{"id":36467,"date":"2022-01-10T10:03:27","date_gmt":"2022-01-10T04:33:27","guid":{"rendered":"https:\/\/mcq-questions.com\/?p=36467"},"modified":"2022-01-11T09:37:11","modified_gmt":"2022-01-11T04:07:11","slug":"mcq-questions-for-class-10-science-chapter-10","status":"publish","type":"post","link":"https:\/\/mcq-questions.com\/mcq-questions-for-class-10-science-chapter-10\/","title":{"rendered":"MCQ Questions for Class 10 Science Chapter 10 Light Reflection and Refraction"},"content":{"rendered":"
Question 1.<\/p>\n
(A) plane mirrors only
\n(B) concave mirrors only
\n(C) convex mirrors only
\n(D) all reflecting surfaces
\nAnswer:
\n(D) all reflecting surfaces<\/p>\n
Explanation:
\nThe laws of reflection hold true for all reflecting surfaces.<\/p>\n
<\/p>\n
Question 2.<\/p>\n
(A) real
\n(B) inverted
\n(C) virtual and inverted
\n(D) virtual and erect
\nAnswer:
\n(D) virtual and erect<\/p>\n
Explanation:
\nWhen an object is kept within the focus of a concave mirror, an enlarged image is formed behind the mirror. This image is virtual and erect.<\/p>\n
Question 3.<\/p>\n
(i) cannot be projected on the screen
\n(ii) are formed by both concave and convex lens
\n(iii) are always erect
\n(iv) are always inverted The correct properties are:<\/p>\n
(A) (i) and (iv)
\n(B) (i) and (ii)
\n(C) (i), (ii) and (iii)
\n(D) (i), (ii) and (iv)
\nAnswer:
\n(C) (i), (ii) and (iii)<\/p>\n
Explanation:
\nA virtual image is formed when reflected rays appear to meet. Such images cannot be obtained on screen. Plane mirrors, convex mirror and concave lens always forms virtual image. They are always erect.<\/p>\n
Question 4.<\/p>\n
(i) actually meet or intersect with each other.
\n(ii) actually converge at a point.
\n(iii)appear to meet when they are produced in the backward direction.
\n(iv) appear to diverge from a point.<\/p>\n
Which of the above statements are correct?<\/p>\n
(A) (i) and (iv)
\n(B) (ii) and (iv)
\n(C) (i) and (ii)
\n(D) (ii) and (iii)
\nAnswer:
\n(C) (i) and (ii)<\/p>\n
Explanation:
\nA real image is formed when light rays actually meet or intersect at a point after reflection or refraction.<\/p>\n
Question 5.<\/p>\n
\n(A) 2,4,1, 3
\n(B) 2,1, 4, 3
\n(C) 1, 2, 4, 3
\n(D) 2,1, 3, 4
\nAnswer:
\n(B) 2,1, 4, 3<\/p>\n
Explanation:
\nAngle 2 is angle of incidence, as it is formed between the incident ray and the normal.Angle is angle of emergence, as it is formed between the emergent ray with normal. Angle 4 is angle of reflection as it is formed between the refracted ray and the normal. 3 shows the lateral displacement. Hence, the correct answer is 2,1,4,3.<\/p>\n
<\/p>\n
Question 6.<\/p>\n
(A) mirror away from the screen
\n(B) screen away from the mirror
\n(C) screen towards the mirror
\n(D) screen towards the building
\nAnswer:
\n(C) screen towards the mirror<\/p>\n
Explanation:
\nThe object is at infinity , so to Obtain sharp image screen should be moved towards mirror.<\/p>\n
Question 7.<\/p>\n
<\/p>\n
(A) P
\n(B) Q
\n(C) R
\n(D) S
\nAnswer:
\n(D) S<\/p>\n
Explanation:
\nAmong the given options, S will the most suitable set up for tracing a ray of light passing through a rectangular glass slab.<\/p>\n
Question 8.<\/p>\n
(A) Laterally inverted and diminished
\n(B) Inverted and diminished
\n(C) Erect and diminished
\n(D) Erect and highly diminished
\nAnswer:
\n(B) Inverted and diminished<\/p>\n
Explanation:
\nWhen the object is at infinity, diminished, inverted and real image is formed.<\/p>\n
Question 9.<\/p>\n
(A) \u2220i is more than \u2220r, but nearly equal to \u2220e
\n(B) \u2220i is less than \u2220r, but nearly equal to \u2220e
\n(C) \u2220i is more than \u2220e, but nearly equal to \u2220r
\n(D) \u2220i is less than \u2220e, but nearly equal to \u2220r
\nAnswer:
\n(A) \u2220i is more than \u2220r, but nearly equal to \u2220e<\/p>\n
Explanation:
\nWhen a ray of light passes through the glass slab, then the angle of incidence is found to be nearly equal to angle of emergence and greater than angle of refraction.<\/p>\n
Question 10.<\/p>\n
(A) Concave mirror as well as convex lens
\n(B) Convex mirror as well as concave lens
\n(C) Two plane mirrors placed at 90\u00b0 to each other
\n(D) Concave mirror as well as concave lens
\nAnswer:
\n(A) Concave mirror as well as convex lens<\/p>\n
Explanation:
\nWhen a point source of light is placed at the focus of concave mirror then all light rays after reflection through mirror will become parallel to the principal axis.
\nWhen this point source of light is placed at the focus of convex lens then after refraction by light rays convex lens will become parallel to the principal axis.<\/p>\n
<\/p>\n
Question 11.<\/p>\n
(A) is less than one.
\n(B) is more than one.
\n(C) is equal to one.
\n(D) can be more than or less than one depending upon the position of the object in front of it.
\nAnswer:
\n(A) is less than one.<\/p>\n
Explanation:
\nConvex mirror is used as rear\u00acview mirror in vehicles. It forms virtual, erect, and diminished images of the objects. Magnification is ratio of height of image to the height of the object, hence, magnification produced by a rear-view mirror fitted in vehicles is less than one.<\/p>\n
Question 12.<\/p>\n
(A) 15 cm in front of the mirror
\n(B) 30 cm in front of the mirror
\n(C) Between 15 cm and 30 cm in front of the mirror
\n(D) More than 30 cm in front of the mirror
\nAnswer:
\n(B) 30 cm in front of the mirror<\/p>\n
Explanation:
\nThe distance of the sun is infinite as compared to the radius of curvature of concave mirror, so, light rays from sun incident parallel all the rays converge at the principal focus. So, the focal length is 15 cm. In case of a concave mirror, the size of image and object will be same if the object is placed at 2f. Hence, in this case object must be placed at 2\/or 2×15 = 30 cm.<\/p>\n
Question 13.<\/p>\n
(A) between the pole and the focus of the reflector.
\n(B) very near to the focus of the reflector.
\n(C) between the focus and centre of curvature of the reflector.
\n(D) at the centre of curvature of the reflector.
\nAnswer:
\n(B) very near to the focus of the reflector.<\/p>\n
Explanation:
\nThe rays of light passing through the principal focus will go parallel to principal axis after reflection thus, forming a concentrated beam of light. So, due to this reason in torches, search lights, and headlights of vehicles, the bulb is placed very near to the focus of the reflector.<\/p>\n
Question 14.<\/p>\n
\n<\/p>\n
(A) Fig. A
\n(B) Fig. B
\n(C) Fig. C
\n(D) Fig. D
\nAnswer:
\n(D) Fig. D<\/p>\n
Explanation:
\nIn case of concave mirror, an incident ray parallel to principle axis passes through F after reflection.<\/p>\n
Question 15.<\/p>\n
(A) Concave lens of focal length 10 cm
\n(B) Convex lens of focal length 20 cm
\n(C) Concave mirror of focal length 10 cm
\n(D) Concave mirror of focal length 20 cm
\nAnswer:
\n(D) Concave mirror of focal length 20 cm<\/p>\n
Explanation:
\nImage formed by the concave mirror in this case is same as when object is at infinity. Due to the great distance, light rays will incident almost parallel to principal axis. After reflection all the rays will converge and meet at principal focus. So, focal length is 20 cm.<\/p>\n
Question 16.<\/p>\n
(A) away from the screen
\n(B) towards the screen
\n(C) to a position very far away from the screen
\n(D) either towards or away from the screen depending upon the position of the object
\nAnswer:
\n(D) either towards or away from the screen depending upon the position of the object<\/p>\n
Explanation:
\nThe incident rays coming from the distant object will be parallel to the principal axis and as we know the rays parallel to the principal axis, after refraction by convex lens, will pass through the principal focus. Hence, a distinct image will be obtained immediately when distance between screen and lens is equal to focal length, so option (D) is correct choice.<\/p>\n
<\/p>\n
Question 17.<\/p>\n
(A) \u2220i > \u2220r > \u2220e
\n(B) \u2220i = \u2220e > \u2220r
\n(C) \u2220i < \u2220r < \u2220e
\n(D) \u2220i = \u2220e < \u2220r
\nAnswer:
\n(B) \u2220i = \u2220e > \u2220r<\/p>\n
Explanation:
\n(Angle of incidence) \u2220i = \u2220e (angle of emergence) because the direction of incident ray and emergent ray is parallel to each other.
\n\u2220e >\u2220r (angle of refraction) because at point of emergence light is entering into optically rarer medium (air) from optically denser medium (glass), so light will bend away from the normal making the angle bigger.<\/p>\n
Question 18.<\/p>\n
(A) Convex mirror of focal length 12 cm
\n(B) Convex lens of focal length 24 cm
\n(C) Concave mirror of focal length 24 cm
\n(D) Convex lens of focal length 12 cm
\nAnswer:
\n(C) Concave mirror of focal length 24 cm<\/p>\n
Explanation:
\nBecause the screen is on the same side of the object which means it cannot be a lens because it happens behind the lenses in such case. Moreover, concave mirror forms real images, that is, image can be obtained on a screen.<\/p>\n
Question 19.<\/p>\n
\n(A) Device X is a concave mirror and device Y is a convex lens, whose focal lengths are 20 cm and 25 cm respectively.
\n(B) Device X is a convex lens and device Y is a concave mirror, whose focal lengths are 20 cm and 25 cm respectively.
\n(C) Device X is a concave lens and device Y is a convex mirror, whose focal lengths are 20 cm and 25 cm respectively.
\n(D) Device X is a convex lens and device Y is a concave mirror, whose focal lengths are 20 cm and 25 cm respectively.
\nAnswer:
\n(D) Device X is a convex lens and device Y is a concave mirror, whose focal lengths are 20 cm and 25 cm respectively.<\/p>\n
Question 20.<\/p>\n
(A) away from the screen
\n(B) towards the screen
\n(C) to a position very far away from the screen
\n(D) either towards or away from the screen depending upon the position of the object.
\nAnswer:
\n(D) either towards or away from the screen depending upon the position of the object.<\/p>\n
Question 21.<\/p>\n
(i) 12.7 cm
\n(ii) 29.7 cm
\n(iii) 57.7 cm
\n(iv) 72.7 cm<\/p>\n
The correct position of the screen was suggested by<\/p>\n
(A) (i)
\n(B) (ii)
\n(C) (iii)
\n(D) (iv)
\nAnswer:
\n(C) (iii)<\/p>\n
Explanation:
\nThe incident rays coming from the distant tree placed will be parallel to the principal axis and as we know the rays parallel to the principal axis, after refraction by convex lens, will pass through the principal focus. Hence, a distinct image will be obtained immediately when distance between screen and lens is equal to focal length. 42.7 cm (position of lens on optical bench) + 15 cm (focal length of lens) = 57.7 (the position of screen on optical bench)<\/p>\n
Question 22.<\/p>\n
(A) erect and laterally inverted
\n(B) erect and diminished
\n(C) inverted and diminished
\n(D) virtual, inverted and diminished
\nAnswer:
\n(C) inverted and diminished<\/p>\n
Explanation:
\nThe image formed by lens will be inverted and diminished.<\/p>\n
<\/p>\n
Question 23.<\/p>\n
(A) lens slightly towards the screen
\n(B) lens slightly away from the screen
\n(C) lens slightly towards the sun
\n(D) lens and screen both towards the sun<\/p>\n
Explanation: Question 24.<\/p>\n (A) both concave. Explanation: Question 25.<\/p>\n (A) A convex lens of focal length 50 cm Explanation: Question 26.<\/p>\n Explanation: Question 27.<\/p>\n <\/p>\n (A) Greater than unity Explanation: Question 28.<\/p>\n (A) A convex lens has 4 dioptre power having a focal length 0.25 m Explanation: or \\(4=\\frac{1}{f}\\) Assertion and Reason Based Mcqs<\/span><\/p>\n Directions : In the following questions, A statement of Assertion (A) is followed by a statement of Reason (R). Mark the correct choice as. Question 1.<\/p>\n Answer: Explanation: Question 2.<\/p>\n Answer: Explanation: <\/p>\n Question 3.<\/p>\n Answer: Explanation: Question 4.<\/p>\n Answer: Explanation: Question 5.<\/p>\n Answer: Explanation: Question 6.<\/p>\n Answer: Explanation: <\/p>\n Question 7.<\/p>\n Answer: Explanation: Question 8.<\/p>\n Answer: Explanation: Case-Based MCQs<\/span><\/p>\n Attempt any 4 sub-parts from each question. Each sub-part carries 1 mark. (A) Between 10cm and 15cm Explanation: Question 2.<\/p>\n (A) positive, negative Explanation: Question 3.<\/p>\n (A) -2.5 Explanation: <\/p>\n Question 4.<\/p>\n (A) Real and inverted Explanation: Question 5.<\/p>\n (A) at infinity Explanation: II. Read the following passage and answer any four questions from Question 1. to Question 5. Question 1.<\/p>\n (A) Convex mirror, it forms virtual image Explanation: Question 2.<\/p>\n (A) At F Explanation: Question 3.<\/p>\n (A) infinity Explanation: <\/p>\n Question 4.<\/p>\n (A) Positive Explanation: Question 5.<\/p>\n (A) erect and enlarged Explanation: III. Read the following passage and answer any four questions from Question 1. to Question 5. Question 1.<\/p>\n (A) convex mirror Explanation: Question 2.<\/p>\n (A) – 4 Explanation: The minus sign in magnification shows that the image formed is real and inverted.<\/p>\n <\/p>\n Question 3.<\/p>\n (A) real in both case Explanation: Question 4.<\/p>\n (A) 45 cm Explanation: Question 5.<\/p>\n (A) virtual, inverted and magnified. Explanation: IV Read the following passage and answer the following questions from Question 1. to Question 4. Question 1.<\/p>\n (A) away from screen Explanation: Question 2.<\/p>\n (A) size of image will decrease Explanation: <\/p>\n Question 3.<\/p>\n (A) intensity of image increases Explanation: Question 4.<\/p>\n (A) a bright image Explanation: V. Read the passage and note the following observations. Answer any four questions from Question 1. to Question 5. Question 1.<\/p>\n (A) 15cm Explanation: Question 2.<\/p>\n (A) (a) Explanation: <\/p>\n Question 3.<\/p>\n (A) In (D) case Explanation: Question 4.<\/p>\n (A) Size of image decreases Explanation: Question 5.<\/p>\n (A) It forms real,inverted and diminished image. Explanation: VI. Study the given diagram and answer any four questions from Question 1. to Question 5. Question 1.<\/p>\n (A) (i) and (ii) <\/p>\n Question 2.<\/p>\n \\(\\text { (Given that } \\sin 45^{\\circ}=\\frac{1}{\\sqrt{2}} ; \\sin 30^{\\circ}=\\frac{1}{2} \\text { ) } \\) Explanation: Question 3.<\/p>\n (A) Speed of light is different in different media. Explanation: Question 4.<\/p>\n (A) Pascal Explanation: <\/p>\n Question 5.<\/p>\n (A) Air Explanation: Light Reflection and Refraction Class 10 MCQ Questions With Answers Question 1. The laws of reflection hold true for: (A) plane mirrors only (B) concave mirrors only (C) convex mirrors only (D) all reflecting surfaces Answer: (D) all reflecting surfaces Explanation: The laws of reflection hold true for all reflecting surfaces. Question 2. When an …<\/p>\n
\nThe candle is at the farthest end of the laboratory. SO, it may be considered at a distance greater that 2Fj and hence the image of formed between F2<\/sub> and ZF2<\/sub>. when the sun will be focussed, the image will be formed as F2<\/sub>. SO, the lens is to the shifted towards the screen.<\/p>\nA spherical mirror and a thin spherical lens have each a focal length of -15 cm. The mirror and the lens are likely to be<\/h2>\n
\n(B) both convex.
\n(C) the mirror is concave and the lens is convex.
\n(D) the mirror is convex, but the lens is concave. R
\nAnswer:
\n(A) both concave.<\/p>\n
\nAs per the sign convention, the focal length of a concave mirror and a concave lens are taken as negative. Hence, both the spherical mirror and the thin spherical lens are concave in nature.<\/p>\nWhich of the following lenses would you prefer to use while reading small letters found in a dictionary?<\/h2>\n
\n(B) A concave lens of focal length 50 cm
\n(C) A convex lens of focal length 5 cm
\n(D) A concave lens of focal length 5 cm
\nAnswer:
\n(C) A convex lens of focal length 5 cm<\/p>\n
\nA magnified image of an object will be obtained when it is placed between the optical centre and focus of a convex lens. Magnification is also higher for convex lenses having shorter focal length. Therefore, for reading small letters, a convex lens of focal length 5 cm should be used.<\/p>\nFigure shows a ray of light as it travels from medium A to medium B. Refractive index of the medium B relative to medium A is<\/h2>\n
\n(A) \\(\\frac{\\sqrt{3}}{\\sqrt{2}}\\)
\n(B) \\(\\frac{\\sqrt{2}}{\\sqrt{2}}\\)
\n(C) \\(\\frac{1}{\\sqrt{2}}\\)
\n(D) \\(\\sqrt{2}\\)
\nAnswer:
\n(A) \\(\\frac{\\sqrt{3}}{\\sqrt{2}}\\)<\/p>\n
\nHere, angle of incidence = 60\u00b0 Angle of refraction = r = 45\u00b0 Refractive index of the medium B relative to medium A
\n=nBA<\/sub>= \\(\\frac{\\sin i}{\\sin r}\\) = \\(\\frac{\\sin 60^{\\circ}}{\\sin 45^{\\circ}}\\)
\n= \\(=\\frac{\\frac{\\sqrt{3}}{2}}{\\left(\\frac{1}{\\sqrt{2}}\\right)}\\)= \\(\\frac{\\sqrt{3}}{\\sqrt{2}}\\)<\/p>\nA light ray enters from medium A to medium B as shown in the figure. The refractive index of medium B relative to A will be<\/h2>\n
\n(B) Less than unity
\n(C) Equal to unity
\n(D) Zero
\nAnswer:
\n(A) Greater than unity<\/p>\n
\nSince, light rays in medium B go towards normal, so it has greater refractive index and lesser velocity of light with respect to medium A. So refractive index of medium B with respect to medium A is greater than unity.<\/p>\nWhich of the following statements is true?<\/h2>\n
\n(B) A convex lens has – 4 dioptre power having a focal length 0.25 m
\n(C) A concave lens has 4 dioptre power having a focal length 0.25 m
\n(D) A concave lens has – 4 dioptre power having a focal length 0.25 m
\nAnswer:
\n(A) A convex lens has 4 dioptre power having a focal length 0.25 m<\/p>\n
\nThe power (P) of a lens of focal length (f) is given by P = 1\/f where f is the focal length meter and power in dioptre.
\nNow, p = 1\/f<\/p>\n
\nor \\(f=\\frac{1}{4} \\mathrm{~m}\\)=0.25m<\/p>\n
\n(A) Both A and R are true and R is the correct explanation of A.
\n(B) Both A and R are true but R is NOT the correct explanation of A.
\n(C) A is true but R is false.
\n(D) A is false and R is True<\/p>\nAssertion (A): Plane mirror may form real image.
\nReason (R): Plane mirror forms virtual image, if object is real.<\/h2>\n
\n(B) Both A and R are true but R is NOT the correct explanation of A.<\/p>\n
\nPlane mirror may form real image, if object is virtual.
\n<\/p>\nAssertion (A): The focal length of the convex mirror will increase, if the mirror is placed in water.
\nReason (R): The focal length of a convex mirror of radius R is equal to, f = \\(\\frac{R}{2}\\)<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nFocal length of the spherical mirror does not depend on the medium in which it is placed.<\/p>\nAssertion (A): The image formed by a concave mirror is certainly real if the object is virtual.
\nReason (R): The image formed by a concave mirror is certainly virtual if the object is real.<\/h2>\n
\n(C) A is true but R is false.<\/p>\n
\nThe image of real object may be real in case of concave mirror.<\/p>\nAssertion (A): An object is placed at a distance of from a convex mirror of focal length its image will form at infinity.
\nReason (R): The distance of image in convex mirror can never be infinity.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nThe distance of image in convex I mirror is always finite.<\/p>\nAssertion (A): If the rays are diverging after emerging from a lens; the lens must be concave.
\nReason (R): The convex lens can give diverging rays.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nIf the rays cross focal point of convex lens, they become diverging.<\/p>\nAssertion (A): Refractive index of glass with respect to air is different for red light and violet light.
\nReason (R): Refractive index of a pair of media depends on the wavelength of light used.<\/h2>\n
\n(A) Both A and R are true and R is the correct explanation of A.<\/p>\n
\nRefractive index of any pair of media is inversely proportional to wavelength of light.
\nHence, \u03bbv<\/sub> < \u03bbr<\/sub>
\nso, \u00b5r<\/sub> < \u00b5v<\/sub>
\nwhere, \u03bbv<\/sub> and < \u03bbr<\/sub> are the wavelengths of violet and red light.\u00b5r<\/sub> and < \u00b5v<\/sub> are refractive index of violet and red light.<\/p>\nAssertion (A): The refractive index of diamond is \u221a6 and refractive index of a liquid is S. If the light travels from diamond to the liquid, it will be initially reflected when the angle of incidence is 30\u00b0. 1
\nReason (R):\\(\\mu=\\frac{1}{\\sin C}, \\)
\nwhere \u00b5 is the refractive index of diamond with respect to liquid.<\/h2>\n
\n(A) Both A and R are true and R is the correct explanation of A.<\/p>\n
\nRefractive index of diamond w.r.t. liquid
\n\\({ }^{d} \\mu_{l}=\\frac{1}{\\sin C}=\\frac{\\mu_{d}}{\\mu_{l}} \\Rightarrow \\frac{\\sqrt{6}}{\\sqrt{3}}=\\frac{1}{\\sin C}\\)
\nsin C = \\(\\frac{1}{\\sqrt{2}}\\) sin 45\u00b0
\n\u2234C = 45\u00b0<\/p>\nAssertion: Light travels faster in glass than in air.
\nReason: Glass is denser than air.<\/h2>\n
\n(D) A is false and R is True<\/p>\n
\nLight travels faster in air than in glass, because glass is denser than air.<\/p>\n
\nI. Following figure illustrates the ray diagram for the formation of image by a concave mirror. The position of the object is beyond the centre of curvature of the concave mirror. On the basis of given diagram answer any four questions from
\n
\nQuestion 1.<\/p>\nIf the focal length of the concave mirror is 10 cm, the image formed will be at a distance .<\/h2>\n
\n(B) Between 10cm and 20cm
\n(C) Beyond 20cm
\n(D) At 20 cm
\nAnswer:
\n(B) Between 10cm and 20cm<\/p>\n
\nIf the focal length of the concave mirror is 10 cm, the image formed will be at a distance between 10 cm and 15 cm.<\/p>\nIn case of concave mirror, the image distance is………. when image is formed in front of the mirror and………..when the image is formed behind the mirror.<\/h2>\n
\n(B) negative, negative
\n(C) negative, positive
\n(D) positive, positive
\nAnswer:
\n(C) negative, positive<\/p>\n
\nIf an image formed behind the concave mirror, the object distance is positive but if an image is formed in front of the mirror, the image distance is negative.<\/p>\nIf the size of the object in the given figure is 5 cm and the magnification produced is -0.5. The size of the image is (in cm)…………..<\/h2>\n
\n(B) -0.1
\n(C) 2.5
\n(D) 0.1
\nAnswer:
\n(A) -2.5<\/p>\n
\nAs we know, magnification,
\nm=\\(\\frac{h_{2}}{h_{1}} \\)
\nh2<\/sub> = \\(\\frac{-(0.5 \\times 5)}{10}\\)
\nh2<\/sub> = -2.5<\/p>\nA negative sign in the magnification value indicate that the image is……….<\/h2>\n
\n(B) Real and erect
\n(C) Virtual and erect
\n(D) Virtual and inverted
\nAnswer:
\n(A) Real and inverted<\/p>\n
\nA negative sign in the magnification value indicate that the image is real and inverted.<\/p>\nAn image formed by concave mirror is virtual, when the object is placed:<\/h2>\n
\n(B) at C
\n(C) Between C and F
\n(D) Between P and F
\nAnswer:
\n(D) Between P and F<\/p>\n
\nAn image formed by concave mirror is virtual, when the object is placed between P and F.<\/p>\n
\nA student wants to project the image of a candle flame on the walls of the school laboratory by using a mirror.<\/p>\nWhich type of mirror should he use and why ?<\/h2>\n
\n(B) Concave mirror, it forms virtual image
\n(C) Concave mirror, it forms real image
\n(D) Convex mirror, it forms real image
\nAnswer:
\n(C) Concave mirror, it forms real image<\/p>\n
\nHe should use a concave mirror as it forms real images.<\/p>\nAt what distance, in terms of focal length of the mirror, should he place the candle flame to get the magnified image on the wall ? Q;<\/h2>\n
\n(B) Between F and C
\n(C) At C
\n(D) At infinity
\nAnswer:
\n(B) Between F and C<\/p>\n
\nHe should place the candle flame between the focus and centre of curvature of the mirror to get the magnified image on the wall.<\/p>\nTo get the diminished image of the candle flame, the object must be placed at:<\/h2>\n
\n(B) at C
\n(C) between F and C
\n(D) At F
\nAnswer:
\n(A) infinity<\/p>\n
\nTo get the diminished image of the candle flame, the object must be placed at infinity.<\/p>\nIf the image formed by this mirror is inverted and real, the magnification will be: A<\/h2>\n
\n(B) Negative
\n(C) Either of them
\n(D) None of the above
\nAnswer:
\n(B) Negative<\/p>\n
\nIf the image formed by this mirror is inverted and real, the magnification will be negative.<\/p>\nA virtual image formed by concave mirror is:<\/h2>\n
\n(B) erect and diminished
\n(C) inverted and diminished
\n(D) inverted and enlarged
\nAnswer:
\n(A) erect and enlarged<\/p>\n
\nA virtual image formed by concave mirror is erect and enlarged.<\/p>\n
\nA student wants to project the image of a candle flame on a screen 60 cm in front of a mirror by keeping the flame at a distance of 15 cm from its pole.<\/p>\nSuggest the type of mirror he should use:<\/h2>\n
\n(B) plane mirror
\n(C) concave mirror
\n(D) none of the above
\nAnswer:
\n(C) concave mirror<\/p>\n
\nHe should use a concave mirror, as it forms a real image on the same side of the mirror.<\/p>\nFind the linear magnification of the image produced.<\/h2>\n
\n(B) + 4
\n(C) – 900
\n(D) + 900
\nAnswer:
\n(A) – 4<\/p>\n
\nObject distance, u = -15 cm
\nImage distance, v = – 60 cm
\nMagnification, m = \\(\\frac{-v}{u}=\\frac{-(-60)}{(-15)}=-4 \\)<\/p>\nWhen object distance is less than focal length the image is and when object distance is more than focal length the image is .<\/h2>\n
\n(B) virtual in both case
\n(C) real, virtual
\n(D) virtual, real
\nAnswer:
\n(D) virtual, real<\/p>\n
\nWhen object distance is less than focal length the image is virtual and when object distance is more than focal length the image is real.<\/p>\nWhat is the distance between the object and its image ?<\/h2>\n
\n(B) 35 cm
\n(C) 75 cm
\n(D) 0 cm
\nAnswer:
\n(A) 45 cm<\/p>\n
\nThe image is formed at a distance of 45 cm from the object.<\/p>\nThe image formed in the above case is:<\/h2>\n
\n(B) real, erect and magnified
\n(C) real, inverted and magnified
\n(D) real, erect and diminished
\nAnswer:
\n(C) real, inverted and magnified<\/p>\n
\nIn this case, the image is formed beyond the centre of curvature. This image is real, inverted and enlarged.<\/p>\n
\nA student focuses the image of a candle flame, placed at about 2 m from a convex lens of focal length 10 cm, on a screen. After that he moves gradually the flame towards the lens and each time focuses its image on the screen.<\/p>\nIn which direction does he move the lens to focus the flame on the screen ?<\/h2>\n
\n(B) towards the screen
\n(C) should not move the screen
\n(D) toward the candle
\nAnswer:
\n(D) toward the candle<\/p>\n
\nLet us assume the screen to lens distance is greater than 20 cm. Since it is required to get image beyond 2F, the object should be F and 2F on other side of the lens. Hence student will move the lens towards candle. (F means a location at a distance from lens that equals the focal length of lens. 2F means distance that equals twice the focal length).<\/p>\nWhat happens to the size of the image of the flame formed on the screen?<\/h2>\n
\n(B) size of image will increase
\n(C) remains unchanged
\n(D) size will become too small
\nAnswer:
\n(B) size of image will increase<\/p>\n
\nSize of the image of the flame increases when object is moving towards lens, from a distance beyond 2F, then 2F, then less than 2F.<\/p>\nWhat difference is seen in the intensity (brightness) of the image of the flame on the screen ?<\/h2>\n
\n(B) intensity of image remains same
\n(C) intensity of image reduces
\n(D) the image disappears
\nAnswer:
\n(A) intensity of image increases<\/p>\n
\nAs the object (candle) is moved towards lens more light intensity is collected by lens, hence brightness of the image increase.<\/p>\nWhat is seen on the screen when the flame is very close (at about 5 cm) to the lens ?<\/h2>\n
\n(B) a magnified image
\n(C) diminished image
\n(D) no image
\nAnswer:
\n(D) no image<\/p>\n
\nWhen the candle is very close about 5 cm, focussing the flame is not possible. Hence student will not get any image on the screen.
\nHe will get diffused light on the screen.<\/p>\n
\nA student focussed the image of a candle flame on a white screen by placing the flame at various distances from a convex lens. He noted his<\/p>\n\n\n
\n Distance of flame from the lens (cm)<\/td>\n Distance of the screen from the lens (cm)<\/td>\n<\/tr>\n \n (A) 60<\/td>\n 20<\/td>\n<\/tr>\n \n (B) 40<\/td>\n 24<\/td>\n<\/tr>\n \n (C) 30<\/td>\n 30<\/td>\n<\/tr>\n \n (D) 24<\/td>\n 40<\/td>\n<\/tr>\n \n (E) 15<\/td>\n 70<\/td>\n<\/tr>\n<\/tbody>\n<\/table>\n From the above table, find the focal length of lens without using lens formula: |AEI<\/h2>\n
\n(B) 30cm
\n(C) 40cm
\n(D) 60cm
\nAnswer:
\n(A) 15cm<\/p>\n
\nu = 30 cm, v = 30 cm This is possible if the object is placed at 2f \u2234 2f = 30 cm, f= 15 cm<\/p>\nWhich set of observations is incorrect?<\/h2>\n
\n(B) (c)
\n(C) (e)
\n(D) (d)
\nAnswer:
\n(C) (e)<\/p>\n
\nu = 15 cm, v = 70 cm is incorrect. This is because if the object is at focus then image is formed at infinity.<\/p>\nIn which case, the size of the object and image will be same:<\/h2>\n
\n(B) In (B) case
\n(C) In (C) case
\n(D) In (A) case
\nAnswer:
\n(C) In (C) case<\/p>\n
\nIn (C) case, because object is at the centre of curvature.<\/p>\nWhat is the change in image observed as the object is moved from infinity towards the concave lens?<\/h2>\n
\n(B) Size of image becomes highly diminished
\n(C) Size of the image remains unchanged
\n(D) Size of the image increases slightly
\nAnswer:
\n(D) Size of the image increases slightly<\/p>\n
\nThe size of the image increases slightly, though it remains diminished in comparison to the size of the object.<\/p>\nWhich of the following statement is false for the formation of images by convex lens?<\/h2>\n
\n(B) It forms virtual erect and enlarged image.
\n(C) It forms virtual, erect, and diminished image.
\n(D) It forms real,inverted and enlarged image.
\nAnswer:
\n(C) It forms virtual, erect, and diminished image.<\/p>\n
\nImage formed by a convex mirror is always diminished and erect.<\/p>\n
\nA very thin narrow beam of white light is made incident on three glass objects shown below.
\nStudy the nature and behaviour of the emergent beam in all the three cases.
\n<\/p>\nFollowing are the possibility of two emergent beams being similar. Choose the correct answer:<\/h2>\n
\n(B) (i) and (iii)
\n(C) (ii) and (iii)
\n(D) No similar emergent beams
\nAnswer:
\n(B) (i) and (iii)
\nExplanation:
\nIn (i) emergent beam is white and laterally displaced. In (ii) emergent beam is a spectrum of seven colours bent in different angles. In (iii) emergent beam from the second prism is white only. Similarity between (i) and (iii) as both emergent rays are white in colour<\/p>\nWhen light enters from air to glass, the angles of incidence and refraction in air and glass are 45\u00b0 and 30\u00b0, respectively. Find the refractive index of glass.<\/h2>\n
\n(A) \u221a2
\n(B) 2\u221a2
\n(C) 1\/(\u221a2)
\n(D) 1
\nAnswer:
\n(A) \u221a2<\/p>\n
\n\\(a_{g}=\\frac{\\sin i}{\\sin r}=\\frac{\\sin 45^{\\circ}}{\\sin 30^{\\circ}} \\)
\n\\(=\\frac{\\frac{1}{\\sqrt{2}}}{\\frac{1}{2}}=\\sqrt{2} \\)<\/p>\nThe light changes its path as its medium changes. Which of the following is an incorrect statement ?<\/h2>\n
\n(B) Light changes its path because light only travels in straight line.
\n(C) Speed of light is dependent on medium through which it is passing.
\n(D) The light chooses the path with minimum time, as it changes its medium.
\nAnswer:
\n(B) Light changes its path because light only travels in straight line.<\/p>\n
\nSpeed of light is different in different media. As the medium changes, the light has to choose a path of minimum time. Hence, the direction of the light changes. This phenomenon is known as refraction of light.<\/p>\nWhat is the unit of refractive index? R<\/h2>\n
\n(B) Joule
\n(C) No unit
\n(D) p m
\nAnswer:
\n(C) No unit<\/p>\n
\nRefractive index being a ratio of two similar quantities hence has no unit.<\/p>\nLight travel fastest in:<\/h2>\n
\n(B) Vacuum
\n(C) Glass
\n(D) diamond
\nAnswer:
\n(B) Vacuum<\/p>\n
\nLight waves travel fastest through a vacuum and air, and slower through other materials such as glass or water.<\/p>\nMCQ Questions for Class 10 Science with Answers<\/a><\/h4>\n","protected":false},"excerpt":{"rendered":"