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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. |
The magnification produced by a spherical lens is `+0.75`. What is the: a) nature of image? b) nature of lens? |
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Answer» a) When the magnification is positive, the nature of image is virtual and erect. In this, case, the magnification is positive, so the nature of image is virtual and erect. b) The value of magnification given here is 0.75(which is less than 1), so, the image is smaller than the object or diminshed. A virtual, erect and diminished image can be formed only by a concave lens, so the nature of lens is concave. |
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| 152. |
A person having a mypoic eye ueses a concave lens of focal length 50 cm. What is the power of the lens? |
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Answer» Here we have a concave lens. Now, the focal length of a concave lens is considered negative, so it is to be written with a minus sign. Thus, Focal length, f=`-50` cm `=-50/100` m `=-0.5` m Now, Power, P`=(1/("f in metres))` `P=1/-0.5` `P=(-1 x 10)/(5)` `P=-2` dioptres (or `-2`D) Thus, the power of this concave lens is, `-2` dioptres which can also be written as, `-2`D. The minus sign with the power indicates that it is diverging lens or concave lens. |
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| 153. |
A lens has a power of -2.5D. What is the focal length and nature of the lens? |
| Answer» The power of this lens has minus sign, so it is a concave lens. Calculate the focal length yourself as shown in the above question. The focal length will be,`-40` cm. | |
| 154. |
Two lenses A and B have focal length of `+20` cm and, `-10` cm, respectively. a) What is the nature of lens A and lens B? b) What is the power of lens A and lens B? What is the power of combination if lenses A and B are held close together? |
| Answer» a) Lens A is convex, Lens B is concave b) Power of lens A`=+5 D`, Power of lens B`=-10 D` c) `-5D` | |
| 155. |
a) Two lenses A and B have power of i) `+2D` and ii) `-4 D` respectively. What is the nature and focal length of each lens? b) An object is placed at a distance of 100 cm from each of the above lenses A and B. Calculate i) image distance, and ii) magnification, in each of the two cases. |
| Answer» a) Lens A is convex, `f=+50 cm`, Lens B is concave, `f=-25 cm` b) Lens A: v`=+100 cm`, `m=-1,` Lens B: `v=-20 cm`, `m=0.2` | |
| 156. |
The power of lens is, `-2D`. Calculate its focal length. |
| Answer» Correct Answer - `-50 cm` | |
| 157. |
What is the nature of a lens whose power of `-4 D` ? |
| Answer» Correct Answer - Concave lens | |
| 158. |
A lens has a focal length of, `-10cm`. What is the power of the lens and what is its nature? |
| Answer» `-10 D`, Concave lens | |
| 159. |
A doctor has prescribed a corrective lens of power, `-1.5D`. Find the focal length of the lens. Is the prescribed lens diverging or converging? |
| Answer» `-66.6` cm, Diverging lens | |
| 160. |
A converging lens is used to produce an image of an object on a screen. What change is needed for the image to be formed nearer to the lens?A. increase the focal length of the lensB. insert a diverging lens between the lens and the screenC. increase the distance of the object from the lensD. move the object closer to the lens. |
| Answer» Correct Answer - c) | |
| 161. |
An object is placed 15 cm from `(alpha)` a converging mirror, and (b) a diverging mirror, of radius of curvature 20 cm. Calculate the image position and magnification in each case. |
| Answer» a) `v=+60 cm, m=-3` b) `v=-8.5 cm,` m`=+0.42`, | |
| 162. |
A converging lens has a focal length of `30 cm.` Rays from a `2.0 cm` high filament that pass through the lens from a virtual image at a distance of `50 cm` from the lens. Where is the filament located? What is the height of the image? |
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Answer» Correct Answer - A::C `1/-50-1/u=1/(+30)` Solving, we get `u=-18.75 cm` `m=v/u=((-50))/((-18.75))=2.67` `I=m(O)=2.67xx2` `=5.33 cm` |
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| 163. |
a) A small object is placed 150 mm away from a diverging lens of focal length 100 mm. i) Copy the figure below and draw rays to show how an image is formed by the lens. ii) Calculate the distance of the image from the lens by using the lens formula. b) The diverging lens in part a) is replaced by a converging lens also of focal length 100 nm. The object remains in the same position and an image is formed by the converging lens. Compare two properties of this image with those of the image formed by the diverring lens in part a) |
| Answer» a) ii) `v=-60 mm` b) `v=+300mm,` The image formed by converging lens is real, inverted and magnified (2 times). It is formed behind the converging lens. On the other hand, the image is formed by diverging lens is virtual, erect and diminished. It is formed in front of the diverging lens. | |
| 164. |
An illuminated object is placed at a distance of 20 cm from a converging lens of focal length 15 cm. The image obtained on the screen is:A. upright and magnifiedB. inverted and magnifiedC. inverted and diminishedD. upright and diminshed |
| Answer» Correct Answer - b) | |
| 165. |
Where must the object be placed for the image formed by a converging lens to be: a) real, inverted and smaller than the object? b) real, inverted and same size as the object? c) real, inverted and larger than the object? d) virtual, upright and larger than the object? |
| Answer» a) Beyond 2F b) At 2F c) Between F and 2F d) Between F and Optical center | |
| 166. |
a) Find the position and size of the virtual image formed when an object 2 cm tall is placed 20 cm from: i) a diverging lens of focal length 40 cm. ii) a converging lens of focal length 40 cm. |
| Answer» a) i) `v=-13.3cm, 1.33 m , 1.33` cm tall ii) `v=-40 cm, 4cm` tall | |
| 167. |
An object is placed at a distance of 100 cm from a converging lens of focal length 40 cm. i) What is the nature of image? ii) What is the position of image? |
| Answer» i) Real and inverted ii) `v=+66.6` cm, The image is formed 66.6 cm behind the convex lens (on its right side) | |
| 168. |
A 2.0 cm tall object is placed 40 cm from a diverging lens of focal length 15 cm. Find the position and size of the image. |
| Answer» `v=-10.90 cm, 0.54 cm` tall | |
| 169. |
An object `5 cm` in length is held `25 cm` away from a converging lens of focal length `10 cm`. Draw the ray diagram and find the position, size and the nature of the image formed. |
| Answer» `v=+16.6` cm, The image is 16.6 cm behind the convex lens, 3.3 cm, Real and inverted | |
| 170. |
An object placed 4 cm in front of a converging lens produces a real image 12 cm from the lens. a) What is the magnification of the image? b) What is the focal length of the lens? Draw a ray diagram to show the formation of image. Mark clearly F and 2F in the diagram. |
| Answer» a) 3 b) `+3` cm | |
| 171. |
a) An object 2 cm tall stands on the principal axis of a converging lens of focal length 8 cm. Find the position, nature and size of the image formed if the object is: i) 12 cm from the lens ii) 6 cm from the lens. b) State one parallel application each of the use of such a lens with the object in position i) and ii) |
| Answer» a) (i) `v=+24 cm`, The image is 24 cm in front of the lens, Virtual and erect , 8 cm b) i) Used in film projector ii) Used as a magnifying glass | |
| 172. |
If the image formed by a lens is always diminished and erect, what is the nature of the lens? |
| Answer» Correct Answer - Concave lens | |
| 173. |
The refracting angle of a glass prism is `30^@.` A ray is incident onto one of the faces perpendicular to it. Find the angle `delta` between the incident ray and the ray that leaves the prism. The refractive index of glass is `mu=1.5.` |
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Answer» Correct Answer - A::D Given , `A=30^@, mu=1.5` and `i_1=0^@` Since, `i_0^@,` therefore, `r_1` is also equal to `0^@.` Further, since, `r_1+r_2=A` `:. r_2=A=30^@` Using, `mu=sin i_2/sin r_2` we have, `1.5=sin i_2/sin30^@` or `sin i_2=1.5sin 30^@` `=1.5xx1/2=0.75` `:. i_2=sin^-1(0.75)=48.6^@` Now, the deviation, `delta=(i_1+i_2)-A` `=(0+48.6)-30` or `delta=18.6^@` |
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| 174. |
A ray of light is incident at an angle of `60^@` on the face of a prism having refracting angle `30^@.` The ray emerging out of the prism makes an angle `30^@` with the incident ray. Show that the emergent ray is perpendicular to the face through which it emerges. |
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Answer» Correct Answer - A::C::D Given `i_1=60^@,A=30^@` and `delta=30^@`. From the relation, `delta=(i_1+i+2)-A` we have, `i_2=0` This means that the emergne ray is perpendicular to the face through which it emerges. |
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| 175. |
If a magnifcation of `-1` is to be obtained by using a converging of focal length 12 cm, then the object must be placed.A. withing 12cmB. at 24 cmC. at 6 cmD. beyond 24 cm |
| Answer» Correct Answer - b) | |
| 176. |
Magnification produced by a concave lens is always:A. more than 1B. equal to 1C. less than 1D. more than 1 or less than 1 |
| Answer» Correct Answer - c) | |
| 177. |
A convex lens produces a magnification of `+5`. The object is placed:A. at focusB. between f and 2fC. at less than fD. beyond 2f |
| Answer» Correct Answer - c) | |
| 178. |
To obtain a magnification of, -0.5 with a convex lens, the object should be placed:A. at FB. between optical center and FC. between F and 2FD. beyond 2F |
| Answer» Correct Answer - d) | |
| 179. |
A convex lens of focal length 15 cm produces a magnification of `+4`. The object is placed:A. at a distance of 15 cmB. between 15 cm and 30 cmC. at less than 15 cmD. beyond 30 cm |
| Answer» Correct Answer - c) | |
| 180. |
In order to obtain a magnification of, `-3`(minus 3) with a convex lens, the object should be placed.A. between optical center nad FB. between F and 2fC. at 2FD. beyond 2F |
| Answer» Correct Answer - b) | |
| 181. |
To obtain a magnification of `-2` with a convex lens of focal length 10 cm, the object should be placed:A. between 5 cm and 10 cmB. between 10 cm and 20 cmC. at 20 cmD. beyond 20 cm |
| Answer» Correct Answer - b) | |
| 182. |
An object is 0.09 m from a magnifying lens and the image is formed 36cm from the lens. The magnification produced is:A. 0.4B. 1.4C. 4D. 4.5 |
| Answer» Correct Answer - c) | |
| 183. |
If the image formed by a convex lens is of the same size as that of the object, what is the position of the image with respect to the lens? |
| Answer» At 2F (at twice the focal length | |
| 184. |
If an object is placed at the focus of a convex lens, where is the image formed? |
| Answer» At infinity (very large distance) | |
| 185. |
a) An object 3 cm high is placed 24 cm away from a convex lens of focal length 8 cm. Find by calculations, the position, height and nature of the image. b) If the object is moved to a point only 3 cm away from the lens, what is the new position, height and nature of the image? c) Which of the above two cases illustrates the working of a magnifiying glass? |
| Answer» `v=+12 cm`, the image is formed 12 cm behind the lens , 1.5 cm high, Real and inverted | |
| 186. |
If an object of 7 cm height is placed at a distance of 12 cm from a convex lens of focal length 8 cm, find the position, nature and height of the image. |
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Answer» First of all we find out the position of the image. By the position of image we mean the distance of image from the lens. Here, Object distance, u=-12 cm (it is to the left of lens) Image distance, v=? (To be calculated) Focal length, f=`+8` cm (It is a convex lens) Putting these values in the lens formula: `1/v-1/u=1/f` We get: `1/v-1/(-12)=1/8` or `1/v+1/12=1/8` or `1/v+1/12=1/8` `1/v=1/8-1/12` `1/v=(3-2)/(24)` `1/v=1/(24)` So, Image distance, `v=+24` cm Thus, the image is formed on the right side of the convex lens. Only a real and inverted image is formed on the right side of a convex lens, so the image formed is real and inverted. Let us calculate the magnification now. We know that for a lens: Magnification, `m=v/u` Here, Image distance, v=24 cm Object distance, u`=-12` cm So, `m=24/-12` or `m=-2` Since the value of magnification is more than 1 (it is 2), so the image is larger than the object or magnified. The minus sign for magnification shows that the image is formed below the principal axis. Hence, the image is real and inverted. Let us calculate the size of the image by using the formula: `m=h_(2)/h_(1)` Here, Magnification, m`=-2` (Found above) Height of object, `h_(1)=+7` cm (Measured upwards) Height of image, `h_(2)=?` (To be calculated) Now, putting these values in the above formula, we get: `-2=h_(2)/(7)` or `h_(2)=-2 xx 7` Thus, Height of image, `h_(2)=-14`cm Thus, the height or size of hte image is 14 cm. The minus sign shows that this height is in the downward direction, that is the image is formed below the axis. Thus, the image is real and inverted. |
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| 187. |
When an object is placed at a distance of 36 cm from a convex lens, an image of the same size as the object is formed. What will be the nature of image formed when the object is formed. What will be the nature of image formed when the object is placed at a distance of: a) 10 cm from the lens? b) 20 cm from the lens? |
| Answer» a) Virtual, erect and magnified b) Real, invered and magnified | |