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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.
| 651. |
Which of the following diagrams shows correctly the dispersion of white light by a prism ?A. B. C. D. |
| Answer» Correct Answer - b | |
| 652. |
A person looking at a mesh of crossed wires is able to see the vertical wires more distincly than the horizontal wires. Why ? How can it be corrected ? |
| Answer» This problem is due to astigmatism of the eye. This defect is removed using a cylindrical lens with appropriate axis and suitable radius of curvature. | |
| 653. |
A short sighted person can see distinctly only those objects which lie between 10 cm and 100 cm from him. The power of the spectacle lens required to see a distant object isA. `+ 0.5 D`B. `- 1.0 D`C. `- 10 D`D. `+ 4.0 ` |
| Answer» Correct Answer - b | |
| 654. |
A myopia person has been using spectacles of power `- 1.0` dioptre for distant vision. During old age, he also needs to use separate reading glasses of power `+ 2.0` dioptre. Explain what may have happened. |
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Answer» As the person is using spectacles of power `- 1.0 dioptre` (i.e., focal length `-100 cm`), the far point of the person is at `100 cm`. Near point of the eye might have been normal (i.e., 25 cm). The objects at infinity produce virtual images at `100 cm` (using spectacles). To see objects between `25 cm` and `100 cm`, the person uses the ability of accommodation of his eye lens. This ability is partially lost in old age. The near point of the eye may recede to `50 cm`. He has, therefore to use glasses of suitable power for reading. Here, `u = -25 cm, v = -50 cm, f = ?` As `(1)/(v)-(1)/(u)=(1)/(f) :. (-1)/(50)+(1)/(25)=(1)/(f)` or `(-1 + 2)/(50) = (1)/(f)` or `f = 50 cm` As `P = (100)/(f)=(100)/(50) = + 2 dioptre`. |
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| 655. |
A ray incident a `15^(@)` on one refracting surface of a prism of angle `60^(@)` , suffers a deviation of `55^(@)` . What is the angle of emergenceA. `95^(@)`B. `45^(@)`C. `30^(@)`D. None of these |
| Answer» Correct Answer - d | |
| 656. |
Where should an object be held so that a concave mirror forms a real, inverted and magnified image ? |
| Answer» The object must lie between principal focus `F` and centre of curvature `C` of the concave mirror. | |
| 657. |
A thin rod of 5 cm length is kept along the axis of a concave mirror of 10 cm focal length such that its image is real and magnified and one end touches to rod. Its magnification will beA. `1`B. `2`C. `3`D. `4` |
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Answer» Correct Answer - B |
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| 658. |
A thin rod of 5 cm length is kept along the axis of a concave mirror of 10 cm focal length such that its image is real and magnified and one end touches the rod. Its magnification will beA. 1B. 2C. 3D. 4 |
| Answer» Correct Answer - b | |
| 659. |
In vacum, to travel distance d, light takes time t and in medium to travel distance 5d, it takes time T. The critical angle of the medium isA. `sin^(-1)((5T)/(t))`B. `sin^(-1)((5T)/(3t))`C. `sin^(-1)((5T)/(T))`D. `sin^(-1)((3t)/(5T))` |
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Answer» Correct Answer - C In vacuum, `c=d/t` In medium, `v=(5d)/(T)` As refractive index, `mu=(c )/(v)=(d//t)/(5d//T)=(T)/(5t)` Also, `sinC=(1)/(mu) (therefore "C is critical angle")` `therefore C=sin^(-1)[(5t)/(T)]` |
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| 660. |
An object placed at 20cm in front of a concave mirror produces three times magnified real image. What is the focal length of the concave mirror?A. 15cmB. 6.6cmC. 10cmD. 7.5cm |
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Answer» Correct Answer - A We know that, linear magnification, `m=(f)/(f-u)` Given object displaced, u=-20cm m=-3 (`therefore "all images aer inverted"`) `So, -3=(f)/(f-(-20))` `-3=(f)/(f+20) Rightarrow -3f-60=f` `4f=-60 Rightarrow f=-(60)/(4)=-15cm` Since, mirror is concace, f=-15cm |
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| 661. |
A concave mirror produces three times magnified real image of an object placed at 10 cm in front of it. Where is the image located ?A. 10 cmB. 15 cmC. 20 cmD. 30 cm |
| Answer» Correct Answer - b | |
| 662. |
Ten identical converging thin lenses,each of focal length `10cm`are in contact .What is the power of the combined lens. |
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Answer» For thin lenses in contact `P=P_(1)+P_(2)+……..` `=10P_(1)=(10xx100)/10` `=100D` |
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| 663. |
A converging lens of focal length `5.0cm` is placed in contact with a diverging lens of focal length `10.0cm.` Find the combined focal length of the system.A. `+ 10.0 cm`B. `- 10.0 cm`C. 5.0 cmD. `- 5 cm` |
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Answer» Correct Answer - A Here, `f_(1)=+5.0 cm` and `f_(2)=-10.0 cm` Therefore, the combined focal length F is given by `(1)/(F)=(1)/(f_(1))+(1)/(f_(2))=(1)/(5.0)-(1)/(10.0)=+(1)/(10.)` `therefore " " F = +10.0 cm` i.e., the combination behaves as a converging lens of focal length 10.0 cm. |
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| 664. |
A lense behaves as a converging lens is air and diverging lens in water. The refractive index of the lens material is -A. Equal to unityB. Equal to 1.33C. Between unity and 1.33D. Greater than 1.33 |
| Answer» Correct Answer - c | |
| 665. |
Two lenses, one diverging of power 2 diopyre and the other converging of power `6 dioptre` are combined together. Calculate focal length and power of the combination. |
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Answer» Correct Answer - 25 cm, 4 D Here, `P_1 = -2 D P_1 = + 6 D` Power of combination, `P = P_1 + P_2 = -2 + 6 = 4 D` `F = (100)/(P) = (100)/(4) = 25 cm`. |
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| 666. |
Real image of an object are formed on the fromed on the two positions of a lens separated by a distance`60cm` `The ratio between the sizes of the two image will be:A. `1//3`B. `3`C. `1//4`D. `1//2` |
| Answer» Correct Answer - C | |
| 667. |
The following figure from shows version of a zoom.The converging lens has a focal length `f_(1)`and the diverging lens has focal length `f_(2)=-|f_(2)|`.the two lens are separated by a variable distance d that it is always less than `f_(1),`also the magnitude of the focal length of the diverging lens satisfies the inequality `|f_(2)|gt(f_(1)-d).` if the rays that emerge form the diverging lens and reach the final image point are extended backward to the left of the diverging lens,they will eventually expand to the original radius `r_(o)`at the same point.`Q`.To determine the effective focal length of the combination lens consider a bundle of parallel rays of radius `r_(0)`entering the emerging lens. the effective focal length `f` is given byA. `(f_(1)|f_(2)|)/(|f_(2)|-f_(1)+d)`B. `(f_(1)f_(2))/(f_(1)-f_(2)-d)`C. `(f_(1))/(f_(2)-f_(1)+d)`D. `(f_(2))/(f_(1)+f_(2)+d)` |
| Answer» Correct Answer - A | |
| 668. |
An object `25cm`high is placed in front of a concave lens of focal length `30cm`.If the height of image fromed is `50cm` The image distance is ,if the image is real and inverted:A. `90cm`B. `45cm`C. `30cm`D. `25cm` |
| Answer» Correct Answer - A | |
| 669. |
STATEMENT-`1` Image fromed by a concave mirror is always smaller in size. `STATEMENT 2` It is always virtual.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 a not a correct explanation for statement -`1`C. Statement -`1`is true,Statement -`2`is FalseD. Statement -`1`is False ,Statement -`2`is True |
| Answer» Correct Answer - B | |
| 670. |
An object `25cm`high is placed in front of a concave lens of focal length `30cm`.If the height of image fromed is `50cm` if the image is erect and virtual then distance between object and image isA. `25cm`B. `15cm`C. `40cm`D. `60cm` |
| Answer» Correct Answer - B | |
| 671. |
A beam of light converges to a point `P`. A lens is placed in the path of the covergent beam `12 cm` from `P`. At what point does the beam converge if the lens is (a) a convex lens of focal length `20 cm` (b) a concave lens of focal length `16 cm` ? |
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Answer» Here, the point `P` on the right the lens acts as a virtual object, `:. u = 12 cm, v = ?` (a) `f = 20 cm` As `(1)/(v)-(1)/(u)=(1)/(f) :. (1)/(v)-(1)/(12)=(1)/(20)` `(1)/(v)=(1)/(20)+(1)/(12)=(3 + 5)/(60) = (8)/(60)` `v = 60//8 = 7.5 cm` Image is at `7.5 cm` to the right of the lens where the beam converges. (b) `f = -16 cm, u = 12 cm, :. (1)/(v)=(1)/(f)+(1)/(u)=-(1)/(16)+(1)/(12)=( - 3 + 4)/(48) = (1)/(48)` `v = 48 cm` Hence, image is at `48 cm`. to the right of the lens, where the beam would converge. |
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| 672. |
An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation ` A. `PQ`is horizontalB. `QR` is horizontalC. `RS` is horizontalD. Either`PQ`or`RS`is horizontal |
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Answer» Correct Answer - B For minimum deviation `QR` will be parallel to base. |
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| 673. |
An equilateral prism is placed on a horizontal surface. A ray PQ is incident onto it. For minimum deviation ` A. PQ is horizontalB. QR is horizontalC. RS is horizontalD. Either PQ or RS is horizontal |
| Answer» Correct Answer - b | |
| 674. |
A point object is placed at the center of a glass sphere of radius 6cm and refractive index 1.5. The distance of virtual image from the surface isA. 2 cmB. 4 cmC. 6 cmD. 12 cm |
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Answer» Correct Answer - C (c ) All rays coming from object at centre arenormal. So, no bending will take place. So, the distance of the virtual image from the surface of sphere is 6cm. |
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| 675. |
Answer the following questions : (a) You have learnt that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain. (b) A virtual image, we always say, cannot be caught on a screen. Yet when we see a virtual image, we bring it to screen i.e. retine of our eye. Is there a contradiction ? ( c) A diver under water looks obliquely at a fisherman standing on the bank of a lake. Would the fisherman look taller or shorter than what he actually is ? (d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decrease ? ( e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter ? |
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Answer» (a) Yes, when rays incident on a plane or a mirror are tending to converge to a point behind the mirror, they are reflected to a point on a screen in front of the mirror. Hence a real image is formed (when the object is virtual). (b) No, there is no contradiction. Eye lens is convergent. It forms a real image of the virtual object (i.e., the virtual image being seen) on the retina. ( c) As fisherman is in air, light travels from rarer to denser medium. It bends towards the normal, appearing to come from a larger distance. Therefore, to the diver under water, fisherman looks taller. (d) Yes, the apparent depth decreases further, when water tank is viewed obliquely compared to the depth when seen near normally. As `mu = (1)/(sin C) :. sin C = (1)/(mu)` As refractive index of diamond is much greater than that of ordinary glass, critical angle `C` for diamond is much smaller `(~~ 24^@)` as compared to that for glass `(~~42^@)`. A skilled diamond cutter exploits the large range of angles of incidence of light `(24^@` to `90^@)` to ensure that light entering the diamond suffers multiple total internal reflections within the diamond. This produces sparkling effect in the diamond. |
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| 676. |
An object approaches a convergent lens from the left of the lens with a uniform speed `5 m//s` and stops at the focus. The image.A. moves away from the lens with an uniform speed 5 m/sB. moves away from the lens with an uniform accelerationC. moves away from the lens with a non-uniform accelerationD. moves towards the lens with a non-uniform acceleration |
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Answer» Correct Answer - C (c ) When an object approaches a convergent lens from the left of the lens with a uniform speed of 5/ms, then the image is move away from the lens with a non-uniform acceleration. |
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| 677. |
An object appraches a convergent lens from the left of the lens with a uniform speed `5m//s` and stops at the focus. The imageA. moves away from the lens with a uniform speed `5m//s`B. moves away from the lens with a uniform accelerationC. moves away from the lens with a non-uniform accelerationD. moves towards the lens with a non-uniform acceleration |
| Answer» Correct Answer - (c) | |
| 678. |
The phenomena involved in the reflected of radiowaves by ionosphere is similar to.A. reflection of light by a plane mirror.B. total internal reflection of light in air during a mirage.C. dispersion of light by water molecules during the formation of a rainbow.D. scattering of light by the particles of air. |
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Answer» Correct Answer - B The phenomenon involved in the reflection of radiowaves by ionosphere is similar to total internal reflection of light in air during a mirage (angle of incidence gt critical angle). Choice (b) is correct. |
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| 679. |
The direction of ray of light incident on a concave mirror is shown by `PQ` while directions in which the ray would travel after reflection is shown by four rays marked `1,2,3` and `4`, Fig. Which of the four rays correctly shows the direction of reflected ray ? .A. 1B. 2C. 3D. 4 |
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Answer» Correct Answer - B In Fig., `PQ` is a ray of light passing through focus, and falling on the surface of a concave mirror. On reflection, from the mirror, the ray becomes parallel to principal axis of the mirror. Choice (b) is correct. |
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| 680. |
A plano convex lens has focal length `f = 20 cm`. If its plano surface is silvered, then new focal length will be.A. 20 cmB. 40 cmC. 30 cmD. 10 cm |
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Answer» Correct Answer - D (d) The focal length of the plano-convex lens is 20 cm with the plane surface bieng silvered. So, the net power of the lens is given by `P=1/fxx2=2/f` Thus, the net focal length is, `f/2=20/2=10 cm` |
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| 681. |
Assertion : A convex lens of glass `(mu = 1.5)` behave as a diverging lens when immersed in carbon disulphinde of higher refractive index `(mu = 1.65)`. Reason : A diverging lens is thinner in the middle and thicker at the edges.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true and reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
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Answer» Correct Answer - B `mu = (mu_(g))/(mu_(c)) = (1.5)/(1.65) lt 1` As `1/f = (mu- 1) (1/(R_(1)) - 1/(R_(2)))` `:. f` becomes negative. Therefore, the lens behaves as a diverging lens. |
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| 682. |
A convex lens of focal legnth `0.2 m` and made of glass `(mu = 1.50)` is immersed in water `(mu = 1.33)`. Find the change in the focal length of the lens.A. `5.8 m`B. `0.58 cm`C. `0.58m`D. `5.8 cm` |
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Answer» Correct Answer - C Using `(1)/(f_(a)) = (.^(a)mu_(g) - 1) ((1)/(R_(1))- (1)/(R_(2)))` Here, `f_(a) = 0.2 m, .^(a)mu_(g) = 1.50` `:. (1)/(0.2) = (1.50 - 1) ((1)/(R_(1)) - 1/(R_(2))) rArr (1)/(R_(1)) - (1)/(R_(2)) = 10` Consider `f_(w)` be the focal length of the lens , when immersed in water. `.^(w)mu_(g) = (.^(a)mu_(g))/(.^(a)mu_(w)) = (1.50)/(1.33) = 1.128` Now, `(1)/(f_(w)) = (.^(w)mu_(g) - 1)((1)/(R_(1)) - (1)/(R_(2))) = (1.28 - 1) xx 10 = 1.28` or `f_(w) = (1)/(1.28) = 0.78` Hence, change in forcal length of the lens is `f_(w) - f_(a) = 0.78 - 0.2 = 0.58 m` |
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| 683. |
Fig shows a small air bubble inside a glass sphere `(mu = 1.5)` of radius 10 cm. the bubble is 4.0 cm below the surface and is viewed normally from the outside. Find the apparent depth of the bubble. |
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Answer» Correct Answer - 3 cm below the surface |
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| 684. |
The radii of curvature of a double convex lens are 30 cm and 60 cm and its refractive index is 1.5. calculate its focal length. |
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Answer» Correct Answer - 40 cm |
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| 685. |
A thin double convex lens has radii of curvature each of magnitude 40 cm and is made of glass with refractive index 1.65. Its focal length is nearlyA. 20 cmB. 31 cmC. 35 cmD. 50 cm |
| Answer» Correct Answer - b | |
| 686. |
A ray of light is incident at an angle of incidence `45^@` on an equilateral prism and emerge at an angle `45^@` then the refractive index of the medium of the prism is:A. `1.96xx10^(8) m//s`B. `2.12xx10^(8) m//s`C. `3.18xx10^(8) m//s`D. `3.33xx18^(8) m//s` |
| Answer» Correct Answer - b | |
| 687. |
A light ray is incident by grazing one of the face of a prism and after refraction ray does not emerge out, what should be the angle of prism while critical angle is `C` ?A. Equal to 2CB. Less than 2CC. More than 2CD. None of the above |
| Answer» Correct Answer - c | |
| 688. |
The angle of incidence for a ray of light at a refracting surface of a prism is `45^(@)`. The angle of prism is `60^(@)`. If the ray suffers minimum deviation through the prism, the angle of minimum deviation and refractive index of the material of the prism respectively, are `:`A. `30^@, sqrt(2)`B. `45^@, sqrt(2)`C. `30^@, (1)/(sqrt(2))`D. `45^@, (1)/(sqrt(2))` |
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Answer» Correct Answer - A Here, `i_1 = 45^@, A = 60^@ , delta_m = ? Mu = ?` When the ray suffers minimum deviation, `i_1 = i_2. As A + delta_m i_1 + i_2 = 90^@` `delta_m = 90^@ - A = 90^@ - 60^@ = 30^@` `mu = (sin(A + delta_m)//2)/(sin A//2) = (sin(60^@ + 30^@)//2)/(sin 60^@//2)` =`(sin 45^@)/(sin 30^@) = (1//sqrt(2))/(1//2) = sqrt(2)`. |
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| 689. |
A ray of light falls on a normally on a refracting face of a prism. Find the angle of prism if the ray just fails to emerge from the prism `(mu=3//2).` |
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Answer» Correct Answer - `41.8^(@)` |
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| 690. |
As shown in Fig. `PQ` is a ray incident on prism `ABC`. Show the corresponding refracted and emergent rays. The critical angle for the material of the prism is `45^@`. What is refractive index of the material of prism ? . |
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Answer» Correct Answer - `sqrt(2)` As `PQ` is normal to `AB`, `:. i_1 = 0, r_1 = 0` The ray goes straight along `PQR. At R, sqrt(i) = 45^@` which is critical angle of material of the prism. Therefore, `QR` suffers total internal reflection at `R` and goes along `RS` `mu = (1)/(sin C) = (1)/(sin 45^@) = (1)/(1//sqrt(2)) = sqrt(2)`. |
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| 691. |
A short sighted person is wearing specs of power `-3.5 D`. His doctor prescribes a correction of `+ 2.5 D` for his near vision. What is focal length of his distance viewing part and near vision. What is focal length of his distance viewing part and near vision part ? |
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Answer» Correct Answer - `-28.5 cm ; 16.7 cm` For distance viewing part, `P_1 = -3.5 D` `:. f_1 = (100)/(P_1)=(100)/(-3.5) cm = -28.5 cm` For near vision part, `P_1 + P_2 = P` `P_2 = P - P_1 = 2.5 D - (-3.5 D) = 6.0 D` `f_2 = (100)/(P_2) = (100)/(6.0) = 16.7 cm`. |
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| 692. |
The refracting angle of a glass prism is `60^@` and `mu` of its material is `1.45`. Calculate angle of incidence at the first that will just reflect internaly the ray at the second face. |
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Answer» Correct Answer - `24.2^@` Here, `A = 60^@, mu = 1.45, i_1 = ?` From `mu = (1)/(sin C)`, `sin C = (1)/(mu) = (1)/(1.45) = 0.6897` `C = sin^-1 (0.6897) = 43.6^@` For total internal reflection at `2 nd` face of prism, `r_2 = C = 43.6^@` From `r_1 +m r_2 = A`, `r_1 = A - r_2 = 60^@ - 43.6^@ = 16.4^@` From `mu = (sin i_1)/(sin r_1)`, `sin i_1 = mu sin r_1 = 1.45 xx sin 16.4^@` =`1.45 xx 0.2823 = 0.4093` `:. i_1 = sin^-1 (0.4093) = 24.2^@`. |
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| 693. |
(a) The far point of a myopic person is `80 cm`. In front of the eye. What is the power of the lens required to enable him to see very distant objects clearly ? (b) In what way does the corrective lens help the person above ? Does the lens magnify very distant objects ? Explain carefully. ( c) The person above prefers to remove his spectacles while reading a book. Explain why ? |
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Answer» (a) To see very distant objects clearly, the person needs a concave lens of focal length `f = -80 cm`. `P = (100)/(f)=(100)/(-80)= -1.25 D`. (b) No, the corrective lens is concave and infact it reduces the size of the image. The eye is able to see distant objects not because the corrective lens magnifies the object, but because it brings the object at the far point `F` of the eye (forming virtual image at F) which can then be focussed on the retina by the eye lens. ( c) The myopic person may have a normal near point `(~~ 25 cm)`. To read a book with his spectacles (for distant vision), he has to hold the book at a distance more than `25 cm` so that concave lens produces the image not closer than `25 cm`. An angular size of object at a distance `gt 25 cm` is less than the angular size at `25 cm`, the person prefers to remove his spectacles while reading. |
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| 694. |
A Galilean telescope has objective and eye`-` piece of focal lengths `200 cm` and `2cm` respectively. The magnifying power of the telescope for normal vision isA. 90B. 100C. 108D. 198 |
| Answer» Correct Answer - b | |
| 695. |
Relative difference of focal lengths of objective and eye lens in the microscope and telescope is given asA. It is equal in bothB. It is more in telescopeC. It is more in microscopeD. It may be more in any one |
| Answer» Correct Answer - b | |
| 696. |
The focal lengths of the objective and eye piece of a microscope are `2 cm and 5 cm` respectively, and the distance between them is `20 cm`. Find the distance of the object from the objective when the final image seen by the eye is `25 cm` from the eye piece. What is the magnifying power ? |
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Answer» Correct Answer - `u_0 = -2.3 cm, m = 41.5` Here, `f_0 = 2 cm, f_e = 5 cm, u_0 = ?, m = ?` As final image is at `25 cm` from eye piece, therefore, `v_e = -25 cm` From `(1)/(u_e)=(1)/(v_e)-(1)/(f_e)=(1)/(-25)-(1)/(5)=(-6)/(25)` `u_e = -(25)/(6) cm` Distance of image from objective, `v_0 = 20 - (25)/(6) = (95)/(6) cm` `(1)/(u_0)=(1)/(v_0)-(1)/(f_0)=(6)/(95) - (1)/(2) = -(83)/(2 xx 95)` `u_0 = -(2 xx 95)/(83) = -2.3 cm` Magnifying power `= m_0 xx m_e = (v_0)/(u_0) (1 + (d)/(f_e))` =`(95//6)/(2 xx 95//83) (1 + (25)/(5)) = 41.5`. |
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| 697. |
The focal lengths of the objective and eye piece of a compound microscope are `4 cm and 6 cm` respectively. If an object is placed at a distance of `6 cm` from the objective, calculate the magnification produced by the microscope. Take distance of distinct vision `= 25 cm`. |
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Answer» Here, `f_0 = 4 cm, f_e = 6 cm`, `d = 25 cm, u_0 = -6 cm, m = ?` If `v_0` is distance of image of object formed by the objective lens, then from `(1)/(v_0)-(1)/(u_0)=(1)/(f_0)` `(1)/(v_0)=(1)/(f_0)+(1)/(u_0)=(1)/(4)+(1)/(-6)=(1)/(12)` `v_0 = 12 cm` `m = (v_0)/(u_0)(1 + (d)/(f_e)) = (12)/(-6)(1 + (25)/(6)) = -2 xx (31)/(6)` `m = -10.33`. |
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| 698. |
The focal lengths of the eye piece and objective of a compound microscope are `5 cm and 1 cm` respectively, and the length of the tube is `20 cm`. Calculate magnifying power of microscope when the final image is formed at inifinity. The least distance of distinct cision is `25 cm`. |
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Answer» Correct Answer - 70 Here, `f_e = 5 cm, f_0 = 1 cm, L = 20 cm , m = ?` `v_e = oo, d = 25 cm` As final image is at infinity, image formed by objective lies at the focus of eye piece. `:. v_0 = L - f_e = 20 - 5 = 15 cm` From `(1)/(u_0)=(1)/(v_0)-(1)/(f_0)=(1)/(15)-(1)/(1)= -(14)/(15)` `u_0 = -(15)/(14) cm` As final image is at infinity, `:. m = (v_0)/(u_0) ((d)/(f_e)) = (15)/(15//14) xx (25)/(5) = 70`. |
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| 699. |
The distance between an object and the screen is 100cm. A lens produces an image on the screen when the lens is placed at either of the positions 40cm apart. The power of the lens is nearlyA. `~~ 3` dioptresB. `~~ 5` dioptresC. `~~ 7` dioptresD. `~~ 9` dioptres |
| Answer» Correct Answer - b | |
| 700. |
Two identical glass `(mu_(g)=3//2)` equi- convex lenses of focal length f each are kept in contact. The space between the two lenses is also filled with water `(mu_(g)=4//3).` The focal length of the combination isA. f/3B. fC. `(4f)/(3)`D. `(3f)/(4)` |
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Answer» Correct Answer - D Using combination of lenses, we get The focal length of the combination `rArr f_("eq")=(3f)/(4)` |
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