InterviewSolution
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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.
| 1. |
If `I_0` is the intensity of the principal maximum in the single slit diffraction pattern. Then what will be its intensity when the slit width is doubled?A. `4I_0`B. `2I_0`C. `(I_0)/(2)`D. `I_0` |
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Answer» Correct Answer - A `I=I_0((sinphi)/(phi))` and ` phi=(pi)/(lamda)(b sin theta)` When the slit width is doubled, the amplitude of the wave at the centre of the screen is doubled, so the intensity at the centre is increased by a factor 4. |
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| 2. |
A thin glass (refractive index 1.5) lens has optical power of `-5D` in air. Its optical power in a liquid medium with refractive index 1.6 will beA. `-1D`B. `1D`C. `-25D`D. `25D` |
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Answer» Correct Answer - B `(1)/(f_a)=((1.5)/(1)-1)((1)/(R)-(1)/(R))` ….(i) `(1)/(f_m)=((mu_g)/(mu_m)-1)((1)/(R_1)-(1)/(R_2))` `(1)/(f_m)=((1.5)/(1.6)-1)((1)/(R_1)-(1)/(R))` ….(ii) Dividing (i) by (ii), `(f_m)/(f_a)=((1.5-1)/((1.5)/(1.6)-1))=-8` `P_a=-5=(1)/(f_a)impliesf_a=-(1)/(5)` `impliesf_m=-8xxf_a=-8xx-(1)/(5)=(8)/(5)` `P_m=(mu)/(f_m)=(1.6)/(8)xx5=1D` |
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| 3. |
A light of wavelength `6000 A` in air, enters a medium with refractive index 1.5 Inside the medium its frequency is….Hz and its wavelength is ….`A` |
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Answer» Correct Answer - A::D Frequency remains the same `f=c/(lamda)=(3xx10^8)/(6000xx10^-10)=5xx10^14 Hz` and `lamda_2=(lamda_1)/(mu)=6000A/(1.5)=4000 A` |
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| 4. |
A point source emits sound equally in all directions in a non-absorbing medium. Two points `P` and `Q` are at the distance of `9 meters` and 25 meters respectively from the source. The ratio of amplitudes of the waves at `P` and `Q` is….. |
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Answer» Correct Answer - B `(I_1)/(I_2)=(A_1^2)/(A_2^2)` ….(i) But `I implies (I_1)/(I_2)=(r_2^2)/(r_1^2)` …(ii) From (i) and (ii) , `(A_1)/(A_2)=(r_2)/(r_1)=(25)/(9)`. |
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| 5. |
A convex lens of focal length 40 cm is in contact with a concave lens of focal length 25 cm. The power of the combination isA. `-1.5`dioptersB. `-6.5` dioptersC. `+6.5` dioptersD. `+6.67` diopters |
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Answer» Correct Answer - A `P=(1)/(f)=(1)/(f_(1))+(1)/(f_(2))=(1)/(0.4)+(1)/(-0.25)=-1.5`diopter. |
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| 6. |
A convex lens if in contact with concave lens. The magnitude of the ratio of their focal length is `(2)/(3)`. Their equivalent focal length is 30 cm. What are their individual focal lengths?A. `-15,10`B. `-10,15`C. `75,50`D. `-75,50` |
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Answer» Correct Answer - A `|f1|/|f2|=(2)/(3)` f1: focal length of convex lens. `(1)/(f)=(1)/(f2)-(1)/(f2)` `implies (1)/(30)=(1)(f1)=(2)/(3f1)` 1f_(1)=10 cm, `f_(2)=-15 cm` |
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| 7. |
Focal length of the plano-convex lens is 15 cm. A small object is placed at A as shown in the figure. The plane surface is silvered. The image will form at ` A. 60 cm to the left of lensB. 12 cm to the left of lensC. 60 cm to the right of lensD. 30 cm to the left of lens |
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Answer» Correct Answer - B The focal length f of the equivalent mirror is `(1)/(f)=(2)/(f1)=(1)/(f_m)=(2)/(15)+(1)/(infty)` `implies f=(15)/(2) cm` Since f has a positive value, the combination behaves as a converging mirror. Here `u=-20cm`, `f=-(15)/(2) cm`, `v=?` According to mirror formula `(1)/(v)+(1)/(u)=(1)/(f)` `implies (1)/(v)-(1)/(-20)=(1)/((-15)/(2)) implies v=-12cm` Negative sign indicates that the image is 12 cm in front of mirror. |
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| 8. |
A diminished image of an object is to be obtained on a screen 1.0 m from it. This can be achieved by appropriately placingA. a concave mirror of suitable focal lengthB. a convex mirror of suitable focal lengthC. a convex lens of focal length less than 0.25 mD. a concave lens of suitable focal length |
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Answer» Correct Answer - C NOTE: A convex mirror and a concave lens always produce virtual image. Therefore, option (b) and (d) are not correct. The image by a convex lens is diminished when the object is placed beyond 2f. Let`u=2f+x` Using `(1)/(v)-(1)/(u)=(1)/(f)` `implies (1)/(v)-(1)/(-(2f+x)=(1)/(f)` `(1)/(v)=(1)/(f)+(1)/(2f+x)=(2f+x-f)/(f(2f+x)=((f+x))/(f(2f+x)` But `u+v=1`(given) `(2f+x)+(f(2f+x))/(f+x)le1` `2f+x[1+(f)/(f+x)]le1` `implies ((2f+x))^2/(f+x) le1` `implies (2f+x)^2lef+x. The above is true for `flt 0.25 m`. |
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| 9. |
A beam of light of wave length 600 nm from a distance source fall on a single slit 1mm wide and a resulting. Diffraction pattern is observed on a screen 2m away. The distance between the first dark fringes on either side of central bright fringe isA. 1.2 cmB. 1.2 mmC. 2.4 cmD. 2.4 mm |
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Answer» Correct Answer - D The distance between the first dark fringe on either side of the central maximum=width of central maximum `=(2Dlamda)/(a)=(2xx2xx600xx10^-9)/(10^-3)=2.4xx10^-3 m=2.4 mm |
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| 10. |
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 ).B. C. D. |
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Answer» Correct Answer - D The electron beam will be diffracted and the maxima is obtained at `y=0`. Also the distance between the first minima on both side will be greater than d. |
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| 11. |
The box of a pin hole camera, of length L, has a hole of radius a . It is assumed that when the hole is illuminated by a parallel beam of light of wavelength `lamda` the spread of the spot (obtained on the opposite wall of the camera) is the sum of its geometrical spread and the spread due to diffraction. The spot would then have its minimum size (say b_(min)) when:A. `a=sqrt(lamdaL)` and `b_(min)=sqrt(4lamdaL)`B. `a=(lamda^2)/L` and `b_(min)=sqrt(4lamdaL)`C. `a=(lamda^(2))/(L)` and ` b_(min)=((2lamda^2)/(L))`D. `a=sqrt(lamdaI)` and `b_(min)=((2lamda)/(L))` |
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Answer» Correct Answer - A Given geometrical spread =a Diffraction spread `=(lamda)/(a)xxL=(lamdaL)/(a)` The sum `b=a+(lamdaL)/(a)` For b be minimum `(db)/(da)=0` `(d)/(da)(a+(lamdaL)/(a))=0` `a=sqrt(lamdaL)` `b_(min)=sqrt(lamdaL)+sqrt(lamdaL)=2sqrt(lamdaL)=sqrt(4lamdaL)` |
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| 12. |
An object 2.4 m in front of a lens forms a sharp image on a film 12 cm behind the lens. A glass plate 1 cm thick, of refractive index 1.50 is interposed between lens and film with its plane faces parallel to film. At what distance (from lens) should object shifted to be in sharp focus of film?A. 7.2 mB. 2.4 mC. 3.2 mD. 5.6 m |
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Answer» Correct Answer - D The focal length of the lens `(1)/(f)=(1)/(v)-(1)/(u)=(1)/(12)+(1)/(240)=(20+1)/(240)=(21)/(240)` `f=(240)/(21)`cm Shift `=t(1-(1)/(mu))implies1(1-(1)/((3)/(2)))=1xx(1)/(3)` Now `v^`=12-(1)/(3)=(35)/(3)`cm Now the object distance u. `(1)/(u)=(3)/(35)-(21)/(240)=(1)/(5)[(3)/(7)-(21)/(48)]` `(1)/(u)=(1)/(5)[(48-49)/(7xx16)]` `u=-7xx16xx5=-560`cm ``=-5.6m` |
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| 13. |
A light ray is incident perpendicularly to one face of a `90^circ` prism and is totally internally reflected at the glass-air interface. If the angle of reflection is `45^circ`, we conclude that the refractive index n A. `ngt(1)/(sqrt2)`B. `ngtsqrt2`C. `nlt(1)/(sqrt2)`D. `nltsqrt2` |
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Answer» Correct Answer - B The incident angle is `45^circ`. Incident anglegtcriticalandgle,igt`i_c` `sinigtsini_c`or `sin45gtsini_c` `sini_c=(1)/(n)` `sin45^circgt(1)/(n)`or`(1)/(sqrt2)gt(1)/(n)impliesngtsqrt2` |
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| 14. |
A ray of light travelling in a transparant medium falls on a surface separating the medium from air at an angle of incidence of `45degree`. The ray undergoes total internal reflection. If n is the refractive in index of the medium with respect to air, select the possible value (s) of n from the following:A. 1.3B. 1.4C. 1.5D. 1.6 |
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Answer» Correct Answer - C::D KEY CONCEPT: For total internal reflection to take place: Angle of incidence `igt` critical angle, `theta_(c)` or `sin45degreegt(1)/(n)` or `(1)/(sqrt2)gt(1)/(n)` or `ngtsqrt2` or `ngt1.414`. Therefore, possible values of n can be 1.5 or 1.6 in the given options. |
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| 15. |
A light ray travelling in glass medium is incident of glass- air interface at an angle of incidence `theta`. The reflected `(R )` and transmitted (T) intensities, both as function of `theta`, are plotted The correct sketch isA. B. C. D. |
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Answer» Correct Answer - C When the light is incident on glass - an interface at an angle less than critical angle a small part of light will be reflected and most part will be transmitted. When the light is incident greater than the critical angle, it gets completed reflected (total internal reflection) These characteristics are depicted in option (c). |
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| 16. |
Which of the following is used in optical fibres?A. total internal reflectionB. scatteringC. diffractionD. refracton |
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Answer» Correct Answer - A In an optical fibre, light is sent through the fiber without any loss by the phenomenon of total internal reflection as shown in the figure. |
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| 17. |
Consider telecommunication through optical fibres. Which of the following statements is not true?A. Optical fibres can be of graded refractive indexB. Optical fibres are subject to electromagnetic interference from outsideC. Optical fibres have extremely low transmission lossD. Optical fibre may have homogeneous core with a suitable cladding |
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Answer» Correct Answer - B Optical fibres form a dielectric wave guide and are free from electromagnetic interference or radio frequency interference. |
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| 18. |
An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation ` A. PQ is horizontalB. QR is horizontalC. RS is horizontalD. Any one will be horizontal |
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Answer» Correct Answer - B NOTE: For minimum deviation, incident angle is equal to emerging angle and QR is parallel to base. |
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| 19. |
A spherical surface of radius of curvature R separates air (refractive index 1.0) from glass (refractive index 1.5). The centre of curvature is in the glass. A point object P placed in air is found to have a real image Q in the glass. The line PQ cuts the surface at a point O, and `PO=OQ`. The distance `PO`A. 5RB. 3RC. 2RD. 1.5R |
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Answer» Correct Answer - A The formula for spherical refracting surface is `(mu_(1))/(u)+(mu_(1))/(v)=(mu_(2)-mu_(1))/R` Here `u=-x`,`v=+x`,`R=+R`,`mu_(1)=1.5` `(-1)/(-x)+(1.5)/(x)=(1.5-1)/R``implies``x=5R` |
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| 20. |
Electrimagnetic waves are transverse is nature is evident byA. polarizationB. interferenceC. reflectionD. differaction |
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Answer» Correct Answer - A The phenomenon of polarisation is shown only by transberse waves. |
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| 21. |
A thin slice is cut out of a glass cylinder along a plane parallel to its axis. The slice is placed on a flat glass plate as shown in Figure. The observed interference fringes from this combination shall be A. straightB. circularC. equally spacedD. having fringe spacing which increases as we go outwards |
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Answer» Correct Answer - A Locus of equal path difference are lines runnning parallel to axis of the cylinder. Hence straight fringes will be observed. |
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| 22. |
Yellow light is used in a single slit diffraction experiment with slit width of 0.6 mm. If yellow light is replaced by X-rays, then the observed pattern will reveal,A. that the central maximum is narrowerB. more number of fringesC. less number fringesD. no diffraction pattern |
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Answer» Correct Answer - D For diffraction pattern to be observed, the dimension of slit should be comparable to the wave length of rays. The wavelength of X-rays `(1-100A) is less than 0.6mm. |
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| 23. |
Two coherent point sources `S_1` and `S_2` are separated by a small distance `d` as shown. The fringes obtained on the screen will be A. pointsB. straight linesC. semi-circlesD. concentric circles |
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Answer» Correct Answer - D It will be concentric circles. |
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| 24. |
A source emits sound of frequency 600Hz inside water. The frequency heard in air will be equal to (velocity of sound in water`=1500(m)/(s)`, velocity of sound in air=300(m)/(s))A. 3000HzB. 120HzC. 600HzD. 6000Hz |
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Answer» Correct Answer - C NOTE: Frequency does not change with change of medium. |
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| 25. |
To deminstrate the phenimenon of interference, we require two sources which emit radiationA. of nearly the same frequencyB. of the same frequencyC. of different wavelengthsD. of the same frequency and having a definite phase relationship |
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Answer» Correct Answer - D For the phenomenon of interference we require two sources of light of same frequency and having a definite phase relationship (a phase relationship that does not change with time) |
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