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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.
851. |
A ray of light is incident at small angle I on the surface of prism of small angle A and emerges normally from the oppsite surface. If the refractive index of the material of the prism is mu, the angle of incidence is nearly equal toA. `A/mu`B. `A/(2mu)`C. `muA`D. `(muA)/(2)` |
Answer» Correct Answer - C As refracted ray emerges normally from opposite surface, `r_(2) = 0` As `A = r_(1) + r_(2) :. r_(1) = A` Now, `mu = (sini_(1))/(sinr_(1)) = (i_(1))/(r_(1)) = (i)/(A)` or `I = muA` |
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852. |
A convex lens of focal length 30 cm forms a real image three times larger than the object on a screen. Object and screen are moved until the image becomes twice the size of the object. If the shift of the object is 6 cm. The shift of screen isA. 36 cmB. 72 cmC. 18 cmD. 9 cm |
Answer» Correct Answer - A | |
853. |
A glass prism has a refractive angle of `90^(@)` and a refractive index of 1.5. A ray is incident at an angle of `30^(@)`. The ray emerges from an adjacent face at an angle ofA. `60^(@)`B. `30^(@)`C. `45^(@)`D. the ray does not emerge |
Answer» Correct Answer - D (d) Critical angle, `theta_(C)=sin^(-1)(1/1.5)=41.8^(@)` Incident angle, `i_(1)=30^(@)` `sinr_(1)=(sini_(1))/(mu)=(sin(30^(@)))/(1.5)` `r_(1)=19.5^(@)` or Angle of emergence, `r_(2)=A-r_(1)=90-19.5=70.5^(@)` Since, `r_(2)gttheta_(C)` `therefore` The ray will not emerge from the prism. |
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854. |
For an equilateral prism, it is observed that when a ray strikes grazingly at one face it emerges grazingly at the other. Its refractive index will beA. `sqrt(3)`B. `(2)/(sqrt(3))`C. 2D. Data not sufficient |
Answer» Correct Answer - C | |
855. |
For an equilateral prism, it is observed that when a ray strikes grazingly at one face, it emerges grazingly at the other face, its refractive index will beA. `(sqrt(3))/(2)`B. `(2)/(sqrt(3))`C. 2D. `(4)/(3)` |
Answer» Correct Answer - C | |
856. |
The refracting angle of a prism is A and refractive index of the material of prism is `cot(A//2)` . The angle of minimum deviation will be |
Answer» The refractive index (`mu`) of a prism of angle A and minimum deviation, `delta_(m)` is given by `mu=sin((4+delta_(m))/2)/sin(A//2)` Given, `mu=cos""A/2` `therefore cot""A/2=sin((A+delta_(m))/2)/sin(A//2)` `(cosA//2)/(sinA//2)=(sin""((A+delta))/2)/sin(A//2)rArrcos""A/2=sin((A+delta_(m))/2)` `therefore sin(90^(@)-A/2)=sin((A+delta_(m))/2)` `rArr90^(@)-A/2=(A+delta_(m))/2rArr180^(@)-A=A+delta_(m)` `rArr` Angle of minimum deviation, `delta_(m)=180^(@)-2A=pi-2A` |
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857. |
The refracting angle of a prism is A and refractive index of the material of prism is `cot(A//2)` . The angle of minimum deviation will beA. `-180cm-3A`B. `-180cm-2A`C. `-90cm-2A`D. `-180cm+2A` |
Answer» Correct Answer - B (b) Refractive index,`mu=sin((A+D_(m))/2)/(sin""A/2)` `rArrcot""A/2=sin((A+D_(m))/2)/(sin""A/2)` `rArr(cot""A/2)/(sin""A/2)=sin((A+D_(m))/2)/(sin""A/2)` `sin((pi)/2-A/2)=sin((A+D_(m))/2)` `pi/2-A/2=A/2+(D_(m))/2` `rArrD_(m)=pi-2A` Angle of minimum deviation, `D_(m)=180^(@)-2A` |
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858. |
For the angle of minimum deviation of a prism to be equal to its refracting angle, the prism must be made of a material whose refractive indexA. lies between `sqrt(2)` and `1`B. lies between `2` and `sqrt(2)`C. is less than `1`D. is greater than `2` |
Answer» Correct Answer - B From prism formula, `mu = (sin(A + delta_m)//2)/(sin A//2)` For `delta_m = A`, `mu = (sin A)/(sin A//2)` =`(2 sin A//2 cos A//2)/(sin A//2) = 2 cos A//2` When `A = 0^@, mu_(max) = 2 cos 0^@ = 2` When `A = 90^@, mu_(min) = 2 cos 45^@ = (2)/(sqrt(2)) = sqrt(2)` Choice (b) is correct. |
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859. |
The refracting angle of a prism is `A`, the refractive index of the material of the prism is `cot((A)/(2))`. The angle of minimum deviation is :A. `180^@ - 2A`B. `90^@ - A`C. `180^@ - 2A`D. `180^@ - 3A`. |
Answer» Correct Answer - A Here, `mu = cot(A//2)` Acc. To prism formula, `mu- (si n((A + delta_m)/(2)))/(sin(A//2)) = (cos A//2)/(sin A//2)` `:. si n((A + delta_m)/(2)) = cos A//2 = sn (90^@ - A//2)` `(A + delta_m)/(2) = 90^@ - A//2` or `delta_m = (180^@ - 2 A)`. |
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860. |
The velocity of image object and mirror both are moving towards each other with velocities `4ms^(-1) and 5ms^(-1)` respectively, isA. `- 14 ms^(-1)`B. `15 ms^(-1)`C. `-9 ms^(-1)`D. `14 ms^(-1)` |
Answer» Correct Answer - A `V_(OM)` = velocity of object with respect to mirror `= - V_(lM)` (Here, `V_(lM)` = velocity of image with respect to mirror) `rArr " " V_(O)-V_(M)=-(V_(l)-V_(M))` `4ms^(-1)-(-5ms^(-1))=-V_(l)+(-5ms^(-1))` `V_(l)=-14 ms^(-1)` |
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861. |
Two thin lenses of focal lengths `f_(1)` and `f_(2)` are in contact. The focal length of this combination isA. `f_(1) + f_(2)`B. `(f_(1)f_(2))/(f_(1) + f_(2))`C. `(1)/(2) (f_(1) + f_(2))`D. `(f_(1) + f_(2))/(f_(1)f_(2))` |
Answer» Correct Answer - D Focal length of combination, `(1)/(F)=(1)/(f_(1))+(1)/(f_(2))` `F=(f_(1)f_(2))/(f_(1)+f_(2))` Power of lens, `P=(1)/(F)=(f_(1)+f_(2))/(f_(1)f_(2))` |
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862. |
A convex lens of refractive index 3/2 has a power of `2.5^(@)`. If it is placed in a liqud of refractive index 2,the new power of the lens isA. 2.5DB. `-2.5D`C. 1.25DD. `-1.25D` |
Answer» Correct Answer - D (d) We have,P=`(µ-1)[1/R_(1)-1/R_(2)]` `2.5=(3/2-1)(1/R_(1)-1/R_(2))` `P=((3//2)/2-1)(1/R_(1)-1/R_(2))` `therefore` P=-1.25D` |
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863. |
A plano - convex lens is made of flint glass. Its focal length isA. inversely proportional to the wavelength of lightB. longer for red than for blueC. longer for blue than for redD. the same for all colours |
Answer» Correct Answer - D Focal length of lens is independent of wavelength. |
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864. |
A convex lens of refractive index 3/2 has a power of `2.5^(@)`. If it is placed in a liqud of refractive index 2,the new power of the lens isA. 2.5 DB. `-2.5 D`C. `1.25 D`D. `- 1.25 D` |
Answer» Correct Answer - D We have, `P=(mu-1)[(1)/(R_(1))-(1)/(R_(2))]` `2.5=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))` `P=((3//2)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))` `therefore " " P=-1.25 D` |
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865. |
Distance of an object from a concave lens of focal length 20 cm is 40 cm. Then linear magnification of the imageA. `= 1`B. `gt 1`C. `lt 1`D. zero |
Answer» Correct Answer - B From a concave lens image is always smaller in size. |
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866. |
The minimum distance between an object and its real image formed by a convex lens isA. 1.5 fB. 2 fC. 2.5 fD. 4f |
Answer» Correct Answer - D `2f+2f=4f` |
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867. |
A plano-convex lens of curvature of 30 cm and refractive index 1.5 produces a real image of an object kept 90 cm from it. What is the magnification?A. 4B. 0.5C. 1.5D. 2 |
Answer» Correct Answer - D We have, `(1)/(v)-(1)/(-90)=(1)/(f)rArr (1.5-1)((1)/(30))=(1)/(60)` `therefore " " v=180 cm` `therefore " " |m|=|(v)/(u)|=(180)/(90)=2.0` |
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868. |
Distance of an object from a concave lens of focal length 20 cm is 40 cm. Then linear magnification of the imageA. 1B. `lt1`C. `lt1`D. zero |
Answer» Correct Answer - B (b) From a concave lens image is always smaller in image. |
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869. |
A convex lens is placed on an optical bench and is moved till it gives a real image of an object at a minimum distance of 80 cm from the latter. Find the focal length of the lens. If the object is placed at a distance of 15 cm from the lens, find the position of the image. |
Answer» Correct Answer - `f = 20`, virtual and erect image at 60 cm from the lens on the same side as the object |
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870. |
A convex lens of focal length 30 cm is placed between a screen and a square plate of area `4cm^(2)`. The image of the plate formed on the screen is `16cm^(2)`. Calculate the distance between the plate and the screen. |
Answer» Correct Answer - 135 cm |
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871. |
The image obtained with a convex lens is erect and its length is `4` times the length of the object. If the focal length of lens is `20 cm`, calculate the object and image distances. |
Answer» Here, `f = 20 cm, m = + 4`, for erect image. From `m = (f)/(u + f)` `4 = (20)/(u + 20) , u = -15 cm` From `m = (f - v)/(f)` `4 = (20 - v)/(20) , v = -60 cm`. |
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872. |
The figure shows an equi-convex lens. What should be the condition of the refractive indices so that the lens becomes diverging? A. `2mu_(3)gtmu_(1)-mu_(2)`B. `2mu_(2)ltmu_(1)+mu_(3)`C. `2mu_(2)gt2mu_(1)_mu_(3)`D. None of the above |
Answer» Correct Answer - B (b) We have, `mu_(2)/v_(1)-mu_(1)/oo=(mu_(2)-mu_(1))/R` and `mu_(3)/f-mu_(2)/v_(1)=(mu_(3)-mu_(2))/(-R)` From Eqs. (i) and (ii) we get `mu_(3)/f=(mu_(2)-mu_(1)-mu_(3)+mu_(2))/R=(2mu_(2)-(mu_(1)+mu_(3)))/R` Lens will become diverging, if f is negative i.e., `2mu_(2)lt(mu_(1)+mu_(3))` |
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873. |
An object is 60 cm in front of a thin lens, the image being 300 cm on the other side of the lens. Calculate the displacement of the image, when the object is moves 20 cm nearer to the lens. |
Answer» Correct Answer - 500 cm |
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874. |
A source of light and a screen are placed 90 cm apart. Where should a convex lens of 20 cm focal length be placed in order to form a real image of the source on the screen ? |
Answer» Correct Answer - 60 cm or 30 cm from the source |
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875. |
When an object is at distances x and y from a lens, a real image and a virtual image is formed respectively having same magnification. The focal length of the lens isA. `(x+y)/2`B. x-yC. `sqrt(xy)`D. x+y |
Answer» Correct Answer - A (a) For real image, u=-x v=+mx `1/(mx)-1/(-x)=1/f` For virtual image, u=-y Then, v=-my `1/(-my)-1/(-y)=1/f` On solving Eqs. (i) and (ii) we get `f=(x+y)/2` |
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876. |
The angle of inductance for an equilateral prism is `60^(@)`. What should be the refractive index of prism so that the ray is parallel to the base inside the prism ?A. `sqrt(2)`B. `sqrt(3)`C. `(4)/(3)`D. `(9)/(8)` |
Answer» Correct Answer - B | |
877. |
The real image which is exactly equal to the size of an object is to be obtained on a screen with the help of a convex glass of focal length 15 cm. For this, what must be in the distance between the object and the screen?A. 15 cmB. 30 cmC. 45 cmD. 60 cm |
Answer» Correct Answer - B (b) Keep the object at a distance 2f from the lens. |
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878. |
When an object is at distances x and y from a lens, a real image and a virtual image is formed respectively having same magnification. The focal length of the lens isA. `(x+y)/(2)`B. `x-y`C. `sqrt(x+y)`D. `x+y` |
Answer» Correct Answer - A | |
879. |
An object is placed at 10 cm from a lens and real image is formed with magnification of 0.5. Then the lens isA. concave with focal length of 10/3 cmB. convex with focal length of 10/3 cmC. concave with focal length of 10 cmD. convex with focal length of 10 cm |
Answer» Correct Answer - B (b) We have `1/f=1/v-1/urArr1/(+5)-1/(-10)=1/(f)" " (absv=absu/2)` `therefore` Focal length,f=`10/3`cm |
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880. |
Where should an object be placed from a converging lens of focal length 10 c, so as to obtain a virtual image of magnification 2 ? |
Answer» Correct Answer - 5 cm |
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881. |
Where should an object be placed from a converging lens of focal length 20 cm, so as to obtain a real image of magnification 2?A. 50 cmB. 30 cmC. `-50 cm`D. `30 cm` |
Answer» Correct Answer - D (d) For real image from converging lens, m=-2 `therefore` Magnification ,m`=f/(f+u)rArr-2=f/(u+f)=20/(u+20)` `rArr` u=-30 cm |
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882. |
Where should an object be placed from a converging lens of focal length 20 cm, so as to obtain a real image of magnification 2? |
Answer» Correct Answer - 30 cm |
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883. |
Where should an object be placed from a converging lens of focal length 20 cm, so as to obtain a real image of magnification 2?A. 50 cmB. 30 cmC. `-50 cm`D. `-30 cm` |
Answer» Correct Answer - D For real image from converging lens, `m = -2` `therefore` Magnification, `m =(f)/(f+u)rArr -2=(f)/(u+f)=(20)/(u+20)` `rArr " " u=-30 cm` |
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884. |
In an astronomical telescope, focal length of eye piece is `5 cm` and focal length of objective is `75 cm`. The final image is formed at the least distance of distinct vision `(=25 cm)` from the eye. What is the magnifying power of the telescope ? |
Answer» Here, `f_e = 5 cm, f_0 = 75 cm, m = ?, d = 25 cm` From `m = -(f_0)/(f_e)(1+(f_e)/(d)) = -(75)/(5) (1+(5)/(25)) = -15 xx 1.2 = -18.0`. |
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885. |
In an astronomical telescope, the focal length of objective lens and eye-piece are 150 cm and 6 cm respectively. In case when final image is formed at least distance of distinct vision. the magnifying power isA. 20B. 30C. 60D. 15 |
Answer» Correct Answer - b | |
886. |
A telescope has an objective of focal length 50 cm and an eye-piece of focal length 5 cm. The least distance of distinct vision is 25 cm. The telescope is focused for distinct vision on a scale 2 m away from the objective. Calculate (i) magnification produced and (ii) separation between objective and eyepiece. |
Answer» Given`f_(o)=50cm and f_(e)=5 cm` For objective,`1/(v_(o))-1/(-200)=1/50` `therefore v_(o)=200/3cm` `m_(o)=(v_(o))/(u_(o))=((200//3))/(-200)=-1/3` eye piece,`1/(-25)-1/(u_(e))=1/5` `therefore u_(e)=-25/6cmandm_(e)=(v_(e))/(u_(e))=(-25)/(-(25//6))=6` (i) Magnification,m=`m_(o)xxm_(e)=-2` (ii) Seperation between objective and eye piece, `L=v_(o)+abs(u_(e))=200/3+25/6=425/6=70.83 cm` Note Here, object is placed at finite distance from the objective. Hence, formulae derived for angular magnification M cannot be applied directly as they have been derived for the object to be at infinity. Here, it will be difficult to find angular magnification. So, only linear magnification can be obtained. |
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887. |
Optical fibers are based on the phenomenon ofA. reflectionB. refractionC. dispersionD. total internal reflection |
Answer» Correct Answer - D | |
888. |
An object is placed in front of a concave mirror of focal length f as shown in figure. Choose the correct shape of the image. A. B. C. D. |
Answer» Correct Answer - B | |
889. |
An object `2 cm` high is placed at a distance of `16 cm` from a concave mirror, which produces a real image `3 cm` high. What is thr focal length of the mirror ? Find the position of the image ? |
Answer» Correct Answer - `f = - 9.6 cm, v = -24 cm` Here, `h_1 = 2 cm, u = -16 cm`, `h_2 = - 3 cm` (because image is real and inverted) As `(- h_2)/(h_1) = (v)/(u)` `:. v = (-h_2)/(h_1) u = (3)/(2) xx (-16) = -24 cm` `(1)/(f)=(1)/(v)+(1)/(u) = (1)/(-24) -(1)/(16) = (-2 -3)/(48) = - (5)/(48)` `f = - (48)/(5) = - 9.6 cm`. |
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890. |
Four convergent lenses have focal lengths `100cm, 10cm, 4cm` and `0.3cm`. For a telescope with maximum possible magnification, we choose the lenses of focal lengthsA. `100 cm, 0.3 cm`B. `10 cm, 0.3 cm`C. `10 cm, 4 cm`D. `100 cm, 4 cm` |
Answer» Correct Answer - a | |
891. |
A convergent doublet of separated lens correct for spherical aberration, are separated by `2m` and has an equivalent focal lenth of `10cm`. Calculate the focal length of its component lenses.A. `f_(1)=18cm, f_(2)=10cm`B. `f_(1)=20cm, f_(2)=28cm`C. `f_(1)=20cm,f_(2)=18cm`D. `f_(1)=24cm,f_(2)=18cm` |
Answer» (c) Since the doublet is corrected for spherical aberration, it satisfies the following condition `f_(1)-f_(2)=d=2cm` `f_(1)=f_(2)+2cm` Let theh equivalent focal length `=F` `F=(f_(1)f_(2))/(f_(1)+f_(2)-d)=10cm` Solving, it `f_(1)=20cm, f_(2)=18cm` |
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892. |
A small fish 0.4 m below the surface of a lake is viewed through a simple converging lens of focal length 3 m.the lens is kep at 0.2 m above the water surface such that the fish lies on the optical axis of the lens. Find the image of the fish seen by the observed. `(mu_(water)=(4)/(3))`A. A distance of 0.2 m from the water surfaceB. A distance of 0.6 m from the water surfaceC. A distance of 0.3 m from the water surfaceD. The same location of fish |
Answer» Correct Answer - D |
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893. |
While determining the refractive index of a liquid experimentally, the microscope was focussed at the bottom of a beaker, when its reading was 3.965 cm. on pouring liquid upto a height 2.537 cm inside the beaker, the reading of the refocussed microscope was 3.348 cm. Find the refractive index of the liquid. |
Answer» Correct Answer - 1.321 |
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894. |
A microscope is focussed on a coin lying at the bottom of a beaker. The microscope is now raised up by `1 cm`. To what depth should the water be poured into the beaker so that coin is again in focus ? (Refration index of water is `4//3`)A. 1 cmB. `4/3` cmC. 3 cmD. 4 cm |
Answer» Correct Answer - d | |
895. |
A microscope is focussed on a coin lying at the bottom of a beaker. The microscope is now raised up by `1 cm`. To what depth should the water be poured into the beaker so that coin is again in focus ? (Refration index of water is `4//3`)A. 1 cmB. `4/3 cm `C. 3 cmD. 4 cm |
Answer» Correct Answer - b | |
896. |
A mark at the bottom of a bekar is focussed by a microscope .When water is poured in the beaker up to a height of`16`cm,the microscope has to be raised through a distance of`4`cm in order to bring the mark to focus again .Calculate the refractive index of water. |
Answer» Let `d` be the real depth of the water.The apparent depth will be `d//mu`.So the microscope has to be raised through a distance `(d-(d)/(mu))=d(1-(1)/(mu))` `d(1-(1)/(mu))=4 or 16(1-(1)/(mu))=4 :.mu=4//3` |
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897. |
When a glass slab is placed on a dot on a paper.It apperas displaced by`4`cm viewed normally.What is the thickness of slab if the refractive index`1.5`. |
Answer» We know that Displacement =`t(1-(1)/(mu))` So`4=t(1-(1)/(mu))` `t=(muxx4)/(mu-1)=(1.5xx4)/(1.5-1)=12cm` |
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898. |
A microscope is focused on a mark on a piece of paper and then a slab of glass of thickness `3cm` and refractive index `1.5` is placed over the mark. How should the microscope be moved to get the mark in focus again ?A. 2 cm upwardB. 1 cm upwardC. `4.5` upwardD. 1 cm downward |
Answer» Correct Answer - B Here, real depth of mark, `x = 3 cm` apparent depth of mark, `y = ?` refractive index, `mu = 1.5` As `mu = (x)/(y) :. y = (x)/(mu) = (3)/(1.5) = 2 cm` Distance through which mark appears to be raised = `x - y = 3 - 2 = 1 cm` `:.` To get the mark in focus again, distance through which microscope be moving upwardsd `= 1 cm`. |
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899. |
An ink mark on a sheet of paper is viewed through a glass slab of thickness `t` and refractive index `mu`. Through what distance the mark appears to be raised ? |
Answer» Normal shift in the position of mark =`t = (1 - (1)/(mu))`. |
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900. |
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 cmB. 2 cmC. 3 cmD. 1.33 cm |
Answer» Correct Answer - D | |