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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.
| 151. |
A normal eye is not able to see objects closer than 25 cm becauseA. the focal length of the eye is 25 cmB. the distance of the retina from the eye lens is 25 cmC. the eye is not able to increase the focal length beyond a limitD. the eye is not able to decrease the focal length beyond a limit |
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Answer» Correct Answer - D (d) A normal eye is not able to see object closer than 25 cm because the eye is not able to decrease the focal length beyond the limit. |
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| 152. |
The passenger side view mirror on an automobile oftern has the notation"Objects seen in mirrros are closer than they appear".Is the image really father away than the objects?A. Yes,the image is smaller and further away than the objectB. No,the image is smaller and closer than the objectC. No,the image is larger and closer than the objectD. Yes,the image is larger than further away the object. |
| Answer» Correct Answer - B | |
| 153. |
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, 56)B. (48, 48)C. (66, 33)D. (78, 39) |
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Answer» Correct Answer - C::D As `(1)/(v)+(1)/(u)=(1)/(f), v = ((f_u)/(u - f))` Putting `u = 42 cm, 48 cm, 66 cm, 78 cm`. `v` comes out to be `56 cm, 48 cm, 37.7 cm and 34.6 cm`, `S0` (a)(b) are correct. And ( c),(d) are incorrectly recorded. |
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| 154. |
In a compound microscope, the focal lengths of two lenses are `1.5 cm` and `6.25 cm` an object is placed at `2 cm` form objective and the final image is formed at `25 cm` from eye lens. The distance between the two lenses isA. `6.00 cm`B. `7.75 cm`C. `9.25 cm`D. `11.0 cm` |
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Answer» Correct Answer - D Here, `f_(o) = 1.5 cm, f_(e) = 6.25 cm, u_(o) = -2 cm` `v_(e) = -25 cm` For objective `(1)/(v_(o))- (1)/(u_(0)) = (1)/(f_(o)) :. (1)/(v_(o)) = (1)/(-2) = (1)/(1.5` or `v_(o) = 6 cm` For eye piece, `:. (1)/(v_(e)) - (1)/(u_(e)) = (1)/(f_(e)) , (1)/(-25) - (1)/(u_(e)) = (1)/(6.25)` `- (1)/(u_(e)) = (1)/(6.25) + (1)/(25)` or `u_(e) = -5 cm` Distance between two lenses `= |v_(e)| + |u_(e)|` `= 6 cm + 5 cm = 11 cm` |
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| 155. |
STATEMENT-`1` In passing through a lens or prism ,the phase difference between two waves doesnot change. `STATEMENT 2` the optical path lengths of all rays are same.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 - C | |
| 156. |
A convex lens is dipped in a liquid whose refractive index is equal to the refractive of the lens. Then its focal length willA. Become infiniteB. Become small, but non–zeroC. Remain unchangedD. Become zero |
| Answer» Correct Answer - a | |
| 157. |
A luminous object is placed at a distance of `30 cm` from the convex lens of focal length `20 cm`. On the other side of the lens, at what distance from the lens a convex mirror of radius of curvature `10 cm` be placed in order to have an upright image of the object coincident with it ?A. 12 cmB. 30 cmC. 50 cmD. 60 cm |
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Answer» Correct Answer - C |
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| 158. |
A convergent beam of light is incident on a convex mirror so as to converge to a distance 12 cm from the pole of the mirror. An inverted image of the same size is formed coincident with the virtual object. What is the focal length of the mirrorA. 24 cmB. 12 cmC. 6 cmD. 3 cm |
| Answer» Correct Answer - c | |
| 159. |
A convergent beam of light is incident on a convex mirror so as to converge to a distance 12 cm from the pole of the mirror. An inverted image of the same size is formed coincident with the virtual object. What is the focal length of the mirror ?A. 24 cmB. 12 cmC. 6 cmD. 3 cm |
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Answer» Correct Answer - C |
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| 160. |
If the luminous intensity of a 100 W unidirectional bulb is 100 candela, then total luminous flux emitted from the bulb isA. 861 lumenB. 986 lumenC. 1256 lumenD. 1561 lumen |
| Answer» Correct Answer - c | |
| 161. |
A point source of light moves in a straight line parallel to a plane table. Consider a small portion of the table directly below the line of movement of the source. The illuminance at this portion varies with its distance r from the source asA. `E prop (1)/(r)`B. `E prop (1)/(r^(2))`C. `E prop (1)/(r^(3))`D. `E prop (1)/(r^(4))` |
| Answer» Correct Answer - C | |
| 162. |
In the above problem, the luminous flux emitted by sun will beA. `4.43 xx 10^(25) lm`B. `4.43 xx 10^(26) lm`C. `4.43 xx 10^(27) lm`D. `4.43 xx 10^(28) lm` |
| Answer» Correct Answer - d | |
| 163. |
A screen receives 3 watt of radiant flux of wavelength 6000 Å . One lumen is equivalent to watt `1.5 xx 10^(-3) "watt"` of monochromatic light of wavelength 5550 Å . If relative luminosity for 6000 Å is 0.685 while that for 5550 Å is 1.00, then the luminous flux of thesource isA. `4 xx 10^(3) lm`B. `3 xx 10^(3) lm`C. `2 xx 10^(3) lm`D. `1.37 xx 10^(3) lm` |
| Answer» Correct Answer - d | |
| 164. |
Light from a point source falls on a small area placed perpendicular to the incident light. If the area is rotated about the incident light by an angle of `60^(@)` , by what fraction will the illuminance changeA. It will be doubledB. It will be halvedC. It will not changeD. It will become one-fourth |
| Answer» Correct Answer - c | |
| 165. |
A point source of 3000 lumen is located at the centre of a cube of side length 2 m . The flux through one side isA. 500 lumenB. 600 lumenC. 750 lumenD. 1500 lumen |
| Answer» Correct Answer - a | |
| 166. |
In the formation of a rainbow light from the sun on water droplets undergoesA. Dispersion onlyB. Only total internal reflectionC. Dispersion and total internal reflectionD. None of these |
| Answer» Correct Answer - c | |
| 167. |
The rainbow formed after or during the rain is due toA. RefractionB. reflectionC. dispersionD. all of these |
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Answer» Correct Answer - D (d) Rainbow is caused due to refraction, reflection and dispersion. |
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| 168. |
Phenomena associated with scattering is/areA. blue colour of the skyB. appearance of reddish sun during sunset and sunriseC. both (a) and (b)D. None of these |
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Answer» Correct Answer - C (c ) The phenomenon based upon the scattering are (i) blue colour of the sky (ii) appearance of reddish sun during sunnset and sunrise. |
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| 169. |
White light is passed through a prism ........... colour shows minimum deviationA. RedB. VioletC. YellowD. green |
| Answer» Correct Answer - a | |
| 170. |
Angle of minimum deviation is equal to the angle prism A of an equilateral glass prism. The angle incidence at which minimum deviation will be obtained isA. `60^(@)`B. `30^(@)`C. `45^(@)`D. `sin^(1)(2//3)` |
| Answer» Correct Answer - A | |
| 171. |
When light is passed through a prismm when………colour shows maximum deviation.A. redB. violetC. yellowD. green |
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Answer» Correct Answer - B (b) When white light is passed through a prismviolet colour is show maximum deviation. |
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| 172. |
The sensation of vision in the retina is carried to the brain byA. Ciliary musclesB. Blind spotC. Cylindrical lensD. Optic nerve |
| Answer» Correct Answer - d | |
| 173. |
A man is suffering from colour blindness for green colour. To remove this defect, he should use goggles ofA. Green colour glassesB. Red colour glassesC. Smoky colour glassesD. None of the above |
| Answer» Correct Answer - d | |
| 174. |
When the power of eye lens increases, the defect of vision is produced. The defect is known asA. ShortsightednessB. LongsightednessC. ColourblindnessD. None of the above |
| Answer» Correct Answer - a | |
| 175. |
When the power of eye lens increases, the defect of vision is produced. The defect is known asA. shortsightednessB. longsightnessC. colourblindnessD. None of the above |
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Answer» Correct Answer - A (a) In shortsightedness, the focal length of eye lens decreases and so the power of eyelens increases. |
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| 176. |
When diameter of the aperture of the objective of an astronomical telescope is increased, itsA. Magnifying power is increased and resolving power is decreasedB. Magnifying power and resolving power both are increasedC. Magnifying power remains the same but resolving power is increasedD. Magnifying power and resolving power both are decreased |
| Answer» Correct Answer - c | |
| 177. |
Can a microscope function as a telescope by inverting it. Can a telescope function as a microscope ? |
| Answer» No, in a telescope, objective lens has much larger focal length than the eye lens. In a microscope, both the lenses have short focal lengths. | |
| 178. |
Calculate the critical angle for glass air surface if a ray of light which is nicident ni air on the glass surface is deviated through `15^(@)`, when the angle of incidence is `45^(@)`. |
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Answer» Correct Answer - `45^@` Here, `i= 45^@, r = 45^@ - 15^@ = 30^@` `mu = (sin i)/(sin r) = (sin 45^@)/(sin 30^@) = (1)/(sqrt(2)) xx 2 = sqrt(2)` `sin C = (1)/(mu) = (1)/(sqrt(2))` `:. C = 45^@`. |
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| 179. |
Determine the critical angle for a glass interface, if a ray of light, which is incident in air on the surface is deviated through `15^@`, when its angle of incidence is `40^@`. |
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Answer» Here, `C = ?, i= 40^(@)`, deviation, `delta = 15^(@)` As ray deviates towards normal, therefore, `r =I - delta = 40^(@) - 15^(@) = 25^(@)` As `mu = (sin i)/(sin r) = (1)/(sin C)` `:. sin C = (sin r)/(sin r) = (sin 25^(@))/(sin 40^(@)) = (0.4226)/(0.6428) = 0.6574` `C = sin^(-1) (0.6574) = 41.1^(@)`. |
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| 180. |
The length of the tube of a microscope is 10 cm . The focal lengths of the objective and eye lenses are 0.5 cm and 1.0 cm . The magnifying power of the microscope is aboutA. 5B. 23C. 166D. 500 |
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Answer» Correct Answer - D (d) We assume that final image is formed at infinity, and so `M_(oo)=((L_(oo)-f_(o)-f_(e))D)/(f_(o)f_(e))=(LD)/(f_(o)f_(e))` Magnifying power of the microscope, `M_(oo)=(10xx25)/(0.5xx1)=500` |
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| 181. |
The magnifying power of a microscope with an objective of `5 mm` focal length is 400. The length of its tube is `20cm.` Then the focal length of the eye`-` piece isA. 200 cmB. 160 cmC. 2.5 cmD. 0.1 cm |
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Answer» Correct Answer - C (c) We assume that final image is formed at infinity, and `M_(prop)=((L_(oo)-f_(o)-f_(e))cdotD)/(f_(o)f_(e))=(LD)/(f_(o)f_(e))` or `400=(20xx25)/(0.5xxf_(e))` `f_(e)=2.5cm` |
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| 182. |
A beam of parallel rays is brought to focus by a planoconvex lens. A thin Concave lens of the same focal length is joined to the first lens. The effect of this isA. The focal point shifts away from the lens by a small distanceB. The focus remains undisturbedC. The focus shifts to infinityD. The focal point shifts towards the lens by a small distance |
| Answer» Correct Answer - c | |
| 183. |
Following data was recorded for values of object distance and the corresponding values of image distance, in the study of real image formation by a convex lens of power `+ 5 D`. One of these observations is incorrect. Identify and give reason : `{:("S.No",1,2,3,4,5,6),("Object distance (cm)",25,30,35,45,50,55),("Image distance (cm)".97,60,37,35,32,30):}` |
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Answer» Here, `P = + 5 D` `f = (100)/(P) = (100)/(5) = 20 D` In observation 3, where `u = 35 cm`, object lies between `f and 2 f`. The image must be beyond `2 f` i.e., `v gt 40 cm` But `v = 37 cm` i.e., image also lies between `f and 2 f`, which is wrong. |
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| 184. |
When does a convex lens behave as a concave lens ? |
| Answer» When the convex lens is held in a transparent medium of refractive index greater than the refractive index of lens material, it would behave as a concave lens. | |
| 185. |
A thin lens focal length `f_(1)` and its aperture has diameter `d`. It forms an image of intensity `I`. Now the central part of the aperture up to diameter`(d)/(2)` is blocked by an opaque paper. The focal length and image intensity will change to |
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Answer» On blocking the central part of the lens, its focal length does not change. It remains `f` only. Intensity of image is directly proportional to the area of the lens through which light passes. Now, initial area `A_1 = pi (d//2)^2 = pi d^2//4` On blocking the central part of the aperture upto diameter `d//2`, the area left out is `A_2 = pi (d//2) - pi (d//4)^2 = pi (d^2)/(4) - pi (d^2)/(16) = (3 pi d^2)/(16)` As `(I_2)/(I_1) = (A_2)/(A_1) = (3 pi d^2 4)/(16 pi d^2) = (12)/(16) = (3)/(4) :. I_2 = (3)/(4) I_1`. |
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| 186. |
A thin lens focal length `f_(1)` and its aperture has diameter `d`. It forms an image of intensity `I`. Now the central part of the aperture up to diameter`(d)/(2)` is blocked by an opaque paper. The focal length and image intensity will change toA. `f//2,I//2`B. `f,I//4`C. `3f//4,I//2`D. `f,3I//4` |
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Answer» Correct Answer - D `Iprop aperture^(2)` `I_(1) prop pid^(2)//4 rArr I_(2)prop(pid^(2)//4-pi(d//4)^(2))` `I_(2)prop(pid^(2))/(4)(1-(1)/(4)) rArr I_(2)prop (3)/(4).(pid^(2))/(4)` `(I_(2))/(I_(1))=(3)/(4) rArrI_(2)=(3)/(4)I_(1)` |
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| 187. |
The speed at which the image of the luminous point object is moving, if the luminous point object is moving at speed ` v_(0)` towards a spherical mirror, along its axis is (Given, `R=` radius of curvature `u=` object distance)A. `v_(l)=-v_(o)`B. `v_(l)=-v_(o)[R/(2u-R)]^(2)`C. `v_(l)=-v_(o)((2u-R)/R)`D. `v_(l)=-v_(o)(R/(2u-R))` |
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Answer» (c) `1/v+1/u=1/f` Differentiating both sides `-1/(v^(2)) (dv)/(dt)=1/(u^(2))(du)/(dt)` `(dv)/(dt)=v_(l)=-(v/u)^(2)(du)/(dt)=-(v/u)^(2)v_(0)` Again `1/v=1/f-1/u=2/R-1/u=(2u-R)/(Ru)` `v=(uR)/(2u-R)` `v_(i)=-(v/u)^(2)v_(o)=-v_(o)(R/(2r-R))^(2)` |
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| 188. |
Through a simple microscope, an object is seen in red light first and then in violet light. In which case is magnifying power more ? |
| Answer» `m = 1 + (d)/(f). As f_r gt f_v`, therefore, magnifying power increases when violet light is used. | |
| 189. |
A myopic person prefers to remove his spectacles while reading a book. Why ? |
| Answer» A myopia person has to use spectacles with concave lens. He many have normal near point `(~~25 cm)`. To read with specs, he has to hold the book at a distance greater than `25 cm`. As angular size of object at a distance `gt 25 cm` is less than the angular size of object at `25 cm`, therefore, the person prefers to remove his spectacles while reading. | |
| 190. |
A hypermetropic eye can see clearly……….but………cannot be seen distinctly. |
| Answer» the far off objects , nearby objects | |
| 191. |
A man can see the objects upto a distance of one metre from his eyes. For correcting his eye sight so that he can see an object at infinity, he requires a lens whose power isA. `+ 0.5 D`B. `+ 1.0 D`C. `+ 2.0 D`D. `- 1.0 D` |
| Answer» Correct Answer - d | |
| 192. |
Does short sightedness (myopia) or long sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accomodation ? If not, what might cause these defects of vision ? |
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Answer» No, a person may have normal ability of accommodation and yet he may be myopic or htpermetropic. Infact, myopia arises when length of eye ball (from front to back) gets elongated and hypermetropia arises when length of eye ball gets shortened. However, when eye ball has normal length, but the eye lens loses partially its power of accommodation, the defect is called presbiopia. |
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| 193. |
A man can see the object between `15 cm` and `30 cm`. He uses the lens to see the far objects. Then due to the lens to see the far objects. Then due to the lens used, the near point will be atA. `(10)/(3) cm`B. `30 cm`C. `15 cm`D. `(100)/(3) cm` |
| Answer» Correct Answer - b | |
| 194. |
For the myopia defect in eye, it can be removed byA. convex lensB. concave lensC. cylindrical lensD. toric lens |
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Answer» Correct Answer - B (b) The defect in the myopia eye is removed by the concave lens. |
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| 195. |
In Galilean telescope, if the powers of an objective and eye lens are respectively `+1.25D` and `-20 D`, then for relaxed vision, the length and magnification will beA. `21.25 cm` and 16B. 7`5 cm` and 20C. `75 cm` and 16D. `8.5 cm` and `21.25` |
| Answer» Correct Answer - c | |
| 196. |
If the distance of the far point for a myopia patient is doubled, the focal length of the lens required to cure it will becomeA. HalfB. DoubleC. The same but a convex lensD. The same but a concave len |
| Answer» Correct Answer - b | |
| 197. |
The far point of a myopia eye is at `40 cm`. For removing this defect, the power of lens required will beA. `40 D`B. `- 4 D`C. `- 2.5 D`D. `0.25 D` |
| Answer» Correct Answer - c | |
| 198. |
A man suffering from myopia can read book placed at `10 cm` distance. For reading the book at a distance of `60 cm` with relaxed vision, focal length of the lens required will beA. 45 cmB. `- 20 cm`C. `- 12 cm`D. `30 cm` |
| Answer» Correct Answer - c | |
| 199. |
A man has a height of 6m. He observes image of 2m height erect, then mirror used isA. concaveB. convexC. planeD. none of these |
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Answer» Correct Answer - B (b) Erect diminshed image is formed by convex mirror. |
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| 200. |
Myopia is due toA. elongation of eye ballB. irregular change in focal lengthC. shortening of eye ballD. older age |
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Answer» Correct Answer - A (a) In myopia, the focal length or radii of curvature of lens reduced or power of lens increases or distance between eye lens and retina increases. |
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