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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 101. |
For a normal eye, the cornea of eye provides a converging power of `40 D` and the least converging power of the eye lens behind the cornea is `20 D`. Using this information, the distance between the retina and the cornea eye lens can be estimated to beA. `1.5 cm`B. `5 cm`C. `2.5 cm`D. `1.67 cm` |
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Answer» Correct Answer - D For a normal eye, when relaxed, the eye lens offers minimum power, which is given as `P = 40 D + 20 D = 60 D`. As rays from infinity are focussed on the retina, therrfore, distance between retina and cornea eye lens `= f = (1)/(P) = (1)/(60)m = (100)/(60)m = 1.67 cm` |
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| 102. |
A parallel beam of fast moving electrons is incident normally on a narrow slit. A fluorescent screen is placed at a large distance from the slit. If the speed of the electrons is increased, which of the following statements is correct?A. The angular width of central maximum will be unaffected.B. Diffraction pattern is not observed on the screen in the case of electrons.C. Th angular width of the central maximum of the diffraction pattern will increase.D. The angular width of the central maximum will decrease. |
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Answer» Correct Answer - D The diffraction pattern of electron beam will be observed on the screen. If the speed of electrons is increased, de Broglie wavelength associated with electron beam decreases `(lambda = h//mv)`. As angular width of central maximum is proportional to `lambda`, therefore, angular width of central maximum will decreses. Choice (d) is correct. |
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| 103. |
In an experiment, electrons are made to pass through a narrow slit of width `d` comparable to their de Broglie wavelength. They are detected on a screen at a distance `D` from the slit (see figure)`. ` Which of the following graphs can be expected to represent the number of electrons `N` detected as a function of the detector position `y` (y=0 corresponds to the middle of the slit ).A. B. C. D. |
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Answer» Correct Answer - D Electron beam serves as a light wave. As width `(d)` of slit is comparable to de-Broglie wavelength, therefore, the electron beam is diffracted on passing through the slit. We obtain a central maximum surrouded by secondary minima and maxima placed alternately. The intensity is represented by `N` and in diffraction pattern, intensity is never zero. Therefore, choice `(d)` refresents the correct graph, in Fig. |
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| 104. |
Assertion: Radio waves can be polarised. Reason: Sound waves in air are longitudinal in nature.A. If both, Assertion and Reason are true and the Reason is the correct explaination of the Assertion.B. If both, Assertion and Reason are true but Reason is not a correct explaination of the Assertion.C. If Assertion is true but the Reason is false.D. If both, Assertion and Reason are false. |
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Answer» Correct Answer - B Radio waves can be polarised because they are transverse in nature. Reason is true but it is not the correct explaination of assertion. |
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| 105. |
What is Rayleigh’s scattering? |
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Answer» The scattering of light by particles in a medium without a change in wavelength is called as Rayleigh’s scattering |
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| 106. |
As per Rayleigh’s scattering law, amount of scattering is: (a) directly proportioanl to fourth power of wavelength (b) inversely proportioanl to fourth power of wavelength (c) inversely proportioanl to square of wavelength (d) directly proportional to square of wavelength |
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Answer» (b) inversely proportioanl to fourth power of wavelength |
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| 107. |
One of the of Young’s double slits is covered with a glass plate as shown in figure. The position of central maximum will,(a) get shifted downwards(b) get shifted upwards (c) will remain the same (d) data insufficient to conclude |
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Answer» (b) get shifted upwards |
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| 108. |
What is Rayleigh’s criterion? |
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Answer» The images of two point objects are just resolved when the central maximum of the diffraction pattern of one falls over the first minimum of the diffraction pattern of the other. |
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| 109. |
Light transmitted by Nicol prism is, (a) partiallypolarised (b) unpolarised (c) plane polarised (d) elliptically polarised |
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Answer» (c) plane polarised |
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| 110. |
Discuss about pile of plates. |
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Answer» The phenomenon of polarisation by reflection is used in the construction of pile of plates. It consists of a number of glass plates placed one over the other. The plates are inclined at an angle of 33.7° (90° – 56.3°) to the axis of the tube. A beam of unpolarised light is allowed to fall on the pile of plates along the axis of the tube. So, the angle of incidence of light will be at 56.3° which is the polarising angle for glass. The vibrations perpendicular to the plane of incidence are reflected at each surface and those parallel to it are transmitted. The larger the number of surfaces, the greater is the intensity of the reflected plane polarised light. The pile of plates is used as a polarizer and also as an analyser. |
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| 111. |
Differentiate between polarised and unpolarised light. |
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| 112. |
The transverse nature of light is shown in, (a) interference (b) diffraction (c) scattering (d) polarisation |
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Answer» (d) polarisation |
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| 113. |
Discuss polarisation by selective absorption. |
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Answer» Selective absorption is the property of a material which transmits w’aves whose electric fields vibrate in a plane parallel to a certain direction of orientation and absorbs all other waves. The polaroids or polarisers are thin commercial sheets which make use of the property of selective absorption to produce an intense beam of plane polarised light. Selective absorption is also called as dichroism. |
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| 114. |
What is angle of polarisation and obtain the equation for angle of polarisation. |
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Answer» The angle of incidence at which a beam of unpolarised light falling on a transparent surface is reflected as a beam of plane polarised light is called polarising angle or Brewster’s angle. It is denoted by ip ip= 90°- Rp |
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| 115. |
The amount of light received by a camera depends upon (a) diameter only (b) ratio of focal length and diameter (c) product of focal length and diameter (d) only one of the focal length |
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Answer» (b) ratio of focal length and diameter The amount of light received by a camera depends on the ratio of the focal length and diameter of the aperture. |
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| 116. |
How is polarisation of light obtained by scattering of light? |
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Answer» The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. This is because of sunlight, which has changed its I direction (having been scattered) on encountering the molecules of the earth’s atmosphere. The electric field of light interact with the electrons present in the air molecules. Under the influence of the electric field of the incident wave the electrons in the molecules acquire components of motion in both these directions. We have an observer looking at 90° to the direction of the sun. Clearly, charges accelerating parallel do not radiate energy towards this observer since their acceleration has no transverse component. The radiation scattered by the molecule is therefore polarized perpendicular to the plane. |
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| 117. |
The light gathering power of a camera lens depends on(a) its diameter only (b) ratio of diameter and focal length (c) product of focal length and diameter (d) wavelength of the light used |
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Answer» (a) its diameter only The light gathering power of a camera lens is proportional to its area or to the square of its diameter |
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| 118. |
Discuss about simple microscope and obtain the equations for magnification for near point focusing and normal focusing. |
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Answer» A simple microscope is a single magnifying (converging) lens of small focal length. The idea is to get an erect, magnified and virtual image of the object. For this the object is placed between F and P on one side of the lens and viewed from other side of the lens. There are two magnifications to be discussed for two kinds of focusing. 1. Magnification in near point focusing m = 1 + \(\frac{D}{f}\) 2. Magnification in normal focusing (angular magnification), m = \(\frac{D}{f}\) |
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| 119. |
The angular magnification of a simple microscope can be increased by increasing (a) focal length of lens (b) size of object(c) aperture of lens (d) power of lens |
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Answer» (b) size of object |
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| 120. |
In refraction, light waves are bent on passing from one medium to the second medium, because in the second medium. (a) the frequency is different (b) the coefficient of elasticity is different (c) the speed is different (d) the amplitude is smaller |
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Answer» (c) the speed is different Speed of light in second medium is different than that in first medium. |
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| 121. |
A glass slab (μ = 1.5) of thickness 6 cm is placed over a paper. Qhat is the shift in the letters?(a) 4 cm (b) 2 cm (c) 1 cm (d) None of these |
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Answer» (b) 2 cm Normal shift, x = t (1- \(\frac{1}{μ}\)) = 6(1- \(\frac{1}{1.5}\)) cm =2 cm |
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| 122. |
A convex lens of 40 cm focal length is combined with a concave lens of focal length 25 cm. The power of combination is(a) -1.5 D (b) – 6.5 D (c) + 6.6 D (d) + 6.5 D |
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Answer» (a) -1.5 D P = P1 + P2 = \(\frac{1}{f_1}\) + \(\frac{1}{f_2}\) = \(\frac{100}{40}\)+ \(\frac{100}{-25}\) = -1.5D |
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| 123. |
An object is placed in front of a convex mirror of focal length of/and the maximum and minimum distance of an object from the mirror such that the image formed is real and magnified.(a) 2ƒ and c (b) c and ∞ (c) ƒ and O (d) None of these |
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Answer» (d) None of these There is no maximum minimum object distance for convex mirror to form real and inverted image. |
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| 124. |
When a plane electromagnetic wave enters a glass slab, then which of the following will not change?(a) Wavelength (b) Frequency(c) Speed (d) Amplitude |
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Answer» (b) Frequency Only the frequency of the electromagnetic wave remains unchanged. |
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| 125. |
For a real object, which of the following can produce a real image? (a) plane mirror (b) concave lens (c) convex lens (d) concave mirror |
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Answer» (d) concave mirror Only concave mirror produces real image provided the object is not placed between its focus and pole. |
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| 126. |
Experience shows that a body irradiated with light with circular polarization acquires a torque. This happens because such a light possesses an angular momentum whose flow density in vacuum is equal to `M = I//omega`, where `I` is the intensity o flight, `omega` is the angualr oscillation frequacny. Suppose light with circualr polarization and wavelength `lambda = 0.70 mu m` falls normally on a uniform black disc of mass `m = 10 mg` which can freely rotate about its axis. How soon will its angualr velocity become equal to `omega_(0) = 1.0 rad//s` provided `I = 10 W//cm^(2)`? |
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Answer» Let `r =` radius of the disc then its moment of inertia about its axis `= (1)/(2)mr^(2)` In time `t` the disc will acquire an angualr momentum `t.pi r^(2).(I)/(omega)` when circularly polarized light of intensity `I` falls on it. By conservation of angualr momentum this must equal `(1)/(2)mr^(2)omega_(0)` where `omega_(0) =` final angualr velocity. Equating `t = (m omega omega_(0))/(2pi I)` But `(omega)/(2pi) = v = (c)/(lambda)` so `t = (mc omega_(0))/(I lambda)` Subsituting the values of the various quantities we get `t = 11.9 hours` |
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| 127. |
When a ray of light enters from one medium to another, then which of the following does not change? (a) Frequency (b) Wavelength (c) Speed (d) Amplitude |
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Answer» (a) Frequency Only frequency remains unchanged |
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| 128. |
An object of length 2.5 cm is placed at the principal axis of a concave mirror at a distance 1.5f. The image height is(a) + 5 m (b) -5 cm (c) – 10 cm (d) + 1 cm |
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Answer» (b) -5 cm m = \(\frac{f}{f-u} = \frac{f}{f-1.5f}\) = -2cm Height image = m x height of object -2 x 2.5 = – 5 cm |
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| 129. |
Up to what maximum potential will a copper ball, remote from all other bodies, be charged when irradiated by electromagnetic radiation of wavelength `lambda = 140 nm`? |
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Answer» When light of sufficiently short wavelength falls on the ball, photoelectrons are ejected and the copper ball gains positive change. The changes ball tends to resist further emission of electorns by attracting them. When the copper ball has enough change even the most energetic electrons are unable to leave it. We can calculate this final maximum potential of the copper ball. It is obviously equal in magnitude (in volt) to the maximum `K.E.` of electrons (in electron volts) initially emitted. Hence `varphi_(max) = (2picancelha c)/(lambdae) - A_(cu)` `= 8.86 - 4.47 = 4.39` volts `(A_(cu)` is the work function of copper.) |
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| 130. |
Which of the following mirror is used by a dentist to examine a small cavity? (a) Concave mirror (b) Convex mirror (c) Combination of (a) and (b) (d) None of these |
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Answer» (a) Concave mirror A concave mirror, because it forms erect and enlarged image when held closer to the cavity. |
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| 131. |
Statement-1 : A dentist uses a concave mirror to examine a small cavity. Statement-2 : A dentist uses a concave mirror so as to form a magnified, virtual image of an object.A. Statement-1 is true, Statement-2 is true. Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is fasle, Statement-2 is true. |
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Answer» Correct Answer - A Both the statement are true and statement-2 is correct explanation of statement-1. |
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| 132. |
Is angular magnification of a telescope equal to ratio of diameters of objective and eye lens ? |
| Answer» Correct Answer - Yes, it is true. | |
| 133. |
What is the eye ring of a telescope or microscope ? |
| Answer» The image of the objective lens formed by the eye piece is called eye ring. This is the ideal position of the eye for observing the image. | |
| 134. |
If a light beam shows no intensity variation when transmitted through a polarised, which is rotated, does it mean that light is unpolarized ? |
| Answer» No, not necessarily. For example, consider light which is made up of electric field components `E_(x), E_(y)` with a `90^(@)` phase difference but equal amplitudes. The tip of electric vector executes uniform circular motion at the frequency of the light itself. This kind of light is said to be circularly polarized. When such light is passed through a polaroid, which is rotated, the transmitted average intensity remains constant. | |
| 135. |
Can electromagnetic waves be polarised ? |
| Answer» Yes, because electromagnetic waves are transverse in nature. | |
| 136. |
For the same objective, the ratio of least separation between two points to be distiguised by a microscope for light of `5000 Å` and electrons acclerated through 100V used as illuminating substance is __________ (neraly) |
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Answer» The limit of resolution of a microscope is `d = (lambda)/(2 sin theta)` where `2 theta` is the angle of cone of light rays entering the objective of the microscope. For light, `lambda_(1) = 5000 Å` For electrons accelerated through `100 V`, de Broglie wavelength, `lambda_(2) = (12.27)/(sqrt(100))Å = 1.227 Å`. ltbr? As `(2 sin theta)` is same in the two cases, therefore, `(d_(1))/(d_(2)) = (lambda_(1))/(lambda_(2)) = (5000)/(1.227) = 4075` |
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| 137. |
Consider sunlight incident on a slit of width `10^(4) Å` . The image seen through the slit shallA. be a fine sharp slit white in colour at the centreB. a bright slit white at the centre diffusing to zero intensities at the edgesC. a bright slit white at the centre diffusing to regions of different coloursD. only be a diffused slit white in colour |
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Answer» Correct Answer - A Here, the width of the slit is `10^(4) Å`, i.e., `10000 Å`. The wavelength of (visible) sunlight varies from `4000 Å` to `8000 Å`. As width of slit `a gt lambda` (wavelength of light), therefore, no diffraction occurs. The image seen through the slit shall be a fine sharp slit white in colour at the centre. Choice (a) is correct. |
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| 138. |
Consider the diffraction pattern for a small pinhole. As the size of the hole is increasedA. the size decreasesB. the intensity increasesC. the size increasesD. the intensity decreases |
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Answer» Correct Answer - A::B As size of hole (a) is increased, width of central maximum of diffraction pattern of hole `(= (2 f lambda)/(a))` decreases. As the same amount of light emergy is now distributed over a smaller area, the intensity increases. Choice (a) and (b) are correct. |
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| 139. |
Consider sunlight incident on a pinhole of width `10^(3)Å`. The image of the pinhole seen on a screen shall beA. a sharp white ringB. different from a geomatrical imageC. a diffused central spot, white in colourD. diffused coloured region around a sharp central white spot |
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Answer» Correct Answer - B::D As width of pinhole is `10^(3) A = 1000 Å` and wavelength of sunlight ranges from `4000 Å` to `8000 Å`, therefore, sunlight is diffracted on passing through the pin hole. The image of the pinhole seen on the screen shall be different from a geometrical image. Infact, the image of pinhole will consists of a diffused coloured region around a sharp central white spot. The choice (b) and (d) are correct. |
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| 140. |
Two polaroids are placed at `90^(@)` to eachother. What happens when `(N - 1)` more polaroids are inserted between them ? Their axes are equally spaced. How does the transmitted intensity behave for large `N` ? |
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Answer» When `(N - 1)` polaroids are inserted between two polaroids total number of polaroids `=(N + 1)`. The axes of all polaroids are equally spaced. If `theta` is the angle between the axes of any two consective polaroids, then `theta + theta + theta + .......N times = 90^(@) = pi//2` `N theta = pi//2` `theta = pi//2 N`. According to Law of Malus, intensity of light passing through a given pair of polaroids is proportinal to `cos^(2) theta`.In the given case, this change is repeated `N` times. If `I_(0)` is intensity of incident light and `I` is intensity of light coming out of last polaroid, then `I = I_(0) (cos^(2) theta)^(N) = I_(0) (cos theta)^(2N)` `= I_(0) "cos"((pi)/(2 N))^(2 N)` When `N` is very large, `theta = (pi)/(2N)` approaches zero and `cos (pi//2 N)` approaches `1`. Hence when `N` is very large, I will approach `I_(0)`. |
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| 141. |
Explain the following features of compton scattering of light by matter: (a) the increase in wavelength `Delta lambda` is independent of the nature of the scattering substance, (b) the intensity of the displaced component of scattered light grows with the increasing angle of scattering and with the diminishing atomic number of the substance, (c) the presence of a non-displaced in the scattered radiation. |
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Answer» (a) Compton scattering is the scattering of light by free electrons. (The free electrons are the electrons whose binding is much smaller than the typical energy transfer to the electrons). For this reason the increase in wavelength `Delta lambda` is independent of the nature of the scattered substance. (b) This is because the effective number of free electrons increase in both cases. with increasing angle of scattering, the energy transfered to electrons increases. With diminishing atomic number of the substance the binding energy of the elecrtons decreases. (c) The presence of a non-dispalced component in the scattered radiation is due to scattering from strong bound (inner) electrons as well as nuclei. For scattering by these the atom essentially recoils as a whole and there is very little enrgy transfer. |
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| 142. |
A screen is placed `80 cm` from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by `10 cm`. Calculate the focal length of the lens used. |
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Answer» Correct Answer - `19.7 cm` Here, `D = 80 cm, d = 10 cm, f = ?` `f = (D^(2) - d^(2))/(4D) = (80^(2) - 10^(2))/(4 xx 80) = (6300)/(320) = 19.7 cm` |
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| 143. |
The near and far points of a person are at `40 cm` and `250cm` respectively. Find the power of the lens he/she should use while reading at `25cm`. With this lens on the eye, what maximum distance is clearly visible?A. `2.5 D`B. `5.0 D`C. `1.5 D`D. `3.5 D` |
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Answer» Correct Answer - C In the eye of the person there is a defect for seeing the near objects as well as far off objects. He intends to read a book keeping it at a distacne `25 cm` from the eye. The lens used should be such that the image of this object by lens should be formed at distance `40 cm` from eye lens, so that the final image formed by eye lens is very clearly seen. Taking refaction from lens used, we have `u = - 25 cm, v = - 40 cm` , The lens formula is `(1)/(f) = -(1)/(u) + (1)/(v)` or `(1)/(f) = -(1)/(-25) + (1)/(-40) = (1)/(25) - (1)/(40) = (3)/(200)` or `f = (200)/(3) cm = (2)/(3)m` Power of lens `P = (1)/(f) = (1)/(2//3) = + 1.5 D` |
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| 144. |
What is the shape of the wavefront in each of the following cases ? (a) light diverging from point source. (b) light emerging out of a convex lens when a point source is placed at its focus. ( c) the portion of the wavefront of light from a distant star intercepted by earth. |
| Answer» The wavefront on earth for the sunlight (treated as point source) will be spherical. However, as distance of earth from the sun is huge `( ~~ 10^(11)m)`, portion of the wavefront on earth can be treated as plane wavefront. | |
| 145. |
A thin prism of refracting angle `2^@` deviates an incident ray through an angle of `1^@`.Find the value of refractive index of material of prism. |
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Answer» Here, `A = 2^(@), delta = 1^(1), mu = ?` As prism is thin, therefore, `delta = (mu -1)A` `1 = (mu - 1)2, mu = 1 + (1)/(2) = (3)/(2) = 1.5` |
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| 146. |
A ray of light incident on an equilateral triangular glass prism of `mu = sqrt(3)` moves parallel to the base of the prism inside it. What is the angle of incidence for this ray ? |
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Answer» When refracted ray is parallel to the bas eof the prism, deviation is minimum. `:. r = A//2 = 60//2 = 30^(@)` From `mu = (sin i)/(sin r)` `sin I = mu sin r = sqrt(3) sin 30^(@) = sqrt(3)//2` `:. I = 60^(@)` |
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| 147. |
The diameter of sun is several hundred times bugger than the moon, still at the time of solar eclips, the entire sun is covered by the moon. How ? |
| Answer» This is because distance of the sum from earth is `1000` times greater than the distance of the moon from earth. At the time of solar eclipse, moon comes between sun and earth. The angles subtended by the sun and the moon at the eye are nearly equal. | |
| 148. |
The refractive index of the material of the prism for a monochromatic beam of light is 2 and its refracting angle is `60^(@)` . The angle of incidence corresponding to which this beam of light suffers minimum deviation is :A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `75^(@)` |
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Answer» Correct Answer - A |
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| 149. |
Light of wavelength `5 xx 10^(-7)m` is diffracted by an aperture of width `2 xx 10^(-3)m`. For what distance travelled by the diffrated beam does the spreading due to diffraction become greater than width of the aperture ? |
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Answer» Here, `lambda = 5 x 10^(-7)m`, `a = 2 xx 10^(-3)m` Fresnel Distance, `Z_(F) = (a^(@))/(lambda) = (2 xx 10^(-3))^(2)/(5 xx 10^(-7))` `= 8m` |
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| 150. |
Light of wavelength `5000Å` is diffrated by an aperture of width `2mm`. For what distance by the diffracted beam does the spreading due to diffraction become greater than the width of the aperture ? |
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Answer» Here, `lambda = 5000Å = 5 xx 10^(-7)m, a = 2mm = 2 xx 10^(-3)m, Z_(F) = ?` As Fresnel distance, `Z_(F) = (a^(2))/(lambda) :. Z_(F) = (2 xx 10^(-3))^(2)/(5 xx 10^(-7)) = (40)/(5) = 8 m` |
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