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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 telescope has an objective lens of `10cm` diameter and is situated at a distance of one kilometre from two objects. The minimum distance between these two objects, which can be resolved by the telescope, when the mean wavelength of light is `5000 Å`, of the order ofA. `5 mm`B. `5 cm`C. `2.5 m`D. `5 m` |
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Answer» Correct Answer - A Here, `D = 10 cm, x = 1 km = 10^(5) cm` Distance between two objects, `d = ?` `lambda = 5000 Å = 5 xx 10^(-5) cm` Limit of resolution `d theta = (d)/(x) = (lambda)/(D)` `:. D = (x lambda)/(D) = (10^(5) xx 5 xx 10^(-5))/(10) = 0.5 cm = 5 mm` |
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| 152. |
How can we increase the resolving power of a microscope ? |
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Answer» (i) By using light of smaller wavelength, (ii) By increasing the aperture of the objective lens. |
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| 153. |
What is the angular resolution of a `10 cm` diameter telescope at a wavelength of `0.6 mu` ? |
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Answer» Here, `d theta = ? D = 10 cm = 0.1 m, lambda = 0.6 mu m = 6 xx 10^(-7)m`. As `dtheta = (1.22 lambda)/(D) :. Dtheta = (1.22 xx 6 xx 10^(-7))/(0.1) = 7.32 xx 10^(-6)` radian |
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| 154. |
What is meant by angular resolution of a telescope ? |
| Answer» Angular resolution of a telescope is the reciprocal of the smallest angular separation of two distant point objects, whose images can just be resolved by the telescope. | |
| 155. |
What is meant by range of a telescope ? |
| Answer» Range of a telescope tells us how far away a star of some standard brightness can be spotted by the telescope. | |
| 156. |
Calculate the numarical aperture of a microscope required to just resolve two points separated by a distance of `10^(-4) cm`. Wavelength of light used is `5.8 xx 10^(-5) cm`. |
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Answer» Correct Answer - `0.29` N.A `= mu = sin theta = ?` `d = 10^(-4)cm, lambda = 5.8 xx 10^(-5) cm` N.A `= mu sin theta = (lambda)/(2 d) = (5.8 xx 10^(-5))/(2 xx 10^(-4)) = 0.29` |
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| 157. |
Calculate the resolving power of a microscope if its numerical aperture is `0.12` and wavelength of light used is `6000 Å`. |
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Answer» Here, `N.A. = mu sin theta = 0.12`, `lambda = 6000 Å = 6 xx 10^(-7) m` R.P. of microscope, `(1)/(d) = (2 mu sin theta)/(lambda) = (2 xx 0.12)/(6 xx 10^(-7)) = 4 xx 10^(5) m^(-1)` |
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| 158. |
By increasing the diameter of the objective of telescope, we can increase its range, why ? |
| Answer» A bigger objective will gather more light. Therefore, even far off stars may produce images of optimum brightness i.e., its range would increase. | |
| 159. |
Why should the objective of a microscope be of small aperture ? |
| Answer» This is because light rays from the nearby tiny object spread over small aperture and the final image formed is very bright. | |
| 160. |
How will you distinguish between a compound microscope and a telescope just by looking at them ? |
| Answer» As aperture of objective of a microscope is much smaller and that of a telescope is much larger, therefore, front end of a microscope is narrow and front end of a telescope is much wider. | |
| 161. |
A narrow beam of plane-polarized light passes through dextrorotatory positive compound placed into a longitudinal magnetic field as shown in Fig. Find the angle through which the polarization plane of the transmitted beam will turn if the length of the tube with the componed is equal to `l`, the specific rotation constant of the compound is equal to `alpha`, the verder constant is `V`, and the magnetic field strength is `H`. |
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Answer» We write `varphi = varphi_("chermical") + varphi_("magnetic")` We look against the transmitted beam and count the positive direction clockwise. The chemical part of the rotation is annulled by reversal of wave vector upon reflection. Thus `varphi_("chermical") = alphal` SInce in effect there is a single transmission. On the other hand `varphi_(mag) =- NHVl` To get the sings right recall that rotatory compounds rotates the plane of vibration in a clockwise direction on looking against the oncoming beam. The sense of rotation of light vibration in Faraday effect is defined in terms of the direction of the field, positive rotation being that of a right handed screw advancing in the direction of the field. This is the opposite of the definition of `varphi_(chemical)` for the present case. Finally `varphi = (alpha - VNH)l` (Note: If plane polarized light is reflected back & both fourth through the same active medium in a magnetic field, the Faraday rotation increases with each traveresal.) |
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| 162. |
A thin, symmetric double-convex lens of power P is cut into three parts A, B and C as shown. The power of (a) A is P (b) A is 2P (c) B is P/2 (d) B is P/4 |
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Answer» Correct Answer is: (a) A is P , (c) B is P/2 |
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| 163. |
A thini, symmetric double convex lens of power P is cut into three parts A, B, and C as shown in Figure. The power of A. `A` is `P`B. `A` is `2P`C. `B` is `P/2`D. `B` is `P/4` |
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Answer» Correct Answer - A::C |
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| 164. |
Which of the following form virtual and erect images for all positions of the object? (a) Convex lens (b) Concave lens (c) Convex mirror (d) Concave mirror |
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Answer» Correct Answer is: (b) Concave lens , (c) Convex mirror |
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| 165. |
If a convergent beam of light passes through a diverging lens, the result (a) may be a convergent beam (b) may be a divergent beam (c) may be a parallel beam (d) must be a parallel beam |
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Answer» Correct Answer is: (a, b, & c) |
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| 166. |
A converging lens of focal length `f_(1)` is placed in front of and coaxially with a convex mirror of focal length `f_(2)`. Their separation is d. A parallel beam of light incident on the lens returns as a parallel beam from the arrangement, Then,A. The beam diameters of the incident and reflected beams must be the same.B. `d=f_(1)-2|f_(2)|`C. `d=f_(1)-|f_(2)|`D. If the entire arrangement is immersed in water, the conditions, will remain unaltered. |
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Answer» Correct Answer - A::B |
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| 167. |
If a convergent beam of light passes through a diverging lens, the resultA. may be a convergent beamB. may be a divergent beamC. may be a parallel beamD. must be a parallel beam |
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Answer» Correct Answer - A::B::C |
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| 168. |
A Point object `P` moves towards a convex mirror with a constant speed `V`, along its optic axis. The speed of the imageA. is always `lt V`B. may be `gt=` or `ltV` depending on the position of `P`C. increases as `P` comes clser to the mirrorD. decreases as `P` comes closer to the mirror |
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Answer» Correct Answer - A::C |
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| 169. |
A point object P moves towards a convex mirror with a constant speed V, along its optic axis. The speed of the image (a) is always < V (b) may be >, = or < V depending on the position of P (c) increases as P comes closer to the mirror (d) decreases as P comes closer to the mirror |
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Answer» Correct Answer is: (a, c) As the object moves from infinity to the pole of the mirror, the virtual image moves from its focus to the pole. |
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| 170. |
A section of a sphere has a radius of curvature of `0.80 m`. Both, inside and ouside surfaces have a mirror like polish. What are the focal lengths of the inside and outside surfaces? |
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Answer» When inside surface has mirror like polish, the mirror is convex, `f = R//2 = + (0.8)/(2) = 0.4 m` When outside surface has mirror like polish, the mirror is concave, `f = R//2 = - (0.8)/(2) = 0.4 m` |
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| 171. |
A coverging beam of light is incident on the concave mirror. Then the reflected light :A. may form a real imaeB. must form a real imageC. may form a virtual imageD. may be a parallel beam |
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Answer» Correct Answer - A::C::D |
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| 172. |
If a converging beam of light is incident on a concave mirror, the reflected light (a) may form a real image (b) must form a real image(c) may form a virtual image (d) may be a parallel beam |
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Answer» Correct Answer is: (a) may form a real image , (c) may form a virtual image , (d) may be a parallel beam |
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| 173. |
Will the reflected rays converge at a point when a parallel beam of light is incident on a concave mirror of large aperture ? |
| Answer» No, this occurs because of spherical aberration of the mirror. Paraxial rays are focussed at the principal focus and marginal rays are focussed between pole and principal focus of the mirror. | |
| 174. |
A `5 cm` long needle is placed `10 cm` from a convex mirror of focul length `40 cm`. Find the position, nature and size of image of the needle. What happens to the size of image when needle is moved farhter away from the mirror ? |
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Answer» Here, `h_(1) = 5 cm, u = -10 cm`, `f = 40 cm` From `(1)/(v) = (1)/(f) - (1)/(u) = (1)/(40) - (1)/(-10) = (5)/(40) = (1)/(8)` `v = 8 cm` Image is virtual, erect and is formed `8 cm` behind the mirror. Magnification, `m = (h_(1))/(h_(2)) = (-v)/(u) = (-8)/(-10) = (4)/(5)` `h_(2) = (4)/(5)h_(1) = (4)/(5) xx 5 = 4 cm` As needle is moved farther away from the mirror, image shifts towards the focus and its size goes on decreasing. |
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| 175. |
A concave mirror of small aperture forms a shrper image. Why ? |
| Answer» This is because a concave mirror of small aperture is free from the defect of spherical aberration. | |
| 176. |
Does size of mirror affect the nature of the image ? |
| Answer» No, size of mirror does not affect the nature of the image except that a bigger mirror forms a brighter image. | |
| 177. |
How will you distinguish between a plane mirror, a convex mirror and a concave mirror without touching them? |
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Answer» In a plane mirror, image formed is virtual, erect and of same size as the object. In a convex mirror, image is erect and smaller in size. In a concave mirror, the size and nature of the image change with change in position of the object in front of it. |
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| 178. |
The eyes of the nocturnal birds like owl are having a large cornea and a large pupil. How does it help them? |
Answer»
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| 179. |
State the fundamental laws on which ray optics is based. |
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Answer» Ray optics is based on the following fundamental laws: i. Light travels in a straight line in a homogeneous and isotropic medium. ii. Two or more rays can intersect at a point without affecting their paths beyond that point. iii. Laws of reflection: a. Reflected ray lies in the plane formed by incident ray and the normal drawn at the point of incidence and the two rays are on either side of the normal. b. Angles of incidence and reflection are equal (i = r). iv. Laws of refraction: a. Refracted ray lies in the plane formed by incident ray and the normal drawn at the point of incidence; and the two rays are on either side of the normal. b. Angle of incidence (90 and angle of refraction (62) are related by Snell’s law, given by, n1 sin θ1 = n2 sin θ2 where, n1, n2 = refractive indices of medium 1 and medium 2 respectively. |
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| 180. |
Assertion : Presbyopia is due to ageing of human beings. Reason : For those persons, ciliary, muscles of the eye become weak.(a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion. (b) If both Assertion and Reason are true but Reason is not the correct explanation of Assertion. (c) If Assertion is true but Reason is false. (d) If Assertion is false but Reason is true. |
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Answer» (a) If both Assertion and Reason are true and Reason is the correct explanation of Assertion. |
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| 181. |
The human eye is ____ in nature. (a) convex (b) concave (c) transparent glass (d) Plano – concave. |
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Answer» Correct answer is (a) convex |
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| 182. |
For normal human eye the value of near point is: (a) 25 cm (b) 25 m (c) 2.5 m (d) 25 mm |
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Answer» Correct answer is (a) 25 cm |
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| 183. |
In all optical instruments, we use.A. (a) ray opticsB. wave opticsC. ( c) physical opticsD. (d) none of these |
| Answer» Correct Answer - A | |
| 184. |
For a normal eye, distance of near point from the eye is.A. `oo`B. `25 cm`C. `25 m`D. none of these |
| Answer» Correct Answer - B | |
| 185. |
A myopia eye can see clearly………but the………cannot be seen distinctly. |
| Answer» the far off objects , nearby objects | |
| 186. |
The lens used for correcting myopia is.A. concaveB. convexC. plano concaveD. none of these |
| Answer» Correct Answer - A | |
| 187. |
List the uses of polaroids |
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Answer» Uses of polaroids: 1. Polaroids are used in goggles and cameras to avoid glare of light. 2. Polaroids are useful in three dimensional motion pictures i.e., in holography. 3. Polaroids are used to improve contrast in old oil paintings. 4. Polaroids are used in optical stress analysis. 5. Polaroids are used as window glasses to control the intensity of incoming light. |
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| 188. |
Calculate time taken by light to travel `1 cm` thickness of glass of `mu = 1.5`. |
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Answer» `t = (x)/(v) = (x)/(c//mu) = (x mu)/(c ) = (10^(-2) xx 1.5)/(3 xx 10^(8))` `= 0.5 xx 10^(-10) s` |
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| 189. |
What is the phase difference corresponding to path difference `lambda` of two waves reaching a point ? |
| Answer» Correct Answer - `2 pi` radian. | |
| 190. |
What is the main condtion to produce interference of light ? |
| Answer» Two source of light must be coherent. | |
| 191. |
A lens immersed in a transparent liquid becomes invisible. Under what condtion does it happen ? |
| Answer» This happens when refractive index of transparent liquid is equal to refractive index of lens material. | |
| 192. |
When a ray of light is incident normally on one refracting surface of an equilateral prism (Refractive index of the material of the prism `=1.5`A. `30^(@)`B. `45^(@)`C. `60^(@)`D. `75^(@)` |
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Answer» Correct Answer - C |
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| 193. |
Statement-1 : A ray of light incident normally on a refracting surface does not suffer any refraction. Statement-2 : The critical angle for total internal refraction is smaller when a ray of light travels from glass to water than when it travels from glass to air.A. Statement-1 is true, Statement-2 is true. Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is fasle, Statement-2 is true. |
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Answer» Correct Answer - C Statement-1 is true, but the statement-2 is false. For a ray incident normally, `/_i = 0 :. /_r = 0`. Also, `sin C = 1//mu`. |
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| 194. |
Statement-1 : Energy is created during constructive interference and it is destroyed during destructive intference. Statement-2 : Because in constructive interference, the fringes are bright and in destructive interference, the fringes are dark.A. Statement-1 is true, Statement-2 is true. Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is fasle, Statement-2 is true. |
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Answer» Correct Answer - D Energy can neither be created nor be destroyed. Energy is simply redistributed in constructive and destructive interference. Statement-1 is false, but statement-2 is true. |
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| 195. |
When light travels from a rarer to denser medium, it loses some speed. Does the reduction in speed imply a reduction in the energy carried by the light wave ? |
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Answer» We know that total average energy density associated with an electromagnetic wave (i.e. light wave) is `u_(av) = (B_(0)^(2))/(2mu_(0)) = (1)/(2) in_(0)E_(0)^(2)` This shows that `u_(av)` depends upon `B_(0) and E_(0)` (i.e. the amplitude of magnetic vector and eletric vector). `u_(av)` is independent of the speed of wave propagation. Therefore, the wave may lose speed, but not the energy. |
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| 196. |
Statement-1 : Light travels faster in glass than in air. Statement-2 : Because air is rarer than glass.A. Statement-1 is true, Statement-2 is true. Statement-2 is correct explanation of Statement-1.B. Statement-1 is true, Statement-2 is true, but Statement-2 is not a correct explanation of Statement-1.C. Statement-1 is true, Statement-2 is false.D. Statement-1 is fasle, Statement-2 is true. |
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Answer» Correct Answer - D Light travels faster in air than in glass, because glass is denser than air. Statement-1 is false, but statement-2 is true. |
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| 197. |
What is total magnification of three lenses of magnification `2, 3, 4` in contact ? |
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Answer» Total magnification, `m = m_(1) xx m_(2) xx m_(3)` `= 2xx 3xx 4xx = 24` |
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| 198. |
Unpolarised light is passed through a polaroid `P_(1)`. When this poalrised beam passes through another polaroid `P_(2)`, and if the pass axis of `P_(2)` makes an angle `theta` with pass axis of `P_(1)`, then write the expression for the polarised beam passing through `P_(2)`. Draw a plot showing the variation of intensity when `theta` varies from 0 to 2 `pi`. |
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Answer» If `I_(0)` is intensity of unpolarised light, then intensity pf polarised light from `P_(1)` is `I_(1) = I_(0)//2` When this is passed through another polaroid `P_(2)`, intensity of polarised light from `P_(2)` is `I_(2) = I_(1) cos^(2) theta` When `theta` is varied from `0 to 2 pi`, we obtain `I_(2)` versus `theta` plot as shown in Fig. |
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| 199. |
The focal length of the objective and eyepiece of a telescope are respectively 100 cm and 2 cm. The moon subtends angle of 0.5° ; the angle subtended by the moon’s image will be(a) 10° (b) 250 (e) 100° (d) 75° |
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Answer» (b) 250 m = \(\frac{B}{a}\) β =\(\frac{f_0}{f_e}\) ; α = x 0.5° = 25° |
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| 200. |
An astronomical telescope has an angular magnification of magnitude 5 for distant object. The separation between the objective and the eyepiece is 36 cm and the final image is formed at infinity. The focal length `f_(0)` of the objective and the focal length `f_(0)` of the eyepiece areA. `f_(0) = 45 cm and f_(e) = - 9 cm`B. `f_(0) = 50 cm and f_(e) = 10 cm`C. `f_(0) = 7.2 cm and f_(e) = 5 cm`D. `f_(0) = 30 cm and f_(e) = 6 cm` |
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Answer» Correct Answer - D Here, Angular magnification of telescope `= (f_(0))/(f_(e)) = 5`, `:. f_(0) = 5 f_(e)` …(i) Now, `f_(0) + f_(e) = 36 cm` …(ii) On solving (i) and (ii), we get `5 f_(e) + f_(e) = 36 cm` `f_(e) = 6 cm, f_(0) = 30 cm` |
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