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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.
| 51. |
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. The dispersive power of crown and fint glasses are `0.02` and `0.04` respectively. An achromtic converging lens of focal length `40 cm` is made by keeping two lenses, one of crown glass and the other of flint glass, in contact with each other. The focal lengths of the two lenses are :A. `20 cm and 40 cm`B. `20 cm and -40 cm`C. `-20 cm and 40 cm`D. `10 cm and -20 cm ` |
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Answer» Correct Answer - B `(omega_(1))/(f_(1))+(omega_(2))/(f_(2))=0` `rArr (omega_(1))/(omega_(2))=(f_(1))/(f_(2))=(1)/(2) ……….(1)` `rArr (1)/(F)=(1)/(f_(1))+(1)/(f_(2))=(1)/(4) ………..(2)` After solving `(1)` &`(2)` `f_(1)=20 cm` `f_(2)=-40 cm` . |
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| 52. |
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 a spherical concave mirror is proportional to :A. `f`B. `f^(2)`C. `1//f`D. none of these |
| Answer» Correct Answer - D | |
| 53. |
If a symmetrical bi-concave thin lens is cut into two identical halves, and they are placed in different ways as shown, then A. three images will be formed in case `(i)`B. tow inamges will be formed in the case `(i)`C. the ratio of focal lengths in `(ii)`&`(iii)` is `1`D. the ratio of focal lengths in `(ii)`&`(iii)` is `2` |
| Answer» Correct Answer - a,c | |
| 54. |
A man washign to get a picture of a Zebra photographed a white donkey after fitting a glass with black streaks onto the objective of his camera.A. The image will look like a white donkey on the photograph.B. The image will look like a Zebra on the photographC. The image will be more intense compared to the case in which no such glass is used.D. The image will be less intense sompnse to the case in which no such glass is used. |
| Answer» Correct Answer - a,d | |
| 55. |
A convex lens made up of glass of refractive index `1.5` is dippedin turn (i) in a medium of refractive index `1.65` (ii) in a medium of refractive index `1.33` (a) Will it behave as converging or diverging lens in the two cases ? (b) How will its focal length changes in the two media ? |
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Answer» Here,`.^(a)mu_(g)=1.5` Let `f_(air)` be the focal length of the lens in air, Then, `(1)/(f_(air))(.^(a)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))` or `((1)/(R_(1))-(1)/(R_(2)))=(1)/(f_(air)(.^(a)mu_(g)-1))` `=(1)/(f_(air)(1.5-1))` or `((1)/(R_(1))-(1)/(R_(2)))=(2)/(f_(air)) .....(i)` (i) When lens is dipped in medium `A` Here, `.^(a)mu_(A)=1.65` Let `f_(A)` be the focal length of the lens, when dipped in medium `A`. Then, `(1)/(f_(A))(.^(A)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))=((.^(a)mu_(g))/(.^(a)mu_(A))-1)((1)/(R_(1))-(1)/(R_(2)))` Using the equation `(i)`, we have `(1)/(f_(A))((1.5)/(1.65)-1)xx(1)/(f_(air))=-(1)/(5.5f_(air))` or `f_(A)=-5.5f_(air)` As the sign of `f_(A)` is opposite to that of `f_(air)` the lens will behave as a diverging lens. (ii) When lens is dipped in medium `B` : Here, `.^(a)mu_(B)=1.33` Let `f_(B)` be the focal length of the lens, when dipped in medium `B`. Then, `(1)/(f_(B))(.^(B)mu_(g)-1)((1)/(R_(1))-(1)/(R_(2)))=((.^(a)mu_(g))/(.^(a)mu_(B))-1)((1)/(R_(1))-(1)/(R_(2)))` Using the equation `(i)`, we have `(1)/(f_(B))((1.5)/(1.33)-1)xx(2)/(f_(air))=(0.34)/(1.33f_(air))` As the sign of `f_(B)` is same as that of `f_(air)` the lens will behave as a converging lens. |
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| 56. |
A transparent thin film of uniform thickness and refractive index `n_(1)``=1.4` is coated on the convex spherical surface of radius R at one end of a long solid glass cylinder of refractive index `n_(2)``=1.5`, as shown in the figure. Rays of light parallel to the axis of the cylinder traversing through the film from air to glass get focused at distance `f_(1)` from the film, while rays of light traversing from glass to air get focused at distance `f_(2)` from the film, Then ` A. `|f_(1)|=3R`B. `|f_(1)|=2.8R`C. `|f_(2)|=2R`D. `|f_(2)|=1.4R |
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Answer» Correct Answer - C `(1)/(f_(film))=(n_(1)-1)((1)/(R)-(1)/(R)) rArr f_(film)=infty` (inf inite) `therefor` No effect of presence of film. For Air to glass Using single spherical Refraction :- `(n_(2))/(v)-(1)/(u)=(n_(2)-1)/(R)` `(1.5)/(v)-(1)/(infty)=(1.5-1)/(R) rArr v=3R` `therefore f_(1)=3R` For Glass to Air `( :-)/` `(1)/(v)-(n_(2))(u)=(1-n_(2))/(-R)` `rArr (1)/(v)-(1.5)/(infty)=(1-1.5)/(-R)` `rArr v=2R` `therefore f_(2)=2R` |
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| 57. |
A ray of light is incident on a parallel slab of thickness t and refractive index n. If the angle of incidence `theta` is small, than the lateral displacement in the incident and emergent ray will beA. `(t theta(n-1))/(n)`B. `(t theta)/(n)`C. `(t thetan)/(n-1)`D. none of these |
| Answer» Correct Answer - A | |
| 58. |
A ray of light is incident on a surface in a direction given by vector `A=2i-2j+k`. The normal to that surface passing through the point of incidence is along the vector `N=j-2k`. The unit vector in the direction of reflected ray is given by `R=aj+bj+ck`. Find three equations in terms of `a,b,c` using which we can find the values of `a,b,& c` |
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Answer» Correct Answer - `a^(2)+b^(2) + c^(2) =1;3a + 4b +2c = ; b - 2c = 4//3` `vec(A)=2hat(i)-2hat(j)+hat(k), vec(N)-hat(j)-2hat(k) vec(R)=ahat(i)+bhat(j)+chat(k) , thereforevec(R)` is a unit vector `thereforea^(2)+b^(2)+c^(2)=1 .......(1)` For all these vectors to be in a plane. `(vec(A)xxvec(N)).vec(R)=0rArr[3hat(i)+4hat(j)+2hat(k)].[ahat(i)+bhat(j)+chat(k)]=0` `rArr 3a+4b+2c=0..............(2)` Now `vec(A).vec(N)=|vec(A)||vec(N)| cos (pi-i)` and `vec(R).vec(N)=|vec(R)||vec(N)| cos i` `therefore(vec(A).vec(N))/(|vec(A)||vec(N)|)=(-vec(R).vec(N))/(|vec(R)||vec(N|))rArr(-2-2)/(3)=(-(b-2c))/(1)rArrb-2c=(4)/(3).........(3)` The equation `(1),(2)` and `(3)` are the required relations. |
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| 59. |
A point object is placed on principal axis of a concave mirror of radius of curvature `20 cm` at a distance `31 cm` from poll of the mirror. `A` glass slab of thikness `3 cm` and refractive index `1.5` is placed between object and mirror as shown in the figure. Find the distance (in `cm`) of final image formed by the system from the mirror. |
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Answer» Correct Answer - 16 Apprent shift in position of the object due to refraction through the slab `=d(1-(1)/(mu_(rel)))` `=3(1-(1)/(3/2))=1 cm ` towards the mirror, Now, for reflection from the mirror, `u=-30 cm f=-(R)/(2)=-10 cm` `therefore` `(1)/(v)=(1)/(f)-(1)/(u)` gives `(1)/(v)=-(1)/(10)-(1)/((-30))` `therefore v=-15 cm` Thus is formed `15 cm` , right of the mirror but because of second refraction through the slab the image is shifted `1 cm` away from the mirror. Hence final image is formed at a distance `16 cm` away from mirror. |
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| 60. |
A point object at 15 cm from a concave mirror of radius of curvature 20 cm is made to oscillate along the principal axis with amplitude 2 mm. The amplitude of its image will beA. `2 mm`B. `4 mm`C. `8 mm`D. `16 mm` |
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Answer» Correct Answer - Converging; real Using mirror formula, `(1)/(-10)=(1)/(v)+(1)/(-15) , implies v=-30cm` . | Axial magnification| `=(V^(2))/(u^(2))=((30)/(15))^(2)=4` amplitude of image `=4xx2=8mm` . |
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| 61. |
Light is incident from glass to are. The variation of the angle of deviation `delta` with the angle of incidence `i` for `0ltilt90^(@)` If the value of `(x+y+z)` is (npi)/(6)` then value of n. |
| Answer» Correct Answer - 5 | |
| 62. |
In the figure, a point object O is placed in air. A spherical boundry separates two media of radius of curvature `1.0 m.` AB is principal axis. The separation between the images formed due to refraction at spherical surface is |
| Answer» Correct Answer - 12 | |
| 63. |
A point object is placed at a distance of `20 cm` from a thin plano-concex lens of focal length `15 cm`. The plane surface of the lens is now silvered. The image created by the is at : A. `60 cm` to the left of the system.B. `60 cm` to the right of the system.C. `12 cm` to the left of the system.D. `12 cm` to the right of the system. |
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Answer» Correct Answer - C By mirror-lens combination formula `(1)/(F)=(1)/(f_(m))-(1)/(f_(L))` `(1)/(F)=(1)/(infty)-(2)/(15)` By mirror formula `(1)/(F)=(1)/(u)+(1)/(v) rArr -(2)/(15)=(1)/(-20)+(1)/(v)` `rArr (1)/(v)=(1)/(20)-(2)/(15)=(3-8)/(60)` `v=-12 cm` negartive means toward left |
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| 64. |
A fluorescent lamp of length `1m` is placed horizontally at a depth of `1.2 m` below a ceiling . `A` plane mirror of length `0.6 m` is placed below the lamp parallel to and symmetric to the lamp at a distance `2.4 m` from it as shown in figure. Find the length in meters (distance between the extreme points of the visible region along `x`-axis ) of the reflected patch of light on the ceiling. |
| Answer» Correct Answer - 3 | |
| 65. |
STATEMENT -1: A white parallel beam of light is incident on a plane glass-vacuum interface as shown. The beam my not undergo dispesion after suffering deviation at the interface (The beam is not incident normally on the interface.) STATEMENT -2: Vacuum has same refractive index for all colours of white light.A. Statement-`1` is True, Statement-`2` is True, Statement-`2` is a correct explanation for Statement -`1`B. Statement-`1` is True, Statement-`2` is True, Statement-`2` is `NOT` a correct explanation for Statement -`1`C. Statement-`1` is True, Statement-`2` is FalseD. Statement-`1` is False, Statement-`2` is true |
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Answer» Correct Answer - B If the angle of incidence is greater than critical angle for all colours, the beam will be torally internally reflrcted. The beam may not always suffer refraction and hence dispersion. Hence both statements are ture. Since statement-`2` has no connection with `TIR` , it is not an an explanation of statement-`1` |
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| 66. |
There is a small black dot at the centre C of a solid glass sphere of refractive index `mu`. When seen from outside, the dot will appear to be locatedA. away from `C` for all values of `mu`B. at `C` for all values of `mu`C. at `C` for `mu=1.5` , but for `C` for `mune1.5`D. at `C` only for `sqrt2lemule1.5` . |
| Answer» Correct Answer - A | |
| 67. |
The wavelength of light in vacuum is `6000 A^(@)` and in a medium it is `4000 A^(@)` The refractive index of the medium is:A. `2.4`B. `1.5`C. `1.2`D. `0.67` |
| Answer» Correct Answer - B | |
| 68. |
In the given figure rasy incident on an interface would converge `10cm` below the interface if they continues to move in straight lines without bending. But due to refraction, the rays will bend and meet some where else. Find the distance of meeting point of refracted rasy below the interface, assuming the rays to be making small angles with the normal to the interface. |
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Answer» Correct Answer - `25 cm` Apparent depth `= (n_(1))/(n_(2)) xx d = (5)/(2) xx 10 = 25cm` |
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| 69. |
Find the focal length of the lens shown in Fig. |
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Answer» `(1)/(f) = (n_("rel") - 1) ((1)/(R_(1)) - (1)/(R_(2))) = ((3)/(2) - 1) ((1)/(-10) - (1)/(10))` `f = -10cm` |
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| 70. |
A prism having an apex angle of `4^@` and refractive index of 1.50 is located in front of a vertical plane mirror as shown in the figure. A horizontal ray of light is incident on the prism. The total angle through which the ray is deviated is: A. `4^(@)` clockwiseB. `178^(@)` clockwiseC. `2^(@)` clockwiseD. `8^(@)` clockwise |
| Answer» Correct Answer - A | |
| 71. |
Light of wavelenght `4000 A` is incident at small angle on a prim of apex angle `4^(@)` . The prism has `n_(v)=1.5` and `n_(r)=1.48` . The angle of dispersion produced by the prism in this light is:A. `0.2^(@)`B. `0.08^(@)`C. `0.192^(@)`D. none of these |
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Answer» Correct Answer - C Dispersion will not occur for a monochromatic light. |
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| 72. |
Find the apparednt distance between the observer and the object shwon in the figure and shift in the position of object. |
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Answer» Correct Answer - `35cm, "Shift" = 5 cm` Apparent shift `= t(1 - (1)/(mu)) = 10(1 - (1)/(2)) = 5cm` towards slab. Apparent distance `= 10 + 20 + 20 - 5 = 35cm` |
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| 73. |
Find the focal length of the lens shwon in figure (a) If the light is incident form left side. (b) If the light is incident form right side. |
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Answer» (a) `(1)/(f) = (n_("net") - 1) ((1)/(R_(1)) - (1)/(R_(2))) = ((3)/(2) - 1) ((1)/(-60) - (1)/(-20))` `f = 60cm` (a) `(1)/(f) = (n_("net") - 1) ((1)/(R_(1)) - (1)/(R_(2))) = ((3)/(2) - 1) ((1)/(20) - (1)/(60))` `f = 60cm` |
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| 74. |
The maximum refractive index of a material, of a prism of apex `90^(@)` , for which light may be transmitted isA. `sqrt3`B. `1.5`C. `sqrt2`D. none of these |
| Answer» Correct Answer - B | |
| 75. |
An extended object of size `2cm` is placed at a distance of `10cm` in air `(n = 1)` form pole, on the other side of refracting surface has refractive index `n = 2`. Find the positon, nature and size of image formed after single refraction through the curved surface. |
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Answer» Correct Answer - `40 cm` form pole in the medium of refractive index `1`, virtual erect and `4cm` in size. For refraction at spherical surface `(mu_(2))/(v) - (mu_(1))/(u) = (mu_(2) - mu_(1))/(R) rArr (2)/(V) - (1)/(-10) = (2-1)/(+20)` `rArr (2)/(v) + (1)/(10) = (1)/(20) rArr (2)/(V) = -(1)/(20) rArr v = -40cm`. (virtual) Using magnification formula. `m = (h_(2))/(h_(1)) = + ((mu_(1))/(mu_(2)))((v)/(u)) rArr (h_(2))/(2) = (1xx(-40))/(2xx(-10)) rArr h_(2) = +4cm` (erect) |
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| 76. |
An image of a candle on a screen is found to be double its size. When the candle is shifted by a distance of 5cm, then the image becomes triple its size. Find the nature and radius of curvature of the mirror. |
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Answer» Since the images formed on screen it is real. Real object and real image implies concave mirror. Applying `m = (f)/(f-u)` or `-2 = (f)/(f-(u))..........(1)` After shifting `-3 = (f)/(f-(u+5)).....(2)` [Why `u + 5` why not `u-5` in a concave mirror, teh size of real image will increases only when the real object is brought closer to the mirror. In doing so, its `x` coordinate will increases] From `(1) & (2)` we get, `f = -30cm`, or `R = -60cm` (concave) adn `R.O.C = 60cm` |
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| 77. |
An object is placed `10cm` away form a glass piecwe `(n = 1.5)` of length `20cm` bounded by spherical surfaces of radii of curvaure `10cm`. Find the position of final image formed after two refractions at the spherical surfaces. |
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Answer» Correct Answer - `50cm` right of `B` For first reaction: `(1.5)/(v_(1)) - (1)/(-10) = (1.5-1)/(10) rARr v_(1) = -30cm` For secound refractions `u = -(30+20) = -50cm` `:. (1)/(v_(2)) - (1.5)/(-50) = (1-1.5)/(-10)` `rArr v_(2) = 50cm` Hence, final image is formed `50cm` right of `B`. |
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| 78. |
In Fig. , find the apparent depth of the object seen below surface AB. |
| Answer» `D_(app) = sum (d)/(mu) = (20)/(((2)/(1.8))) + (15)/(((1.5)/(1.8))) = 18 + 18 = 36cm`. | |
| 79. |
Find the apparent depth of object `O` below surface `AB`, seen by an observer in mediumn of refractive index `mu_(2)` |
| Answer» `d_(pp) = (t_(1))/(mu_(1)//mu_(2))` | |
| 80. |
A biconcave lens made of transparent material of refractive index `1.25` is immersed in water of refractive index `1.33`. Will the lens behave as a convering or diverging lens? Given reason. |
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Answer» `(1)/(f)=((mu_(i))/(mu_(s))-1) [(1)/(R_(1))-(1)/(R_(2))]` For concave lens `[(1)/(R_(1))-(1)/(R_(2))]=-ve` here `=1.25` `s=1.32` Then `((mu_(i))/(mu_(s))-1)=-ve` Then `(1)/(f)rarr(-ve)(-ve)` Thus it will behave like converging lens. |
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| 81. |
When a lens of power `P` (in air) made of material of refractive index `mu` is immersed in liquid of refractive index `mu_(0)` Then the power of lens is:A. `(mu-1)/(mu-mu_(0))P`B. `(mu-mu_(0))/(mu-1)P`C. `(mu-mu_(0))/(mu-1)P`. `(P)/(mu_(0))`D. none of these |
| Answer» Correct Answer - D | |
| 82. |
Two symmetric double convex lenses A and B have same focal length but the radii of curvature differ so that `R_(A) = 0.9R_(B)`. If refractive index of A is 1.63 find the refractive index of B.A. `1.7`B. `1.6`C. `1.5`D. `4//3` |
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Answer» Correct Answer - C `(1)/(f)=(mu-1)((1)/(R_(1))-(1)/(R_(2))) (1)/(f)=(1.63-1)((2)/(R_(A)))=(n_(B)-1)((2)/(R_(B)))` `n_(B)-1=0.63xx(R_(B))/(R_(1))=(0.63)/(0.9)=0.7 , n_(B)=1.7` |
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| 83. |
I is the image of a point object `O` formed by spherical mirror, then which of the following statements is incorrect :A. If `O` and `I` are on same side of the principal axis, the they have to be on opposite sides of the mirror.B. If `O` and `I` are on opposite side of the principal axis, then they have to be on same side of the mirror.C. If `O` and `I` are on opposite side of the principal axis, then they can be on opposite side of the mirror as well.D. If `O` is on principal axis then `I` has to lie on principal axis only. |
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Answer» Correct Answer - Converging; real `(I)/(O)=-(v)/(u)` If `O` and `I` are on same sides of `PA`. `(I)/(O)` will be positive which implies `v` and `u` will be of oppostite signs. Similarly if `O` and `I` are on app. Sidess, `(I)(O)` will be-ve which implies `v` and `u` will have same sign. If `O` is on `PA` , `I=(-(V)/(u))(O)=0` implies I will also be on. `P`,`A`. |
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| 84. |
An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between 1/v versus 1/u isA. B. C. D. |
| Answer» Correct Answer - B | |
| 85. |
An object is placed at a distance u from a concave mirror and its real image is received on a screen placed at a distance of v from the mirror. If f is the focal length of the mirror, then the graph between 1/v versus 1/u isA. B. C. D. |
| Answer» Correct Answer - A | |
| 86. |
A medium has `n_v = 1.56, n_r=1.44`. Then its dispersive power is:A. `3//50`B. `6//25`C. `0.03`D. none of these |
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Answer» Correct Answer - D `omega=(n_(v)-n_(r))/(((n_(v)+n_(r))/(2))-1)=(6)/(25)` |
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| 87. |
A monochromatic beam of light is incided at `60^(@)` in one face of an equilateral prism of refractive index `n` and emerges from the opposite face making an angle `theta(n)` with the normal (see the figure). For `n=sqrt3` the value of `theta` is `60^(@)` and `(dtheta)/(dn)=m`. The value of `m` is |
| Answer» Correct Answer - a,c | |
| 88. |
Find the angle of incidence of the ray shown in Fig. for which it passes through the pole, given that `MI||CP`. |
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Answer» `/_ MlC = /_ CIP = theta` `MI || CP /_ MI theta = /_ ICP = theta` `CI = CP` `/_ CIP = /_ CPI = theta` `:.` In `Delta CIP` all angle are equal `3 theta = 180^(@) rArr theta = 60^(@)` |
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| 89. |
In the shown figure `M_(1)` and `M_(2)` are two concave mirrors of the same focal length `10 cm` . `AB` and `CD` are their pricipal axes respectively. `A` point object `O` is kept on the line `AB` at a distance `15 cm` from `M_(1)` . The distance between the mirror is `20 cm` . Considering two successive reflections first on `M_(1)` and then on `M_(2)` . The distance of final image from the line `AB` is: A. `3 cm`B. `1.5 cm`C. `4.5 cm`D. `1 cm` |
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Answer» Correct Answer - B For `M_(1)` `v=(uf)/(u-f) =(-15xx(-10))/(-15-(-10))=-30 cm` For `M_(2) u=10 cm` `therefore v=(10xx(-10))/(10-(-10))=-5 cm` magnification `m=(-v)/(u)=-((-5)/(10))=(1)/(2)` so, distance of image from `CD=(1)/(2)xx3=(3)/(2) cm` `therefore` distance of image from `AB=3-(3)/(2)=(3)/(2) cm` |
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| 90. |
A coin is placed `10cm` in front fo a concave mirror. The mirror produces a real image that has diameter `4` times that of the coin. What is the image distance. |
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Answer» `m = (d_(2))/(d_(1)) = -(v)/(u)` `rArr -4 = - (v)/(u) rArr v = 4u` `= 4xx(-10)` `= -40cm` |
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| 91. |
Find the position of the final image after three successive reflections taking the first reflection on `m_(1)` . |
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Answer» `1^(st)` reflection at `m_(1)` `u = -15cm` `f = -10cm` `(1)/(v) + (1)/(u) = (1)/(f) = (-3+2)/(30) = -(1)/(30)` `v = -30cm` `2^(nd)` reflection at plane mirror: `u = 5cm` `v = -5cm` For `III` reflection on curved mirror again: `u = -20cm` `v = (uf)/(u-f) = ((-20)xx(-10))/(-20+10) = (200)/(-10) = 20cm` Image is `20cm` right of `m_(1)` |
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| 92. |
Find the position of the final image after three successive reflections taking the first reflection on `m_(1)` . |
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Answer» `I` reflection: Forcus of mirror `= -10cm rArr u = -15 cm` Applying mirror formula: `(1)/(v) + (1)/(u) = (1)/(f) rArr v = -30cm` For `II` reflection on plane mirror: `u = -10 cm` `:. V = 10cm` For `III` reflection on curved mirror again: `u = -50cm` `f = -10cm` Applying mirror formula: `(1)/(v) + (1)/(u) = (1)/(f)` `v = -12.5cm` |
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| 93. |
Find the lateral magnification produced by the combination of lenses shown in Figure. |
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Answer» `(1)/(f) = (1)/(f_(1)) + (1)/(f_(2)) = (1)/(10) - (1)/(20) = (1)/(20) rArr f = +20` `,. (1)/(v) - (1)/(-10) = (1)/(20) rArr (1)/(v) = (1)/(20) - (1)/(10) = (-1)/(20) = -20cm` `:. M = (-20)/(-10) = 2` |
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| 94. |
An object of height `h_(0)=1 cm` is moved along principal axis of a convex lens of focal length `f=10 cm`. Figure shows variation of magnitude of height of image with image distance `(v)` . Find `v_(2)-v_(1)` in `cm` . |
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Answer» Correct Answer - 10 `(h_(i))/(h_(0))=(f-v)/(f) rArrh_(i)=-(v)/(f)h_(0)+h_(0)rArr |h_(i)|=-(v)/(f)h_(0)+h_(0)-inftylevlef` `|h_(i)|=(v)/(f)h_(0)-h_(0) flevle-infty` So `h_(2)=h_(2)=1 cm` From second eq. `v_(2)=2f` Or When `vrarr0,urarr0&h_(i)rarrh_(0)soh_(2)=h_(0)=1 cm` image of same height is obtained when `v=2f` so `v_(2)=2f` |
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| 95. |
A thin convex lens made from crown glass `(mu = 3//2)` has focal length `f`. When it is measured in two different liquids having refractive indiced `4//3` and `5//3`, it has the focal length `f_(1)` and `f_(2)` respectively. The correct ralation between the focal lengths isA. `f_(1)=f_(2)ltf`B. `f_(1)gtf` and `f_(2)` becomes negativeC. `f_(2)gtf` and `f_(1)` becomes negativeD. `f_(1)` and `f_(2)` both become negative |
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Answer» Correct Answer - A::B::C::D `(1)/(f)=((3)/(2)-1)((1)/(R_(1))-(1)/(R_(2)))` `(1)/(f)=(1)/(2x)rArr f=2x` here `((1)/(x)=(1)/(R_(1))-(1)/(R_(2)))` `(1)/(f_(1))=((3//2)/(4//3)-1)(1)/(x)` `(1)/(f_(2))=((3//2)/(5//3)-1)((1)/(x)): rArr f_(2)` is iegative `(1)/(f_(1))=(1)/(8x)=(1)/(4(2x))=(1)/(4f)` `rArr f_(1)=4f` Analytically, If a lense is inserted in a denser sourrounding the sing of focal length changes changes and if lens is inserted in a rarer sourrounding, the sign of focal length remain same. If lense is inserted in rarer medium the focal length increases. |
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