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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. |
The resolving power of an astronomical telescope is 0.2 seconds. If the central half portion of the objective lens is covered, the resolving power will beA. `0.1 sec`B. `0.2 sec`C. `1.0 sec`D. `0.6 sec` |
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Answer» Correct Answer - A `R.P = (a)/(1.22lambda)` |
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| 2. |
A lady cannot see objects closer than `40cm` from the left eye and closer than `100cm` from the right eye. While on a mountaining trip, she is lose from her team. She tries to make an astronomical trip, from her reading glasses to look from her teammates. (a) Which glass should she use as the eyepiece? (b) What magnification can she get with relaxed eye? |
| Answer» Correct Answer - B | |
| 3. |
A compound microphone has an objective of focal length 1 cm and an eyepiece of focal length 2.5 cm. An object has to be placed at a distance of 1.2 cm awy from the objective for normal adjustment. a.Find the angular magnification. b.Find the length of the microscope tube. |
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Answer» a. If the first image is formed at a distance v form the objective we have `1/v-1/((-1.2cm))=1/(1cm)` or v=6cm The angular magnification in normal adjustment is `m=v/u D/(f_e)=-(6cm)/(1.2cm)=(25 cm)/(25cm)=-50. b. for normal adjustment the first image mut be in the focal planeof the eyepiece. the length of the tube is therefore `L=v+f_e=6cm+2.5cm=8.5cm`. |
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| 4. |
A small object is placed at distance of 3.6 cm from a magnifier of focal length 4.0 cm. a. find the position of theimage. B. find the linear magnification c. Find the angular magnification. |
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Answer» a. Using `1/v-1/u=1/f` 1/v=1/u+1/f=1-(-3.6cm)+1/(4.0cm)` or `v=-36cm` b. Linear magnification `=v/u` `9-36cm)/(-3.6cm)=10` c. If th object placed at a distance `u_0 form the lens, the angle subtended by the object on thelens is `beta=h/u_0` where h is the height of the object. The maximum angle subtended on the uN/Aided eye is `alpha=h/D` Thus, the angular magnification is `m=beta/alpha=D/u_0=(25cm)/(3.6cm)7.0` |
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| 5. |
In previous question, the magnifying power of the microscope isA. `10`B. `15`C. `20`D. `30` |
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Answer» Correct Answer - C `m = (v_(0))/(u_(0)) (1+(D)/(f_(e))) = (10)/(-2.5) (1+(25)/(25//4))` =`- 4xx5 =- 20` |
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| 6. |
A nearsighted man can clearly see objects up to a distance of 1.5m. Calculate the power of the lens of the spectacles necessary for the remedy of this defect. |
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Answer» The lens should form a virtul image of a distance object at 1.5 m form the lens. Thus, it should be a divergent lens andits focal length should be -1.5 m. Hence f-1.5 m `or P=1/f=-11.5m^-1=0.67D.` |
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| 7. |
The magnifying power of a compound microscope isA. objective magnification`//`eyepiece magnificationB. objective magnification`xx`eyepiece magnificationC. eyepiece magnification`//`objective magnificationD. objective magnification`+`eyepiece magnification |
| Answer» Correct Answer - B | |
| 8. |
(a) An object is seen through a simple microscope of focal length `12cm`. Find the angular magnification produced if the image is formed at the near point of the eye which is `25cm` away from it. (b) A `10D` lens is used as a magnifier. Where shold the object be placed to obtain maximum angular magnification for a nirmal eye (near point `= 25cm)`? |
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Answer» (a) `f = 12cm, D = 25 cm` `m = 1+(D)/(f) = 1+(25)/(12) = 3.08` (b) `P = 10 D` `f = (100)/(P) = (100)/(10) = 10 cm` For maximum angular magnification, image is formed at near point. `u =- u, v =- D =- 25 cm, f = 10 cm` `(1)/(v) - (1)/(u) = (1)/(f)` `(1)/(-25) - (1)/(-u) = (1)/(10)` `(1)/(u) = (1)/(10) +(1)/(25) = (5+2)/(50)` `u = (50)/(7) cm` |
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| 9. |
An object is seen thorugh a simple microscope of focal length 12 c. Find the angular mgnification produced if the image is formed at the near pointofhe eye which is 25 cm away from it. |
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Answer» The angual rmagnification produced by a simple microscope when the image is formed at the near point of the eye is given by `m1+D/f` Here f=12 cm, D=25cm. Hence `m=1+25/12=2.08` |
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| 10. |
(a) An object is seen through a simple microscope of focal length `12cm`. Find the angular magnification produced if the image is formed at the near point of the eye which is `25cm` away from it. (b) A `10D` lens is used as a magnifier. Where should the object be placed to obtain maximum angular magnification for a normal eye (near point `= 25cm)`? |
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Answer» Maximum angular magnification is achieved when the fiN/Al image is formed at the near point. Thus, `v=-25cm. The focal length is `f=1/10 m=10 cm.` We have `1/v-1/u=1/f` `or -1/(25cm)-1/u=1/(10cm)` or `1/u=-1/(25cm)-1/(10cm)` `or u=-50/7cm=-7.1cm` |
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| 11. |
Which of the following is correct regarding telescope?A. In Astronomical telescope, aperture and focal length of objective is large than eye-piece.B. In Terrestrial telescope, three convex lenses are used and final image is errect.C. In Galilean telescope, two lenses, one convex (objective) and one concave (eye-piece) are used.D. All options are correct. |
| Answer» Correct Answer - D | |
| 12. |
A man is looking at a small object placed at his near point. Without altering the position of lhis eye or the object, he puts a simple microscope of magnifying power 5x before his eyes. The angular magnification achieved isA. 5B. 2.5C. 1D. 0.2 |
| Answer» Correct Answer - C | |
| 13. |
To increase the angular magnification of a simple microscope, one should increaseA. the focal length of the lensB. the power of the lensC. the apeture of the lensD. the object size |
| Answer» Correct Answer - B | |
| 14. |
A simple magnifying lens is used in such a way that an image is formed at `25 cm` away from the eye. In order to have 10 times magnification, the focal length of the lens should beA. `5 cm`B. `2 cm`C. `25 mm`D. `0.1 mm` |
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Answer» Correct Answer - C `m = 1+(D)/(f_(e)) rArr 10 = 1+(25)/(f_(e))` `f_(e) = (25)/(9)cm ~~ 25mm` |
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| 15. |
Choose the correct option regarding compound microscope:A. Compound microscope consists of two convergent lenses. Lens (of focal length `f_(0)`) facing the object is called objective or field lens while the lens (of focal length `f_(e)`) facing the eye, eye-piece or ocular. The objective has a smaller aperture and smaller focal length than eye-piece.B. For a microscope, magnifying power or angular magnification is minimum when final image is at `oo` and maximum when final image is at `D`.C. For a given microscope, magnifying power `(MP)` for normal adjustment remians parctially unchanged if objective and ey-piece are interchanged. In normal setting, `f_(0)` and `f_(e)` should be smaller to have grater `M.P.`D. All options are correct. |
| Answer» Correct Answer - D | |
| 16. |
Choose the correct option:A. When the eye is focused on distant object, the focal length of eye-lens has its maximum value which is equal to its distance form the retina.B. When the eye is focused on a closer object, the focal length of the eye-lens decreases. The process of adjusting focal length is called accommodation. The focal length cannot be adjusted beyond a limit to from the image on the retina. Thus there is a minimum distance for the clear vision of an object.C. The nearest point for which the image can be focused on the retina is called the near point of the eye. The distance of the near point form the eye is called the least distance for clear vision. The farthest point up to which an eye can clearly see is called the far point.D. All options are correct. |
| Answer» Correct Answer - D | |
| 17. |
Choose the correct option:A. The far and near point for normal eye are usually taken to be infinite and `25cm` respectively.B. In eye, convex eye-lens forms real, inverted and diminished image at the retina.C. The human eye is most sensitive to yellow-green light wavelength `5550 Å` and least to violet red.D. All options are correct. |
| Answer» Correct Answer - D | |
| 18. |
Choose incorrect option:A. The size of image on the retina is roughly proportional to the angle subtended by the object on the eye. This angle is knows as visual angle. The optical instruments are used to increase the visual angle.B. In normal adjustement, eye is least strained and final image is formed at infinity.C. The magnifying power is written with a unit `X`.D. All options are correct. |
| Answer» Correct Answer - D | |
| 19. |
The resolving power of acompound microscope can be increased if weA. use infrared light for illuminating the object under observation instead of visible lightB. use ultraviolet light for illuminating the object under observation instead of visible lightC. use an objective of large focal lengthD. none of the above |
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Answer» Correct Answer - B `R.P.` of microscope `= (2mu sin theta)/(lambda)` |
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| 20. |
How can we increase the resolving power of a microscope ?A. decrease the focal length of the objectiveB. increase the focal length of the objetciveC. decrease the aperture-diameter of the objectiveD. increase the aperture-diameter of the objective |
| Answer» Correct Answer - D | |
| 21. |
Choose the correct option:A. The limit of resolution of eye is one minute i.e. two objects will be visible distinctly to the eye if the angle subtended by them at the eye is lesser than one minute.B. The persistance of vision is `(1//10)` second i.e if time interval between two consecutive light pulses is lesser than `0.1` sec, eye cannot distinguish them separately. This fact is taken into account in motion pictures.C. According to Rayleigh criterion, resolving limit of eye `theta = 1.22 lambda//d`, where `lambda` is wavelength of light values of `lambda` and `d` for eye, it turns to be one minute.D. All options are correct. |
| Answer» Correct Answer - D | |
| 22. |
The focal length of a normal eye lens is aboutA. `1mm`B. `2cm`C. `25cm`D. `1m` |
| Answer» Correct Answer - B | |
| 23. |
The human eye has a lens which hasA. soft protion at its centreB. hard surfaceC. varying refractive indexD. constant refractive index |
| Answer» Correct Answer - C | |
| 24. |
The focal length of a normal eye lens is aboutA. 1 mmB. 2 cmC. 25 cm to 300 cmD. 1 m |
| Answer» Correct Answer - B | |
| 25. |
A man weasring glasses of focal length +1 m cannot clearly see beyond 1 mA. if he is farsightedB. if he is nearsightedC. if his vision is normalD. in each of these cases |
| Answer» Correct Answer - D | |
| 26. |
The maximum focal length of the eye lens of a person is greater than its distance from the retina. The eye isA. always strained in looking at an objectB. strained for objects at large distance onlyC. strained fro objects at short distance onlyD. unstrained for all distances |
| Answer» Correct Answer - A | |
| 27. |
The maximum focal length of the eye lens of a person is greater than its distance from the retina. The eye isA. always strained in looking at an objectB. strained for objects at large distances onlyC. strained for objects at short distances onlyD. unstrained for all distances |
| Answer» Correct Answer - A | |
| 28. |
The distance of the eye lens from the retina is x. For a normal eye, the maximum focal length of the eye lensA. `=x`B. `ltx`C. `gtx`D. `=2x` |
| Answer» Correct Answer - A | |
| 29. |
A young boy can adjust the power of his eye lens between 50 D and 60 D. His far point is infinity. A. What is the distance of his retina from the eye-lens? b. What is the near point? |
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Answer» a. When the eye is fuly relaxed, its focal length is largest and the power of the eye lens is minimum. This power is 50 D according to the given data. The focal lengts is `1/50m=2cm.` As the far point is at infinity, the parallel rays coming from infinity are focussed on the retiN/A in the fuly relaxed condition. HEnce, the distance of the retinN/A from the lens equals the focal length which is 2 cm. b. when the eye is focussed at the near point, the power is maximum which is 60 D. the focal lenth in this case is `f=1/60m=5/3cm. The image is formed on the retiN/A and thus v=2 cm. We have, `1/v-1/u=1/f` or `1/u=1/v-1/f=1/(2cm)-3/(5cm)` or `u=-10cm` The near point is at 10 cm |
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| 30. |
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? |
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Answer» If an object is placd at 25 cm from the correcting lensl it should produce the virtual image at 40 cm. Thus, `u=-25cm, v=-40cm` `1/f=1/v-1/u` `=1/(40cm)+1/(25cm)` or `f=200/3cm=+2/3m` `P=1/f+1.5D` The uN/Aided eye can see a maximum distance of 250 cm. Suppose the maximum distance for clear vision is d when the lens is used. Then the object at a distance d is imaged by the lens ast 250 cm. We have `1/v-1/u=1/f` `or - 1/(250cm)-1/d=3/(200cm)` `or d=-53cm` Thus the person will be able to see up to a maximum distance of 53 cm. |
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| 31. |
The near point and the far point of a child are at 10 cm and 100 cm. If the retina is 2.0 cm behind the eye-lens, what is the range of the power of the eye-lens? |
| Answer» Correct Answer - D | |
| 32. |
Mark the correct options (i) If the far point goes ahead, the power of the divergent lens should be reduced (ii) If the near point goes ahead, the power of the convergent lens should be reduced (iii) If the far point is `1m` away from the eye, divergent lens should be used (iv) If the near point is `1m` away from the eye, divergent lens should be usedA. `(i),(ii)`B. `(i),(iii)`C. `(ii),(iv)`D. `(i),(iv)` |
| Answer» Correct Answer - B | |
| 33. |
Two convex lenses of focal length `0.01` , and `0.06` m, respectively are arranged to from a microscope. A small object is placed `0.012` m from object glass. If the image seen appears to be `0.25`m from the eye-piece, what is the distance between the object glass and eye-piece? |
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Answer» Correct Answer - `0.1084m` |
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| 34. |
Wavelength of light used in an optical instrument are `lambda_(1) = 4000 Å` and `lambda_(2) = 5000Å` then ratio of their respective resolving powers (corresponding to `lambda_(1)` and `lambda_(2)`) isA. `16 : 25`B. `9 : 1`C. `4 : 5`D. `5 : 4` |
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Answer» Correct Answer - D `R.P = (a)/(1.22lambda)` `R.P. prop (1)/(lambda)` `((R.P.)_(1))/((R.P.)_(2)) = (lambda_(2))/(lambda_(1)) = (5000)/(4000) = (5)/(4)` |
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| 35. |
A telescope with magnification M = 15 was submerged in water so that the inside of the telescope is filled up with water. To make the system work as a telescope within the former dimensions, the objective was removed. What was the magnification of the telescope after the change? `mu` of the material of the eye-piece=`1.5` and `mu` of water `=(4)/(3)` |
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Answer» Correct Answer - 3 |
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| 36. |
A planet is observed by an astronomical refracting telescope having an objective of ofcal length 16m and an eyepiece of focal length 2 cm. Then,A. `(i)`,onlyB. `(ii)`, onlyC. `(i),(ii)` onlyD. all |
| Answer» Correct Answer - D | |
| 37. |
An astronomical telescope has an angular magnification of magnitude 5 for distant object. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length `f_(0)` of the objective and the focal length `f_(0)` of the eyepiece areA. `45cm` and `-9 cm`B. `50 cm` and `10 cm`C. `7.2cm` and `5 cm`D. `30cm` and `6cm` |
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Answer» Correct Answer - D `m =- 5 =- (f_(0))/(f_(e)) rArr f_(0) = 5f_(e)` `L = f_(0) +f_(e) = 36` `sigma f_(e) = 36 rArr f_(e) 6cm, f_(0)=5f_(e) = 30cm` |
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| 38. |
A simple telescope, consisting of an objective of focal length `60cm` and a single eye lens of focal length `5cm` is focused on a distant object in such a way that parallel rays emerge form the eye lens. If the object makes an angle of `2^(@)` at the objective, the angular width if the image isA. `10^(@)`B. `24^(@)`C. `50^(@)`D. `48^(@)` |
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Answer» Correct Answer - B `m =- (f_(0))/(f_(e)) =- (60)/(5)=- 12` `m = (theta)/(theta_(0)) rArr 12 = (theta)/(2^(@)) rArr theta = 24^(@)` |
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| 39. |
Astigmatism for a human eye can be removed by usingA. concave lensB. convex lensC. cylindrical lensD. prismatic lens |
| Answer» Correct Answer - C | |
| 40. |
The velocity of light in vacuum is `3xx10^(8)ms^(-1)`. What is the velcoity of light in glass if the index of refraction of glass if `1.5`? |
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Answer» Correct Answer - `2xx10^(6)ms^(-1)` |
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| 41. |
An achromaic convergent system of focal length `+20cm` is made of two lenses (in contact) of materials having dispersive powers in the ratio `1:2`.Their focal lenths must be respectivelyA. `10cm, -20cm`B. `20cm,10cm`C. `-10cm,-20cm`D. `20cm,-10cm` |
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Answer» Correct Answer - A `(omega_(1))/(f_(1)) +(omega_(2))/(f_(2)) = 0 rArr (omega)/(f_(1)) +(2omega)/(f_(2)) = 0` `f_(2) =- 2f_(1)` `(1)/(f_(1)) +(1)/(f_(2)) = (1)/(F) rArr (1)/(f_(1))-(1)/(2f_(1)) = (1)/(20)` `(1)/(2f_(1)) = (1)/(20) rArr f_(1) = 10cm, f_(2) =- 20cm` |
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| 42. |
The muiscles of a normal eye are least strained when the eye is focussed on an objectA. far away from the eyesB. very close to the eyeC. at abot 25 cm from the eyeD. at about 1 m from the eye |
| Answer» Correct Answer - A | |
| 43. |
The focal lengths of the objective and eye piece of an astronomical telescope are `25 cm and 2.5 cm` respectively. The telescope is fucussed on an object `1.5 m` from objective, the final image being formed `25 cm` from eye of the observer. Calculate the length of the telescope. |
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Answer» Correct Answer - `31.85xx10^(-2)m,16.2` |
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| 44. |
In an astronomical telescope in normal adjustment a straight black line of length `L` is drawn on inside part of objective lens. The eye piece forms a real image of this line. The length of this image is `I`. The magnification of the telescope isA. `(L)/(l)`B. `(L)/(l)+1`C. `(L)/(l)-1`D. `(L+l)/(L-l)` |
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Answer» Correct Answer - A In normal adjustment `L = f_(0)+f_(e)` Treating line on obhective as object and eye-piece the lens `(1)/(v)-(1)/(u)=(1)/(f) rArr (1)/(v) - (1)/(-(f_(O)+f_(e)) = (1)/(f_(e))` `v = ((f_(O)+f_(e))f_(e))/(f_(O))` Magnification `= |(v)/(u)| = (f_(e))/(f_(O)) = ("image size")/("object size") = (l)/(L)` `(f_(O))/(f_(e)) = (L)/(l) =` magnification of telescope in normal adjustment |
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| 45. |
Chromatic aberrations in the formation of image by a lens arise becausesA. of non-paraxial raysB. radii of curvatures of the two sides are not the sameC. of the defect in grindingD. the focal length varies with wavelength |
| Answer» Correct Answer - D | |
| 46. |
A person wears glasses of power `-2.5D`. The defect of the eye and the far point of the person without the glasses are, respectivelyA. farsightedness,`40cm`B. nearsightedness, `40 cm`C. astigmatism, `40 cm`D. nearsightedness, `250 cm` |
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Answer» Correct Answer - B `f = (100)/(P)cm = (100)/(-2.5)=- 40cm` `(1)/(-(F.P.)) - (1)/(-("distance of object")) = (1)/(f)` `(1)/(-(F.P.)) - (1)/(oo)=(1)/(-40) rArr F.P. = 40cm` `F.P. = 40 cm`, nersightedness |
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| 47. |
A far-sighted man can see clearly at a distance of `1.0m`. The power of the lens that would make him see clearly at a distance of `0.2m` isA. `4 D`B. `-4 D`C. `6 D`D. `-6 D` |
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Answer» Correct Answer - A `(1)/(-(N.P.)) - (1)/(-("distance of object")) = (1)/(f)` `(1)/(-1.0)-(1)/(-0.2) =- 1+5 = (1)/(f) rArr (1)/(f) = P = 4D` |
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| 48. |
How should people wearing spectacles work with a microscopeA. they cannot use the microscope at allB. they should keep on wearing their spectaclesC. they should take off spectaclesD. `(2)` and `(3)` are both way |
| Answer» Correct Answer - C | |
| 49. |
An eye specialist prescribes spectacles having combination of convex lens of focal length 40cm in contact with a concave lens of focal length 25cm. The power of this lens combination in diopters isA. `+1.5`B. `-1.5`C. `+6.67`D. `-6.67` |
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Answer» Correct Answer - B `(1)/(F)=(1)/(f_(1)) +(1)/(f_(2)) = (1)/(40) +(1)/(-25)` `=(5-8)/(200) =- (3)/(200)` `P = (100)/(F(cm))D = (100)/(-200//3) =- 1.5D` |
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| 50. |
Two places A and B on the same longitude are h km apart along the surface of the earth, A is at the equator. The shadow of a vertical pole is zero at A at some hour of observation and the shadow of an identical pole at B is just half its length at the same hour of observation. Show that these observations enable us to find the radius of the earth. |
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Answer» Correct Answer - `R=(h)/(tan^(1)(1)/(2))` |
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