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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.
| 701. |
In a primary rainbow, what is the order of colours ? And what is true for secondary rainbow ? |
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Answer» Outer arc of primary rainbow is red and inner arc is violet. In secondary rainbow, the order of colours is just the reverse of colours of primary rainbow. |
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| 702. |
Light of wavelength `600 nm` is incident normally on a slit of width `3 mm`. Calculate linear width of central maximum on a screen kept `3 m` away from the slit. |
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Answer» Correct Answer - `1.2 mm` Here, `lambda = 600 nm = 6 xx 10^(-7)m`, `a = 3 mm = 3 xx 10^(-3)m` `D = 3 m , (2x) = ?` `2x = (2 lambda D)/(a) = (2 xx 6 xx 10^(-7) xx 3)/(3 xx 10^(-3))` `= 12 xx 10^(-4)m = 1.2 mm` |
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| 703. |
Half the surface of a transparent sphere of refractive index 2 is silvered. A narrow, parallel beam of light is incident on the unsilvered surface, symmetrically with respect to the silvered part. The light finally emerging from the sphere will be a (a) parallel beam (b) converging beam (c) slightly divergent beam (d) widely divergent beam |
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Answer» Correct Answer is: (a) parallel beam |
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| 704. |
Figure illustrates an arrangement employed in observations of diffraction of light by ultrasound. A plate light wave with wavelength `lambda = 0.55 mum` passes through the water-filled tank `T` in which a standing ultrasonic wave is sustained at a frequency `v = 4.7MHz`. As a result of diffraction of light by the optically inhomogemeous periodic structure a diffraction spectrum can be observed in the foacl plane of the objective `O` with focal length `f = 35 cm`. The separation between neighbouring maxima is `Deltax = 0.60mm`. Find the propegation velocity of ultrasonic oscillations in water. |
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Answer» When standing ulta sonic waves are sustained in the tank it behaves like a grating whose grating element is `d=(v)/(v) =` wavelength of the ultrasonic `v=` velocity of ultrasonic. Thus for maxima `(v)/(v)sin theta_(m) = m lambda` On the other hand `f tan theta_(m) = m Deltax` Assuming `theta_(m)` ti be small (beacuse `lambda lt lt (v)/(v))` we get `Deltax = (f tan theta_(m))/(m) = (f tan theta_(m))/((v)/(v lambda) sin theta_(m)) = (lambda v f)/(v)` or `v =(lambda v f)/(Deltax)` Putting the values `lambda = 0.55mum, v = 4.7MHz` `f = 0.35m `and `Deltax = 0.60 xx 10^(-3)m` we easily get `v=1.51km//sec`. |
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| 705. |
Why do we not encounter diffraction effects of light in everyday observations ? |
| Answer» This is because objects around us are much bigger in size compared to the wavelength of visible light `(~= 10^(-6)m)`. | |
| 706. |
A lens shown in Fig. 6(b).75 is made of two different materials. A point object is placed on the principal axis of the lens. How many images will be obtained ? |
| Answer» As the lens is made of two different materials, it has two refractive indices and hence two different focal lengths. Hence two distinct images will be obtained. | |
| 707. |
A convex lens forms the image of the sun at a distance of `10 cm`. Where will be the image when (i) another lens of same power but dounle the aperture is used ? (ii) another lens of same aperture but double the power is used ? |
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Answer» (i) At `10 cm` only, because aperture does not affect `f`. (ii) `5 cm` because when `P` is doubled, `f` is halved. |
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| 708. |
Assertion : The sun looks bigger in size at sunrise and sunset than during day. Reason : In detraction light rays bend around the edges of the obstacle. (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. |
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Answer» (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. |
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| 709. |
Assertion : No diffraction is produced in sound waves near a very small opening. Reason : For diffraction to take place, the aperture of opening should be of same order as wavelength of waves.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 - A Wavelength of sound is of the order of one metre , and size of opening is much smaller. Therefore, sound is not diffracted. Choice (a) is correct. |
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| 710. |
Assertion: Colours can be scan in thin layers of oil on the surface water. Reason: White light is composed of several colours. (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. |
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Answer» (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. |
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| 711. |
Assertion : Colours can be seen in thin layers of oil on the surface of water. Reason : White light is composed of several colours.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 Colours are seen in thin layer of oil on the surface of water, because of interference of light. Reason is otherwise correct, but not a true explaination of the assertion. |
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| 712. |
If a plane glass slab is placed on letters of different colours, then red coloured letter appears to be raised minimum, why ? |
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Answer» The apparent normal shift `= t(1 - (1)/(mu))` As `mu`, is lowest among visible colours, the red coloured letter appears to be raised minimum. |
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| 713. |
Two plane mirrors are inclined to each other at an angle of 60°. A point object is placed in between them. The total number of images produced by both the mirror is(a) 2 (b) 4 (c) 5 (d) 6 |
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Answer» (c) 5 Number of images formed, θ =\(\frac{360°}{θ}\) – 1 = \(\frac{360}{60}\) = 1 = 5. |
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| 714. |
Two mirrors are kept at 60° to each other and a body is placed at middle. The total number of images formed is (a) six (b) four (c) five (d) three |
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Answer» (c) Five Number of images formed, n =\(\frac{360}{θ}\) -1 = \(\frac{360}{60}\)-1 = 5 |
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| 715. |
What are the two main applications of parabolic mirrors ? |
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Answer» As is known, the shape of parabolic mirror is such that all light rays falling on it in a direction parallel to the principal aixs are reflected through a single image point, irrespective of their distance from the principal axis. Two main applications of these mirrors are : (i) In solar-thermal electric plant, focussed rays from the sun are used to generate steam, which drives a turbine connected to an electric generator. (ii) In automobile head lights, a high intensity light bulb is held at the focus of parabolic mirror, and strong light emerges in a direction parallel to the principal axis of the mirror. |
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| 716. |
A tank is filled with water to a height of `12.5 cm` The apparent depth of a needle lying at the bottom of the tank is measured by a microscope to be `9.4cm`. What is the refractive index of water ? If water is replaced by a liquid of refractive index `1.63` upto the same height, by what distance would the microscope have to be moved to focus on the needle again ? |
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Answer» Here, real depth `= 12.5 cm` , apparent depth `= 9.4 cm , mu = ?` As `mu = ("real depth")/("apparent depth") :. mu = (12.5)/(9.4) = 1.33` Now, in the second case, `mu = 1.63`, real depth `= 12.5 cm` , apparent depth, `y = ?` `:. 1.63 = (12.5)/(y)` `y = (12.5)/(1.63) = 7.67 cm` `:.` Distance through which microscope has to be moved up `= 9.4 - 7.67 = 1.73 cm` |
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| 717. |
At what distance should an object be placed from a convex lens of focal langth `15 cm` to obtain an image three times the size of the object ? |
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Answer» Here, `u = ?, f = 15cm, m = +-3 = (v)/(u)` `:. v = +-3 3u` From `(1)/(v) - (1)/(u) = (1)/(f)` `(1)/(+- 3u) - (1)/(u) = (1)/(15)` `:. u = 20 cm`, when `m` is negative, `u = 10cm`, when `m` is positive, i.e., for virtual image, `u = 10 cm` and for reqal image, `u = 20cm` |
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| 718. |
A telescope has an objective of focal length `30 cm` and an eye piece of focal length `3.0 cm`. It is focussed on a scale distant `2.0 m`. For seeing with relaxed aye, calculate the separation between the objective and eye piece. |
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Answer» Correct Answer - `37.97 cm` Here, `f_(0) = 30 cm, f_(e) = 3 cm, u_(0) = - 200 cm`. Separation between the lenses `= (v_(0) + |u_(0)|)` From `(1)/(v_(0)) - (1)/(u_(0)) = (1)/(f_(0))` `(1)/(v_(0)) = (1)/(f_(0)) + (1)/(u_(0)) = (1)/(30) - (1)/(200) = (20 - 3)/(600) = (17)/(600)` `v_(0) = (600)/(17)cm = 35.29 cm` From `(1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e))` `(1)/(u_(e)) = (1)/(v_(e)) - (1)/(f_(e)) = (1)/(-25) - (1)/(3) = (-28)/(75)` `u_(e) = (-75)/(28) = - 2.68 cm` `:.` Separation between the lenses `= 35.29 + 2.68 = 37.97 cm` |
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| 719. |
A gaint refrecting telescope at an observatory has an objective lens of focal length `15 m`. If an eye piece lens of focal length `1 cm` is used, find the angular magnification of the telescope. If this telescope is used to view the moon, what is the diameter of image of moon formed by objective lens ? The diameter of the moon is `3.42 xx 10^6 m` and radius of lunar orbit is `3.8 xx 10^8 m`. |
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Answer» Correct Answer - `1500 ; 13.5 cm` Here, `f_(0) = 15 cm, f_(e) = 1 cm = (1)/(100)m` Angular magnification `= (f_(0))/(f_(e)) = (15)/(1//1100) = 1500` Diameter of image of moon formed by objective lens `= alpha f_(0) = (3.42 xx 10^(6))/(3.8 xx 10^(8)) xx 15 m = 0.135 m` `= 13.5 cm` |
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| 720. |
A refracting telescope has an objective of focal length `1 m` and an eye piece of focal length `20 cm`. The final image of the sun `10 cm` in diameter is formed at a distance of `24 cm` from eye piece. What angle does the sun subtend at the objective ? |
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Answer» Correct Answer - `0.0455 rad` Here, `f_(0) = 1 m = 100 cm, f_(e) = 20 cm`. If `h_(1) =` size of image formed by objective lens and `h_(2) =` size of image formed by eye piece `tan alpha = (h_(1))/(f_(0)) = (h_(1))/(100)` `m_(e) = (h_(2))/(h_(1)) = 1 + (d)/(f_(e))` `(10)/(h_(1)) = 1 + (24)/(20) = (44)/(20) h_(1) = (200)/(44) = (50)/(11) cm` `alpha ~= tan alpha = (h_(1))/(100) = (50)/(1100) = (1)/(22)rad = 0.0455 rad` |
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| 721. |
An astronimical telescope is to be designed to hve a magnifying power of 50 in normal adjustment. If the length of the tube is 102 cm, fid the powers of the objective and the eyepiece. |
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Answer» Correct Answer - `1 D` ; `50 D` Here, `m = - 50, L = 102 cm` As `m = -(f_(0))/(f_(e)) = - 50, f_(0) = 50 f_(e)` Also, `L = f_(0) + f_(e) = 102` , `50 f_(0) + f_(e) = 102, f_(e) = 2 cm` `f_(0) = 50 f_(e) = 50 xx 2 = 100 cm` `P_(0) = (100)/(f_(0)) = (100)/(100) = 1 D, P_(e) = (100)/(f_(e)) = (100)/(2) = 50 D` |
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| 722. |
What is angle of deviation due to refraction? |
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Answer» The angle between the incident and deviated light is called angle of deviation. When light travels from rarer to denser medium it deviates towards normal. The angle of deviation in this case is, d = i – r |
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| 723. |
What is the vlaue of refractive index of a medium of polarising angle `60^(@)` ?A. `1.732`B. `1`C. `1.414`D. `2` |
| Answer» Correct Answer - A | |
| 724. |
A plane electromagnetic wave propagates in a medium moving with constant velocity `Vlt lt c` relative to an interial frame `K`. Find the velocity of that wave in the frame `K` if the refractive index of the medium is equal to `n` and the propagation direction of the wave coincides with that of the medium. |
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Answer» In vaccum inertial frames are all equivalent, the velocity of light is `c` in any frame. This equivalence of inertial frames does not hold in material media and here the frame I which the medium is at rest is singled out. It is in his frame that the velocity of light is `(c )/(n)` where `n` is the refractive in desc of light for that medium. The velocity of light in the frame in which the medium is moving is then by the law of addition of velocities `((c)/(n)+v)/(1+(c)/(n).v//c^(2)) = ((c)/(n)+v)/(1+(v)/(cm)) = ((c)/(n)+v) (1-(v)/(cn)+.......)` `=(c)/(n)+v-(v)/(n^(2)) +.... = (c)/(n)+v (1-(1)/(n^(2)))` This is the velocity of light in the medium in a frame in which the medium is moving with velocity `v lt lt c`. |
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| 725. |
A mark is made on the bottom of beaker and a microscope is focussed on it. The microscope is raised through `1.5 cm`. To what height water must be poured into the beaker to bring the mark again into focus ? Given that `mu` for water is `4//3`. |
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Answer» Correct Answer - `6 cm` Here, apparent shift, `d = 1.5 cm` Let `t` be the height which water must be poured into the beaker. As `d = t (1 - (1)/(mu))` `:. 1.5 = t (1 - (1)/(4//3)) = (t)/(4) , t = 4 xx 1.5 = 6 cm` |
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| 726. |
A small ink dot on a paper is seen through s glass slab of thickness `4 cm` and refractive index `1.5`. The dot appears to be raised byA. `1 cm`B. `2 cm`C. `3 cm`D. `1.33 cm` |
| Answer» Correct Answer - D | |
| 727. |
When an object is placed at a distance of `60 cm` from a convex spherical mirror, the magnification produced is `1//2`. Where should the object be placed to get a magnification of `1//3` ? |
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Answer» Here, `u_(1) = -60 cm, m_(1) = (1)/(2)` `u_(2) = ?, m_(2) = (1)/(3)` As `m_(1) = -(v_(1))/(u_(1)) :. (1)/(2) = -(v_(1))/(-(-60)), v_(1) = 30 cm` From `(1)/(f) = (1)/(v_(1)) + (1)/(u_(1)) = (1)/(30) - (1)/(60) = (1)/(60)` `f = 60 cm` Again, `m_(2) = -(v_(2))/(u_(2)) = (1)/(3), v_(2) = -(u_(2))/(3)` From `(1)/(v_(2)) + (1)/(u_(2)) = (1)/(f)` `(1)/(u_(2)) = (1)/(f) - (1)/(v_(2)) = (1)/(60) + (3)/(u_(2))` or `(1)/(u_(2)) - (3)/(u_(2)) = (1)/(60), u_(2) = -120 cm` |
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| 728. |
When an object is held between pole and focus of a concave mirror, the image formed is.A. real and invertedB. real and enlargedC. virtual and diminishedD. virtual and enlarged |
| Answer» Correct Answer - D | |
| 729. |
Explain why (a) A diamond glitters in a brightly lit room, but not in a dark romm. (b) A crack in window pane appears silvery. (c ) The bubbles of air rising up in a water tank appear silvery when viewed from top. |
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Answer» (a) A daimond glitters in a brightly lit room because light entering the diamond from any face suffers multiple reflections and does not come out. The diamond appears illuminated from inside. This would not happen in a dark room. (b) A crack in a window pane appears silvery on account of total internal reflection on light in the crack. (c ) The bubbles of air rising up in a water tank appear silvery when viewed from top again on account of total internal reflection of light from the bubble. |
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| 730. |
Define radius of curvature of a spherical mirror. |
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Answer» Radius of the sphere of which a mirror is a part is called as radius of curvature of the mirror. |
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| 731. |
Two waves of amplitudes `3mm and 2 mm` reach a point in the same phase. What is the resultant amplitude ? |
| Answer» Resultant amplitude `= 3 + 2 = 5 mm`. | |
| 732. |
Two waves of amplitude `5 mm and 7 mm` reach a point in opposite phase. What is the resultant amplitude ? |
| Answer» Resultant amplitude `= 7 - 5 = 2 mm` | |
| 733. |
If two waves of same frequency and same amplitude superimpose and produce third wave of same amplitude, then waves differ in phase by –A. `pi//3`B. `2pi//3`C. `pi//2`D. `pi` |
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Answer» Correct Answer - B |
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| 734. |
Phase difference between two waves having same frequency `(v)` and same amplitude `(A)` is `2pi//3`. If these waves superimpose each other, then resultant amplitude will be–A. `2A`B. `0`C. `A`D. `A^(2)` |
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Answer» Correct Answer - C |
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| 735. |
Laser light is considered to be coherent because it consists ofA. Many wavelengthsB. Uncoordinated wavelengthC. Coordinated waves of exactly the same wavelengthD. Divergent beam |
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Answer» Correct Answer - C |
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| 736. |
Light of wavelength `lambda` is incident on a slit of width d and distance between screen and slit is D. Then width of maxima and width of slit will be equal if D is ––A. `(d^(2))/(lambda)`B. `(2d)/(lambda)`C. `(2d^(2))/(lambda)`D. `(d^(2))/(2lambda)` |
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Answer» Correct Answer - D |
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| 737. |
Assertion : The frequencies of incident, reflected and refracted beam of monochromatic light are same. Reason : The incident, reflected and refracted rays are coplanar.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 Assertion is true, because frequency in all the three beams is dependent on source of light. The reason is also true, but it is not a correct explaination of the assertion. |
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| 738. |
A beam of monochromatic light is refracted from vacuum into a medium of refractive index 1.5. The wavelength of refracted light will be(a) same (b) dependent on intensity of refracted light (c) larger (d) smaller |
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Answer» (d) smaller As light enters the medium, its wavelength decreases and becomes equal to \(\frac{λ}{μ}\) |
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| 739. |
When monochromatic light travels from one medium to another, its wavelength changes, but its frequency remains the same. Why ? |
| Answer» Frequency is the characteristic of the source. That is why it remains the same. But wavelength is charactristic of the medium. So wavelength and veloctiy both change. | |
| 740. |
Radius of curvature of an equiconvex lens is `0.2 m`. Its refractive index is `1.5`. Calculate its focal length. If two such lenses are kept separated with common principal axis by a distance of `0.2 m`, what will be the effective focal length of the combination ? |
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Answer» Here, `R_(1) = 0.2 m = 20 cm`, `R_(2) = -20 cm, mu = 1.5` , `f = ? D = 0.2 m = 20cm, F = ?` As, `(1)/(f) = (mu - 1)((1)/(R_(1)) - (1)/(R_(2)))` `:. (1)/(f) = (1.5 - 1)((1)/(20) + (1)/(20)) = 0.5 xx (1)/(10) = (10)/(20)` `:. f = 20 cm` Now, `f_(1) = f_(2) = 20 cm` As, `(1)/(F) = (1)/(f_(1)) + (1)/(f_(2)) - (d)/(f_(1)f_(2))` `:. (1)/(F) = (1)/(20) + (1)/(20) - (20)/(20 xx 20) = (1)/(20), F = 20 cm` |
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| 741. |
An object is placed at a distance of `16 cm` from a convex mirror of focal length `20 cm`. Locate the position and nature of the image. |
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Answer» Here, `u = -16 cm, f = +20 cm, v = ?` `(1)/(v) = (1)/(f) - (1)/(u) = (1)/(20) - (1)/(-16) = (16 + 20)/(320) = (36)/(320)` `v = (320)/(36) = (80)/(9) = 8.9 cm` As `v` is positive, therefore, image is virtual, erect and is formed behind the mirror. |
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| 742. |
The focal length of an equiconvex lens placed in air is equal to radius of curvature of either surface. Is it true ? |
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Answer» Yes, `f = R` only when `mu = 1.5`. This follow from `(1)/(f) = (mu -1)((1)/(R_(1)) - (1)/(R_(2)))`. If `mu != 1.5`, the `f != R` |
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| 743. |
For the same angle of incidence, the angles of refraction in media `P, Q and R` are `35^(@) ,25^(@), 15^(@)` resp. In which medium will the velocity of light be minimum ? |
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Answer» For medium `R`, angle of refraction (r ) is minimum. Therefore, `mu = (sin i)/(sin r)` is maximum . As `mu = ( C)/(v)`, therefore, `v` is minimum, i.e., velocity of light is minimum in medium `R`. |
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| 744. |
Where should an object be placed from a convex lens to from an image of the same size ? Can it happen in case of a concave lens ? |
| Answer» The object has to be placed at a distance `2 f` from the lens. No, it cannot happen in case of a concave lens, as image formed is always smaller in size. | |
| 745. |
Statement I : Light reflected from any surface obeys the law of reflection. Statement II: Distance measured in the direction of light are taken as positive.A. If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I.B. If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I.C. If Statement - I is true but Statement - II is false.D. If Statement - I is false but Statement - II is true. |
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Answer» Correct Answer - B |
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| 746. |
Statement I : Magnification of a convex mirror is always positive. Statement II: `m = -v/u` and convex mirror always forms virtual image.A. If both Statement- I and Statement- II are true, and Statement - II is the correct explanation of Statement– I.B. If both Statement - I and Statement - II are true but Statement - II is not the correct explanation of Statement – I.C. If Statement - I is true but Statement - II is false.D. If Statement - I is false but Statement - II is true. |
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Answer» Correct Answer - A |
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| 747. |
A beam of plane-polarized light falls on a polarizer which rotates about the axis of the ray with angular velocity `omega = 21rad//s`. Find the energy of light passing through the Polarizer per one revolution if the flux of energy of the incident ray is equal to `Phi_(0) = 4.0mW`. |
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Answer» When the polarizer rotates with angualr velocity `omega` its instantaneous principle direction makes angle `omegat` from a reference direction which we choose to be along the direction of vobration of the plane polarized incident light. The transmitted flux at this instant is `Phi_(0)cos^(2)omegat` and the total energy passing through the polarizer per revolution is `underset(0)overset(T)int Phi_(0)cos^(2)omegat dt, T = 2pi//omega` `= Phi_(0)(pi)/(omega) = 0.6mJ`. |
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| 748. |
Out of speed, frequency and wavelength, name the parameter (s) which remain the same on refraction. |
| Answer» Only frequency remains the same on refraction. | |
| 749. |
Out of speed, frequency and wavelength, name the parameter (s) which remain the same on reflection. |
| Answer» All the three parameters remain the same on reflection. | |
| 750. |
For a given incident ray, when a plane mirror is turned……….. The reflected ray turns through………. . |
| Answer» an angle `theta` , an angle `2 theta` | |