.In a compound microscope, magnification produced by objective is 5 and that produced by eye- piece is 50. What is the total magnification produced by the microscope?
Solution:-
Magnification produced by the objective is
M1 = 5
Magnification produced by the eye-piece is
M2= 50
The total magnification ‘M’ produced by the objective and eye-piece of the compound microscope is equal to the product of M1 and M2. That is
M= M1M2
Putting the values of M1 and M2, we get
M=5*50= 250
M= 250
11. Why an objective of large aperture is used in a reflecting mirror telescope?
As we have read that the objective of a telescope should be of long focal length and of large aperture. As it is difficult to make a large lens free from chromatic and spherical aberrations. It is easier to make a mirror of large aperture than a large lens. Moreover, concave mirrors are fee from chromatic aberration. Spherical aberration can also be eliminated in reflecting telescopes by using parabolic mirrors.
12. An octagonal mirror is rotating with frequency f. calculate the time of relation of one its sides?
The time taken by mirror to complete f revolution = 1 second Time taken by mirror to complete one revolution= 1.f second Or Time taken by 8 faces to complete one revolution =1/f second The time taken by one face of the mirror= 1/8f Ans
13. What are different type of optical fiber?
There are three type of optical fibres, Which are grouped by the way they propagates light. These are named as (i) Single mode step index (ii) Muti mode step index (iii) Multi mode grade.
14. How signal is transmitted and converted into sound in an optical fibres?
To transmit information by light xaves, whether it is an audio signal, a telescope signal or a computer data signal it is necessary to modulate the light waves. The most common method is digital modulation in which a laser is flashed on and off at any extremely fast rate. A plus of light represent number 1 and the absence of light represent 0. Any information can be represented by a particular pattern or code of these l s and 0 s. The receiver is programmed to decode the l s and 0s. It receives into sound, pictures or data as required.
15. How power is lost in optical fibr through dispersion?
If the light signal is not perfectly monochromatic and contains different wavelengths, dispersion takes place in the fibre. A narrow beam of wavelengths is refracted in different directions and the light spread. All the rays travel along the fibre by multiple reflections. Thus, all light of different wavelengths reaches the other end at different times. The signal received is therefore, distorted or faulty. Hence, the power is lost during the transmission in optical through dispersion.
16. Why do we use photodiode to transmit TV signal through optical fibre?
At the end of fibre, a photodiode is used to transmit TV signal because it converts the light signals. Which are then amplified and decoded and reconstructs the signals originally transmitted.
17. Why do astronomers prefer reflecting telescope over refracting telescope?
Refracting telescope has an important defect. This defect is called chromatic aberration. Because of this problem, refracting telescope cannot be focused properly. One the other hand, reflecting telescopes do not suffer from this problem.
18. what is the effect on the image if half of a converging lens is covered?
So far as the size of the image is concerned, the image will remain unchanged but its brightness will be reduced due to less intensity (half of the intensity). This is because half of the lens is covered, then only half of the aperture will be used to collect the rays.
19. Show why a pinhole placed in front of a lens leads to good image even if the image is not quite in focus?
The pinhole placed in front of the lens enables us to use only central portion of the lens. Since a lens of small aperture gives the image completely free from spherical aberration and thus the image formed will be very well defined and sharp.
20. Why is convex lens of small focal length preferred for a magnifying glass?
We know that the magnifying power of a magnifying glass or simple microscope is given by M = 1+ d/f This relation shows that smaller the focal length, greater will be the magnification (I.e. M α l/f) Therefore, a lens of smaller focal length is used to increases the magnifying power of magnifying glass.
21. Under what conditions does a double convex lens acts as a diverging lens?
A double convex lens acts as diverging lens under the following conditions. (i) If an object is placed between the focus and the optical centre of a convex lens. The rays are diverged and virtual image is formed. (ii) If the double convex lens is placed in a medium whose refractive in dex is greater than the refractive index of the material of the lens.
22. How can a real image be be distinguished from a virtual image? Can each type of image be projected on a screen? Explain
Real and virtual image:- A real image is formed by the actual intersection of the reflected rays, so it can be projected on the screen. On the other hand, a virtual image is formed by the imaginary intersection of the reflected or refracted rays, so it can not be projected on the screen. A real image is usually inverted while a virtual image is erect. Hence, if a image can bi projected on a screen it is real otherwise virtual.
23. What is difference between (i) chromatic aberration and (ii) spherical aberration?
i) Chromatic aberration:- It is the defect of a lens due to which lens along with diffraction also disperses white light into component colours. A complete image thus formed will have all the colours of the spectrum. The lens focuses all the dispersed colours at one point. Due to this defect, the image will be coloured and not well defined. (ii) Spherical aberration:- This is a defect of lens in which the outer rays after passing through the lens converge more than the inner rays (central rays).Thus the outer incident rays come to a focus closer to the lens than the central (inner) rays, producing more than one focus. In this way, the image no longer remains sharp and well defined but becomes blurred. This defect is mainly caused due to the large size of aperture of lenses.
24. Find the magnifying power of a convex lens of 25 cm. focal length?
Data :- The focal length of convex lens = f=25cm The least distinct of distinct vision = d = -25cm Magnifying power of lens = M.p. = ? Calculations :- Using the formula for magnifying power M.P. = (1- 25/25)1-1=0 M.P.=o Ans
25. When light enters glass from air, its speed becomes less. Is it due to change in its frequency or wavelength?
Speed of light changes with wave length:- The speed of light (i.e, v = fλ) in various media depends upon the wavelength of light and not on the frequency. When light enters glass from air, its wavelength decreases, thus its speed becomes less.
26. Why does a diamond sparkle more than a glass imitation of the a same shape and size?
The refractive index of diamond is very high and its critical angle is quite small (i.e. 240). Hence when a beam of light enters it, it is totally reflected a number of times inside it and it emerges in random directions causing the sparkling of the diamond. But in case of a glass imitation refractive index is low, so it does not sparkle like a diamond.
27. Why is a beam of white light not dispersed into component colors when it passes perpendicularly through a pane of glass?
When a beam of white light falls perpendicularly on a pane of a glass, the angle of incident is zero, therefore the angle of refraction is also zero and it passes through without suffering any deviation. Since the colors are separated due to their different deviations, therefore in this case they will not be separated and no dispersion will take place.
28. Why are danger signals in red when the eye is more sensitive to yellow and green?
Danger signals are in red :- Light of any colors, When travels through space, is scattered from dust particles ect, present in the atmosphere and its intensity decreases with the distance. The scattering of light depends upon its wavelength which is given by the relation. Scattering α 1/λ4 i.e. The larger the wave length, the smaller the scattering. As the wavelength of red color is larger than all the other color, so it can travel with a small loss of intensity and can be seen at longer distance. That is why danger signals are In red.
29. Why object of short focal length is preferred in microscope?
The magnifying power of microscope (e.c compound microscope) can be written as
M = L/f0 (1+ d/fe) ……………………… (1)
Where f0 = focal length of objective lens
fe= focal length of eye piece
L= Length of compound microscope
d= Least distance of distinat vision
It is clear from the above relation that smaller the focal length, greater will be the magnifying power (i.e. M α 1/f). therefore, in order to increase the magnifying power of microscope and objective of short focal length is used.
30. Explain resolving power of an optical instrument?
The ability of an instroument to travel (show) the minor details of the object under examination is called its resolving power.
The purpose of an microscope or telescope is not only to magnify an object but to reveal (show) it in great detail. The amount of detail revealed by an optical instrument depends on what is called resolving power.
Mathematically it is expressed as
α min=1.22 λ/D
This equation shows that greater the diameter (aperture) of the convex lens and smaller value of the wave length of light, higher will be resolving power of the optical instrument.
31. Define refractive index?
The index of refraction can be defined as the ratio of speed of length ‘c’ in air or vacuum to the speed of length ‘v’ in the given medium. Mathematically, it is written as Index of refraction = speed of light in air or vacuum/speed of light in medium Or n = c/v
32. What is Snell’s law?
The ratio of sine of angle of incidence to the sine of angle of refraction is equal to the ratio of refractive indices of two media is called snell’s law.
Mathematically, it is expressed as
sinθ1/sinθ2 = n2/n1
or n1sinθ1= n2sinθ2
33. Write different types of optical fibres?
There are three types of optical fibres depending upon the way they propagate light. These types are given below. (1) single mode step index fibre. (2) Multimode step index (3) Multimode graded index Here the word mode is used for method by which light is propaged withen the fibre or various paths that light can follow in traveling down the fibre.
34. Define magnifying power?
It is defined as the ratio of the angle subtended by the image as seen through the optical device to the angle subtended by the object at the naked eye, provided both the object and the image are placed as the same distance from thr eye.
It can be expressed as
Magnifying power = Angle subtended by image/Angle subtended by the object
Or M= θl/θ0 = B/α
35. What is an optical fibre?
Optical fibre is a thin glass or plastic material in the shop of a flexible rod through which light signals are transmitted. The propagation of light in the optical fibre occurs by total internal reflection and by continuous refraction. The fibre optic technology is being widely used now a days in communication system due to its much wider bandwidth capacity that is information carrying capacity. The fibre optic system has much thinner and light ( ) weight cables having diameter of about 6.00 mm.
36. Define total internal reflection?
When the rays of light pass from a denser medium to a rarer medium then angle of refraction for each ray is greater than the critical angle, no refraction will take place and the total light rays will be reflected back into the same denser medium from the boundary. This phenomenon is called the total internal reflection.
37. What are the conditions necessary for total internal reflection?
There are two conditions of total internal reflection (1) The ray of light should travel from denser medium to rarer medium. (2) The angle of incidence should be greater than the critical angle of the denser medium.
38. Define critical angle and its formula?
When a ray of light travel from a denser medium (glass) to a rarer medium (air) it bends away from the normal. The angle of incidence for which angle of refraction is 900 is called critical angle. It is denoted by θc Formula:- According to snell’s law
n1sinθ1 = n2sinθ2
θ1 = θc,θ2=900
when
Therefore,
n1sinθc = n2sin 900
or n1sinθc= n2 (sin 900= 1)
or sin θc = n2/n1
39. Define least distance of distinct vision?
The minimum distance from the eye at which an object appears to be distinct is called the least distance of distinct vision or near point. This distance is denoted by ‘d’. its value is about 25 cm from the eye of a normal persion (i.e. d =25cm)
40. Define power of a lens and its unit?
Power of a lens:-
The reciprocal of the focal length of a lens expressed in metres is called the power of a lens. It is denoted by D. That is
D= 1/f Unit of power:-
The unit of power of a len is dioptre. Definition of dioptre:-
One dioptre is the power of a lens of one meter focal length.
Thus,
D= 1/focal length of the lens in meter=1/f
The power of a convex lens is positive while that of concave is nagitive.
41. What is simple microscope?
An ordinary convex lens which is held closed to the eye to magnify the image of the object is called a simple microscope. When this lens is placed between eye and object, it helps us to see the details of an object by bringing it closer than 25 cm. Its magnifying power is given M = (l+ d/f)
42. What is compound microscope?
Compound microscope is an optical instrument which is used for higher magnification of a near small object.
Magnifying power of a compound microscope is given by the formula
M = q/p ( 1+ d/fe)
43. An astronomical telescope of having magnifying power of 5 consist two thin lens 24 cm apart. Find the focal lengths of the lens?
The total length of the blue is given by
L = f0 + fe
putting the value, we have
24 = f0+fe ………………………… (2)
Magnifying power = M= f0/fe or 5 = fo/fe Or fo= 5fe ………………………….(3)
putting the value of f0 from equ (3) in equ (2) we get
24 = 5fe+ fe
or 6fe = 24
fe = 4cm …………………(4)
Putting the value of fe in equ (2)
24= fo + 4
or f0 =24 -4= 20 cm
F0 = 20 cm
44. Distinguish between convex lens and concave lens?
Convex lens. A lens which converges a beam of parallel rays to a point after refraction is called convex lens(or conversing lens)
A lens which is thicker in the center (middle) and thinner at the edges is called a convex lens or converging lens.
This lens produces a real image whose sign is positive. Concave lens :-
A lens which diverges a beam of parallel rays after refraction is called concave lens(or diverging lens)
A lens which is thinner at the center and thicker at the edges is called a concave or diverging lens.
A concave lens produces a virtual image whose sign is taken as negative.
45. Name the instrument which is used to find the refractive index of transparent material?
The instrument which is used to find the refractive index of transparent material is called spectrometer.
The refractive (u) of prism can be found by using the formula
u = sin(A+Dm)/2/ sin A/2
where ‘A’ is the angle of prism and Dm is the angle of minimum deviation.
46. What is spectrometer?
It is an optical instrument which is used for careful study of the spectra from different sources of light.
It consists of three main parts.
(i) A collimator
(ii) A Turntable
(iii) A Telescope Uses:-
(1) It is used to fined the wavelength of light.
(2) It is also used to fined the refractive index of transparent material (i.e. prism)
(3)It is used to study the spectra of different sources of light.
47. What is collimator?
A collimator is a main part in the construction of spectrometer. It consists of a metal tube having an adjustable slit at one end of a convex lens at the other end. It provides a parallel beam of light from the source.
48. How the power is last in optical fibre through scattering and absorption.
When a light signal travels along fibres by multiple reflections, some light is absorbed due to impurities in the glass. The groups of atoms formed at joints where fibres are joined together also scatter some light. Power losses due to scattering and absorption can be reduced by careful manufacturing.
49. Explain the difference between (a) signal mode (b) Multimode step index fibre?
a) Single Mode Step index fibre:- Single mode step fibre has a very thin core about 5 mm diameter and has a relatively large cladding or glass or plastic. A strong monochromatic light source or laser source is used to send light signals through it. It can carry more than l4 TV channels or 14000 phone calls. (b) Multimode step index fibre:- A multimode step index fibre has a core of relatively larger diameter such as 50 um. It is mostly used to carry white light signals but due to dispersion effect, it is useful only for short distances. The central core of this fibre has a constant refractive index (1.52) . The refractive index then decreases from 1.52 to n = 1.48 at the boundary of cladding and remains constant throughout the thickness of the cladding. This is called step index multimode fiber.
50. Describe multimodi graded index fibre?
A multimode graded index fibre has a central core whose diameter ranges from 50 mm to 1000 mm. Its core has relatively high refractive index which decreases gradually from the middle to the outer surface of the fibre. There is no particular boundary between central core and cladding . This type of fibre is called a multimode graded- index fibre. It is useful for long distance applications and white light is used to carry the signals.
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