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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. |
Image of an object approaching a convex mirror of radius of curvature 20m slong its optical axis is observed to move from `(25)/(3)`m to `(50)/(7)`m in 30 seconds. What is the speed of the object in km per hour? |
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Answer» Correct Answer - 3 `R=20 cm, f=10 m` For mirror, `(1)/(v)+(1)/(u)=(1)/(f)` `(1)/(25//3)+(1)/(u_(1))=(1)/(10)` `rArr u_(1)=-50 cm` `& (1)/(50//7)+(1)/(u_(2))=(1)/(10)` `rArr (1)/(u_(2))=(1)/(25) rArr u_(2)=-25 cm` So, speed `=|(Deltau)/(Deltat)|=(35)/(30) m//sec.=(5)/(6)m//sec` . `&` in `km//hr=(5)/(6)xx(18)/(5)=3 km//hr` . |
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| 2. |
The `x-y` plane is the boundary between two transparent media. Medium-`1` with `z gt0` has refractive index `sqrt2` and medium `-2` with `zlt 0` has a refractive index `sqrt3`. `A` ray of light in medium `-1` given by the vector `A=6sqrt3 i+8sqrt3j-10k` is incident on the plane of separation. If the unit vector in the direction of refracted ray in medium `-2` is `(1)/(5) (ai+bj-(5)/(sqrt2)k)` then the value of ab. |
| Answer» Correct Answer - 6 | |
| 3. |
The refractive index of a glass is 1.520 for red light and 1.525 for blue light. Let `D_1` and `D_2` be angles of minimum deviation for red and blue light respectively in a prism of this glass. Then,A. `D_(1)` can be less than or greater than `D_(2)` depending upon the angle of prismB. `D_(1)gtD_(2)`C. `D_(1)ltD_(2)`D. `D_(1)=D_(2)` |
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Answer» Correct Answer - B Angle of minimum deviation `D=(mu-1)A` `because mu_(blue)gtmu_(red)` `therefore D_(2)gtD_(1)` |
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| 4. |
Refractive index of glass for red and violet colours are 1.50 and 1.60, repectively. Find: a. The refractive index for yellow colour, approximately b. Dispersive power of the medium. |
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Answer» (a) `mu_(Y) = (mu_(Y) + mu_(R))/(2) = (1.50+ 1.60)/(2) = 1.55` (b) `omega = (mu_(Y) + mu_(R))/(mu_(y) - 1) = (1.60+ 1.50)/(1.55 - 1) = 0.18` |
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| 5. |
The focal lenghts of a convex lens for red, yellow and violet rays are `100cm, 99cm` and `98 cm` respectively. Find the dispersice power of the material of the lens. |
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Answer» Correct Answer - `(99)/(4900)` Dispersive power `= (delta_(*V) - delta_(R))/(delta_(Y)) = ((mu_(V) - mu_(R)))/((mu_(Y) - 1)) = ((mu_(V) - 1) - (mu_(R) - 1))/((mu_(Y) - 1))` We know that, `(1)/(f) = (mu-1) ((1)/(R_(1)) - (1)/(R_(2))) = (mu-1)K` where `K = (1)/(R_(1)) - (1)/(R_(2))` So, `(mu_(V) - 1)K = (1)/(98) .....(i)` `(mu_(R) - 1) K = (1)/(100)...(ii)` and `(mu_(Y) - 1)K = (1)/(99) ...(iii)` `:.` Dispersice power, `omega = ((1)/(98) - (1)/(100))/((1)/(99)) = (99)/(4900)` |
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| 6. |
Find the focal length of a convavo-convex lens (Positive me menisus ) with `R_(1) = 15cm`, and `R_(2) = 25cm`. The refractive index of the lens material `n = 1.5`m `(1)/(f) = (1.5-1)((1)/(15)-(1)/(25)) = 0.5((10-6)/(150)), :. F = (300)/(4) = 75cm` |
| Answer» `(1)/(f) = (1.5-1) ((1)/(15)-(1)/(25)) = 0.5((10-6)/(150)), :. f = (300)/(4) = 75cm` | |
| 7. |
A student performed the experiment of determination of focal length of a concave mirror by `u-v` method using an optical bench of length 1.5 meter. The focal length of the mirror used is 24 cm. The maximum error in the location of the image can be 0.2 cm. The 5 sets of `(u,v)` values recorded by the student (in cm) are: `(42,56),(48,48),(60,40),(66,33),(78,39)` . The data set (s) that cannot come from experiment and is (are) incorrectly recorded, is (are)A. `(42,560`B. `(48,48)`C. `(66,33)`D. `(78,39)` |
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Answer» Correct Answer - C::D `(1)/(v)+(1)/(u)=(1)/(f)` or `(1)/(-|v|)+(1)/(-|u|)=(1)/(-|f|)rArr |v|=(|u||f|)/(|u|-|f|)` For `|u|=42,|f|=24 ,|v|=((42)(24))/(42-24)=56 cm` so `(42,56)` is correct observation For `|u|=48` or `|u|=2f` so `(48,48)` is correct observation `|v|=((66)(24))/(66-24)~~36 cm` which is not in the permissible limit so `(66,33)` , is incorrect recorded For `|u|=78, |f|=24 cm` `|v|=((78)(24))/(78-24) ~~32 cm` which is also not in the permissible limit. so `(78,39)` , is incorrect recorded. |
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| 8. |
In an optics experiment, with the position of the object fixed, a student varies the position of a convex lens and for each position, the screen is adjusted to get a clear image of the object. A graph between the object distance u and the image distance v, from the lens, is plitted using the same scale for the two axes. A straight line passing through the origin and making an angle of `45^circ` with x-axis meets the experimental curve at P. The coordinates of P will be.A. `((f)/(2),(f)/(2))`B. `(f,f)`C. `(4f,4f)`D. `(2f,2f)` |
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Answer» Correct Answer - B `v=u` and `(1)/(v)+(1)/(u)=(1)/(f)` `(2)/(v)=(1)/(f)` `rArr v=2f, u=2f` |
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| 9. |
An equilateral prism deviates a ray through `40^@` for two angles of incidence differing by `20^@`. The possible angles of incidences isA. `40^(@)`B. `50^(@)`C. `20^(@)`D. `60^(@)` |
| Answer» Correct Answer - a,d | |
| 10. |
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 m`B. `2.4 m`C. `3.2 m`D. `5.6 m` |
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Answer» Correct Answer - C `(1)/(f)=(1)/(12)+(1)/(240)=(20+1)/(240)` `f=(240)/(21)m` shift `=1(1-(2)/(3))=(1)/(3)` Now `v=12-(1)/(3)=(35)/(3) cm` `therefore (21)/(240)=(3)/(35)-(1)/(u)` `(1)/(u)=(3)/(35)-(21)/(240)=(1)/(5)((3)/(7)-(21)/(48))` `(5)/(u)=|(144-147)/(48xx7)|` `u=560 cm =5.6 m` |
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| 11. |
A `2.5` dipote lens forms a virtual image which is 4 times the object placed perpendicually on the principle axis of the lens. Find the required distance of the object form the lens. |
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Answer» Correct Answer - `30cm` `f = (1)/(p) = (1)/(2.5)m = 40cm, m = (v)/(u) = 4, v = 4u` Using lens formula `(1)/(40) = (1)/(4u) - (1)/(u) = (1-4)/(4u), (1)/(40) = (-3)/(4u) rArr u = -30 cm` So, required distance `= 30 cm` |
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| 12. |
Which type of lens is used to correct astigmatisum ? |
| Answer» Cylendircal lens is used to correct astigmatisum ? | |
| 13. |
The dispersion of light in a medium implies that :A. lights of different wavelengths travel with different speeds in the mediumB. lights of different frequencies travel with different speeds in the mediumC. the refractive index of medium is different for different wavelengthsD. all of the above. |
| Answer» Correct Answer - C | |
| 14. |
An object `O` is kept in air and a lens of focal length `10 cm` (in air) is kept at the botton of a container which is filled upto a height `44 cm` by water. The refractive index of water is `4//3` and that of glass is `3//2`. The botton of the container is closed by a thin glass slab of refractive index `3//2`. Find the distance (in `cm`) of the final image formed by the system from botton of container (refer to figure shown below). |
| Answer» Correct Answer - 90 | |
| 15. |
See the figure Find the distance fo final image formed by mirror |
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Answer» Shift `= 3(1 - (1)/(3//2))` For mirror object is at a distance `= 21 - 3 (1 - (1)/(3//2)) = 20cm` Object is at the centre of curvature of mirror. Hence the light rays will retrace and image will be formed on the object itself. |
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| 16. |
A convex lens of focal length `25 cm` and a concave lens of focal length `20 cm` are mounted coaxially separated by a distance `d cm` . If the power of the combination is zero, `d` is equal toA. `45`B. `30`C. `15`D. `5` |
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Answer» Correct Answer - D `(1)/(f)=(1)/(f_(1))+(1)/(f_(2))-(d)/(f_(1)f_(2))=0` `= (1)/(25)+(1)/(-20)-(d)/(-500)=0=(20-25)/(500)=-(d)/(500)` `d=5 cm` |
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| 17. |
Consider a concave mirror and a convex lens (refractive index 1.5) of focal length `10 cm` each separated by a distance of `50 cm` in air (refractive index = 1) as shown in the Fig. An object is placed at a distance of `15 cm` from the mirror. Its erect image formed by this combination has magnification `M_1`. When this set up is kept in a medium of refractive index `7//6`, the magnification becomes `M_2`. The magnitude `((M_2)/(M_1))` is : . |
| Answer» Correct Answer - b | |
| 18. |
Find the maximum angle that can be made in glass medium `(mu=1.5)` if a light ray is refreacted from glass to vacuum. |
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Answer» `1.5sin C = 1 sin 90^(@)`m where `C` = critical angle. `sin C = 2//3` `C = sin^(-1) 2//3` |
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| 19. |
Find the behaviour of concave lens placed in a rarrer medium. |
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Answer» Factor(A) is negative, becuyase the lens is concave. Factor(B) is positive, because the lens is placed in a rarrer medium. Therefore the focal length of the lens, which depneds on the product of these factors, is negative and hence the lens will behave as diverging lens. |
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| 20. |
The sun subtends an angle of `(1//2)^(@)` on earth. The image of sun is obtained on the screen with the help of a convex lens of focal length 100 cm the diameter of the image obtained on the screen will beA. `1`B. `3`C. `5`D. `9` |
| Answer» Correct Answer - D | |
| 21. |
Locate the image fo the point `P` as seen by the eye in the figure. |
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Answer» Correct Answer - `0.9cm` above `P` Apparent shift `= 1.4 (1 - (1)/(1.4)) + 2 (1 - (1)/(1)) + 1.3 (1 - (1)/(1.3)) + 2(1 - (1)/(1))+ 1.2 (1 - (1)/(1.2)) + 2 (1 - (1)/(1))` `= 0.4 + 0.3 + 0.2 = 0.9cm` towards the eye. So image is formed `0.9cm` above `P`. |
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| 22. |
The minimum distance between a real object and its real image formed by a thin converging lens of focal length `f` isA. `4f`B. `2f`C. `f`D. `f//2` |
| Answer» Correct Answer - B | |
| 23. |
All the listed things below are made of flint glass. Which one of these have greatest dispersive power (`omega`).A. prismB. glass slabC. biconvex lensD. all have same `omega` |
| Answer» Correct Answer - D | |
| 24. |
In the above question the radius of curvature of the curvature of the curved surface of plano-convex lens is :A. `(280)/(9) cm`B. `(180)/(7) cm`C. `(39)/(3) cm`D. `(280)/(11) cm` |
| Answer» Correct Answer - D | |
| 25. |
A plano-concave lens is placed on a paper on which a flower is drawn. How far above its actual position does the flower appear to be? A. `10 cm`B. `15cm`C. `50cm`D. none of these |
| Answer» Correct Answer - A | |
| 26. |
The image for the converging beam after refraction through the curved surface is formed at A. `x=40 cm`B. `x=(40)/(3)cm`C. `x=-(40)/(3)cm`D. `x=(180)/(7) cm` |
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Answer» Correct Answer - D `(1)/(v)-(3)/(2xx30)=(1-(3)/(2))/(+20) (1)/(v)=-(1)/(40)+(1)/(20)=+(1)/(40) v=40 cm` . |
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| 27. |
A parallel beam of light is incident normally on the flat surface of a hemisphere of radius `6 cm` and refractive index `1.5`, placed in air as shown in figure (i). Assume paraxial ray approxiamtion. .A. The rays focused at `12 cm` from the point `P` to the right, in the situation as shown in figure `(i)`B. The rays focused at `16 cm` from the point `P` to the right, in the situation as shown in figure `(i)`C. If the rays are incident at the curved surface (figure `(ii)`) then are focused at distance `18 cm` from point `P` to the right.D. If the rays are incident at the curved surface (figure `(ii)`) then are focused at distance `14 cm` from point `P` to the right. |
| Answer» Correct Answer - A::D | |
| 28. |
A person with a defective sight is using a lens having a power of +2D. The lens he is using isA. concave lens with `f=0.5 m`B. convex lens with `f=2.0 m`C. concave lens with `f=0.2 m`D. convex lens with `f=0.5 m` |
| Answer» Correct Answer - D | |
| 29. |
The focal lengths of the objective and eye`-` lens of a microscope are `1cm` and `5 cm` respectively. If the magnifying power for the relaxed eye is `45`, then the length of the tube isA. `30 cm`B. `25 cm`C. `15 cm`D. `12 cm` |
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Answer» Correct Answer - C By using `m_(infty)=((L_(infty)-f_(0)-f_(e)).D)/(1xx5)` `rArr45((L_(infty)-1-5)xx25)/(1xx5)rArr L_(infty)=15 cm` |
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| 30. |
By properly combining two prisms made of different materials, it is possible to (choose the incorrect option)A. have dispesion without average deviationB. have deviation without dispersionC. have both dispersion and average deviationD. have neither dispersion nor average deviation |
| Answer» Correct Answer - a,b,c | |
| 31. |
The convex lens is used in-A. MicroscopeB. TelescopeC. ProjectorD. all of the above. |
| Answer» Correct Answer - D | |
| 32. |
The focal length of the objective lens of a compound microscope isA. arbitraryB. less than the focal length of eyepieceC. equal to the focal length of eyepiecsD. greater than the focal length of eyepiece |
| Answer» Correct Answer - B | |
| 33. |
Resolving power of a microscope depends uponA. the focal length and aperture of the eye lensB. the focal lengths od the objective and eye lensC. the apertures of the objective and the eye lensD. the waveleght of light illuminating the object |
| Answer» Correct Answer - D | |
| 34. |
The magnifying power of a simple microscope can be increased, if we use eye-piece ofA. shorter focal length is usedB. longer focal length is usedC. shorter diameter is usedD. longer diameter is used |
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Answer» Correct Answer - D `m=1+(D)/(f)` |
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| 35. |
The resolving power of a telesope is more when its objective lens hasA. greater focal lengthB. smaller focal lengthC. greater diameterD. smaller diameter |
| Answer» Correct Answer - D | |
| 36. |
In a compound microscope, the intermediate image isA. vitual, erect and magnifiedB. real, erect and magnifiedC. real, inverted and magnifiedD. vitual,erect and reduced |
| Answer» Correct Answer - C | |
| 37. |
Find the equivalent focal length of the system for paraxal rays parallel to axis |
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Answer» `(1)/(f_(eq)) = (1)/(f_(1)) + (1)/(f_(2)) - (d)/(f_(1) f_(2)) = (1)/(10) - (20)/(10(-10)) = (1)/(5)` `rArr f_(eq) = 5cm` |
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| 38. |
A prism having refractive index `sqrt2` and refractive angle `30^@` has one of the refractive surfaces polished. A beam of light incident on the other surfaces will trace its path if the angle of incidence isA. `0^(@)`B. `30^(@)`C. `45^(@)`D. `60^(@)` |
| Answer» Correct Answer - C | |
| 39. |
If the focal length of objective and eye lens are `1.2 cm` and `3 cm` respectively and the object is put `1.25cm` away from the objective lens and the final image is formed at infinity. The magnifying power of the microscope isA. `150`B. `200`C. `250`D. `400` |
| Answer» Correct Answer - B | |
| 40. |
An object and a plane mirror are shown in figure. Mirror is moved with velocity `V` as shown. The velocity of image is : .A. `2 V sin theta`B. `2V`C. `2V cos theta`D. none of these |
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Answer» Correct Answer - A `vec(V)_(I,M) = -vec(V)_(O,M)` (normal to plane mirror) `rArr vec(V)_(1) - vec(V)_(M) = -(vec(V)_(0) - vec(V)_(M))` `rArr V_(1) - V sin theta = - (0 - V sin theta)` `rArr V_(1) = 2V sin theta` |
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| 41. |
An equiconvex lens, with redii of curvature of magnitude `10 cm` each, is put over a liquid layer poured on top of a plane mirror. `A` small needle, with its tip on the principal axis of the lens, is moved along the axis until its inverted real image coincides with the needle itself. The distance of the needle, from the lens, is measured to be `15 cm` . On removing the liquid layer, and repeating the experiment the distance is measured to be `10 cm` . Given that the two values of the distance measured represent the force length values in the two cases, calculate the refractive index of the liqiud. |
| Answer» Refractive index of the liquid `=4//3` | |
| 42. |
An equiconvex lens of refractive index `n_(2)` is placed such that the refractive index of the surrouding media is as shown. The lens : A. must be diverging if `n_(2)` is less than the aritmetic mean of `n_(1)` and `n_(3)`B. must be converging if `n_(2)` is greater than the arithmetic mean of `n_(1)` and `n_(3)`C. may be diverging if `n_(2)` is less than the arihmetic mean of `n_(1)` and `n_(3)`D. will neither be diverging nor converging if `n_(2)` is equal to arithmetic mean of `n_(1)` and `n_(3)` |
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Answer» Correct Answer - a,b,d `f=(n_(1)R)/(2n_(2)-n_(1)-n_(3))` or `(n_(3)R)/(2n_(2)-n_(1)-n_(3))` if `n_(2)lt(n_(1)+n_(3))/(2)rArrf is -ve rArr` lens is diverging if `n_(2)gt(n_(1)+n_(3))/(2) rArrf is+ve rArr` lens is converging. If `n_(2)=n_(1)+n_(2) rArr f=infty` neither converging nor diverging. |
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| 43. |
See the figure Find the deciation caused by a prisim having refreacting angle `4^(@)` and refreactive index `(3)/(2)` |
| Answer» `delta = ((3)/(2) - 1) xx 4^(@) = 2^(@)` | |
| 44. |
In Figure., find the position of final image formed. |
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Answer» For converging lens `u = -15cm, f = 10cm, v = (fu)/(f+u) = 30cm` For diverging lens `u = 5cm` `f = -10cm, v = (fu)/(f+u) = 10cm` |
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| 45. |
In Figure., find the position of final image formed. |
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Answer» For lens, `(1)/(v) - (1)/(u) = (1)/(f)` `(1)/(v) - (1)/(-15) = (1)/(-10) rArr v = +- 30cm` Hence it is object for mirror `u = -15cm` `(1)/(v) + (1)/(-15) = (1)/(-10) rArr v = +30` Hence final image will from at a distance `30cm` formt the lens towards left. |
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| 46. |
What should be the value of distance d so that final image is formed on the object itself. (Focal length of the lenses are written on the lenses.) A. `10 cm`B. `20 cm`C. `5 cm`D. none of these |
| Answer» Correct Answer - A | |
| 47. |
The values of `d_(1)` & `d_(2)` for final rays to be parallel to the principal axis are: (focal lengths of the lenses are written above the respective lenses in the given figure) A. `d_(1)=10 cm, d_(2)=15 cm`B. `d_(1)=20 cm, d_(2) =15 cm`C. `d_(1)=30 cm, d_(2)=15 cmD. none of these |
| Answer» Correct Answer - a,b,c | |
| 48. |
A thin symmetrical double convex lens of power `P` is cut into three parts, as shown in the figure. Power of `A` is A. `2 P`B. `(P)/(2)`C. `(P)/(3)`D. `P` |
| Answer» Correct Answer - B | |
| 49. |
The image of a white object in with light formed by a lens is usually colored and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As `R` .`I`. `mu` of lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red farthest from it as shown in figure. As a result of this, in case of convergent lens if a screen is placed at `F_(v)` center of the image will be violet and focused while sides are red and blurred. While at `F_(R)` , reverse is the case, `i.e` ., center will be red and focused while sides violet and blurred. The differece between `f_(v)` and `f_(R)` is a measure of the longitudinal chromatic aberration `(L.C.A),i.e.,` `L.C.A.=f_(R)-f_(v)=-df` with `df=f_(v)-f_(R)` ...........`(1)` However, as for a single lens, `(1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))]` .............`(2)` `rArr -(df)/(f^(2))=dmu[(1)/(R_(1))-(1)/(R_(2))]` ...............`(3)` Dividing E1n. `(3)` by `(2)` : `-(df)/(f)=(dmu)/((mu-1))=omega, [omega=(dmu)/((mu-1))] "dispersive power" , .........(4)` And hence, from Eqns. `(1)` and `(4)` , `L.C.A.=-df=omegaf` Now, as for a single lens neither `f` nor `omega` zero, we cannot have a single lens free from chromatic aberration. Condition of Achromatism : In case of two thin lenses in contact `(1)/(F)=(1)/(F_(1))+(1)/(F_(2)) i.c,. -(dF)/(F^(2))=(df_(1))/(f_(1)^(2))-(df_(2))/(f_(2)^(2))` The combination will be free from chromatic aberration if `dF=0` `i.e., (df_(1))/(f_(1)^(2))+(df_(2))/(f_(2)^(2))=0` which with the help of Eqn. `(4)` reduces to `(omega_(1)f_(1))/(f_(1)^(2))+(omega_(2)f_(2))/(f_(2)^(2))=0 , i.e., (omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0 ........(5)` This condition is called condition of achromatism (for two thin lenses in contact ) and the lens combination which satisfies this condition is called achromatic lems, from this condition, `i.e.,` form Eqn. `(5)` it is clear the in case of achromatic doublet : Since, if `omega_(1)=omega_(2), (1)/(f_(1))+(1)/(f_(2))=0 i.e., (1)/(F)=0` or `F=infty` `i.e.,` combination will not behave as a lens, but as a plane glass plate. `(2)` As `omega_(1)` and `omega_(2)` are positive quantities, for equation `(5)` to hold, `f_(1)` and `f_(2)` must be of opposite nature, `i.e.,` if one of the lenses is converging the other must be diverging. `(3)` If the achromatic combination is convergent, `f_(C)ltf_(D)` and as `(f_(C))/(f_(d))=(omega_(C))/(omega_(D)), omega_(C)ltomega_(d)` `i.e.,` in a convergent achromatic doublet, convex lens has lesses focal legth and dispersive power than the divergent one. Chromatic aberration in the formation of image by a lens arises because :A. of non-paraxial rays.B. the radil of curvature of the two sides are not same.C. of the defect in grinding.D. the focal length varies with wavelength. |
| Answer» Correct Answer - D | |
| 50. |
The image of a white object in with light formed by a lens is usually colored and blurred. This defect of image is called chromatic aberration and arises due to the fact that focal length of a lens is different for different colours. As `R` .`I`. `mu` of lens is maximum for violet while minimum for red, violet is focused nearest to the lens while red farthest from it as shown in figure. As a result of this, in case of convergent lens if a screen is placed at `F_(v)` center of the image will be violet and focused while sides are red and blurred. While at `F_(R)` , reverse is the case, `i.e` ., center will be red and focused while sides violet and blurred. The differece between `f_(v)` and `f_(R)` is a measure of the longitudinal chromatic aberration `(L.C.A),i.e.,` `L.C.A.=f_(R)-f_(v)=-df` with `df=f_(v)-f_(R)` ...........`(1)` However, as for a single lens, `(1)/(f)=(mu-1)[(1)/(R_(1))-(1)/(R_(2))]` .............`(2)` `rArr -(df)/(f^(2))=dmu[(1)/(R_(1))-(1)/(R_(2))]` ...............`(3)` Dividing E1n. `(3)` by `(2)` : `-(df)/(f)=(dmu)/((mu-1))=omega, [omega=(dmu)/((mu-1))] "dispersive power" , .........(4)` And hence, from Eqns. `(1)` and `(4)` , `L.C.A.=-df=omegaf` Now, as for a single lens neither `f` nor `omega` zero, we cannot have a single lens free from chromatic aberration. Condition of Achromatism : In case of two thin lenses in contact `(1)/(F)=(1)/(F_(1))+(1)/(F_(2)) i.c,. -(dF)/(F^(2))=(df_(1))/(f_(1)^(2))-(df_(2))/(f_(2)^(2))` The combination will be free from chromatic aberration if `dF=0` `i.e., (df_(1))/(f_(1)^(2))+(df_(2))/(f_(2)^(2))=0` which with the help of Eqn. `(4)` reduces to `(omega_(1)f_(1))/(f_(1)^(2))+(omega_(2)f_(2))/(f_(2)^(2))=0 , i.e., (omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0 ........(5)` This condition is called condition of achromatism (for two thin lenses in contact ) and the lens combination which satisfies this condition is called achromatic lems, from this condition, `i.e.,` form Eqn. `(5)` it is clear the in case of achromatic doublet : Since, if `omega_(1)=omega_(2), (1)/(f_(1))+(1)/(f_(2))=0 i.e., (1)/(F)=0` or `F=infty` `i.e.,` combination will not behave as a lens, but as a plane glass plate. `(2)` As `omega_(1)` and `omega_(2)` are positive quantities, for equation `(5)` to hold, `f_(1)` and `f_(2)` must be of opposite nature, `i.e.,` if one of the lenses is converging the other must be diverging. `(3)` If the achromatic combination is convergent, `f_(C)ltf_(D)` and as `(f_(C))/(f_(d))=(omega_(C))/(omega_(D)), omega_(C)ltomega_(d)` `i.e.,` in a convergent achromatic doublet, convex lens has lesses focal legth and dispersive power than the divergent one. Chromatic aberration of a lens can be corrected by :A. providing different suitable curvatures of its two surfaces.B. proper polishing of its two surfaces.C. suitably conbining it with another lens.D. reducing its aperture. |
| Answer» Correct Answer - C | |