{"id":117186,"date":"2022-04-04T12:41:46","date_gmt":"2022-04-04T07:11:46","guid":{"rendered":"https:\/\/www.mapsofindia.com\/my-india\/?p=117186"},"modified":"2022-04-04T12:41:46","modified_gmt":"2022-04-04T07:11:46","slug":"chapter-10-light-reflection-and-refraction-questions-and-answers-ncert-solutions-for-class-10-science","status":"publish","type":"post","link":"https:\/\/www.mapsofindia.com\/my-india\/education\/chapter-10-light-reflection-and-refraction-questions-and-answers-ncert-solutions-for-class-10-science","title":{"rendered":"Chapter 10 – Light Reflection and Refraction Questions and Answers: NCERT Solutions for Class 10 Science"},"content":{"rendered":"

Question 1
\nDefine the principal focus of a concave mirror.<\/h2>\n

Answer:
\nThe principal focus of a concave mirror is a point on its principal axis to which all the light rays which are parallel and close to the axis, converge after reflection from the concave mirror.<\/h3>\n

Question 2
\nThe radius of curvature of a spherical mirror is 20 cm. What is its focal length?<\/h2>\n

Answer:
\nFocal length = 12 x Radius of curvature = 12 x 20 cm = 10 cm<\/h3>\n

Question 3
\nName a mirror that can give an erect and enlarged image of an object.<\/h2>\n

Answer:
\nConcave mirror.<\/h3>\n

Question 4
\nWhy do we prefer a convex mirror as a rear-view mirror in vehicles ?<\/h2>\n

Answer:
\nWe prefer a convex mirror as a rear-view mirror in vehicles because of two reasons :
\n1. A convex mirror always produces an erect image of the objects.
\n2. The image formed in a convex mirror is highly diminished or much smaller than the object, due to which a convex mirror gives a wide field of view of the traffic behind. A convex mirror enables the driver to view such larger area of the traffic behind him.<\/h3>\n

Question 1
\nFind the focal length of a convex mirror whose radius of curvature is 32 cm.<\/h2>\n

Solution:
\nR = +32 cm and f=R2=+322=+16cm<\/h3>\n

Question 2
\nA concave mirror produces three times magnified (enlarged) real image of an object placed at 10 cm in front of it. Where is the image located ?<\/h2>\n

Solution:
\nBecause the image is real, so magnification m must be negative.<\/h3>\n

\"\"
\nThus the image is located at a distance of 30 cm from the mirror on the object side of the mirror.<\/h3>\n

Question 1
\nA ray of light travelling in air enters obliquely into water. Does the light ray bend towards the normal or away from the normal ? Why ?<\/h2>\n

Answer:
\nThe light-ray bends towards the normal because the ray of light goes from a rarer medium to a denser medium.<\/h3>\n

Question 2
\nLight enters from air to glass having refractive index 1.50. What is the speed of light in the glass ? The speed of light in vacuum is 3 x 108 ms-1.<\/h2>\n

Solution:
\nRefractive index of glass, n8 = 1.50<\/h3>\n

\"\"
\nQuestion 3
\nFind out, from Table 10.3, the medium having highest optical density. Also find the medium with lowest optical density.<\/h2>\n

Answer:
\nFrom table 10.3, diamond has highest refractive index (= 2.42), so it has highest optical density.
\nAir has lowest refractive index (= 1.0003),
\nso it has lowest optical density.<\/h3>\n

Question 4
\nYou are given kerosene, turpentine and water. In which of these does the light travel fastest ? Use the information given in Table 10.3.<\/h2>\n

Answer:
\nFor kerosene, n = 1.44
\nFor turpentine, n = 1.47
\nFor water, n = 1.33
\nBecause water has the lowest refractive index, therefore light travels fastest in this optically rarer medium than kerosene and turpentine oil.<\/h3>\n

Question 5
\nThe refractive index of diamond is 2.42. What is the meaning of this statement?<\/h2>\n

Answer:
\nBy saying that the refractive index of diamond is 2.42, we mean that the speed of light in diamond is lower by a factor of 2.42 relative to that in vacuum.<\/h3>\n

Question 1
\nDefine 1 dioptre of power of a lens.<\/h2>\n

Answer:
\nOne dioptre is the power of a lens whose focal length is 1 metre.<\/h3>\n

Question 2
\nA convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the convex lens if the image is equal to the size of the object ? Also, find the power of the lens. , Sol. Here, u \u2014 +50 cm ..<\/h2>\n

Solution:
\nHere \u03bd = +50cm
\nBecause the real image is of the same size as the object,<\/h3>\n

\"\"
\nQuestion 3
\nFind the power of a concave lens of focal length 2 m.<\/h2>\n

Solution:
\nBecause the focal length of a concave lens is negative,
\ntherefore f = -2 m<\/h3>\n

\"\"
\nQuestion 1
\nWhich one of the following materials cannot be used to make a lens ?
\n(a) Water
\n(b) Glass
\n(c) Plastic
\n(d) Clay<\/h2>\n

Answer:
\n(d) Clay<\/h3>\n

Question 2
\nThe image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object ?
\n(a) Between the principal focus and the centre of curvature
\n(b) At the centre of curvature
\n(c) Beyond the centre of curvature
\n(d) Between the pole of the mirror and its principal focus.<\/h2>\n

Answer:
\n(d) Between the pole of the mirror and its principal focus.<\/h3>\n

Question 3
\nWhere should an object be placed in front of a convex lens to get a real image of the size of the object ?
\n(a) At the principal focus of the lens (b) At twice the focal length
\n(c) At infinity
\n(d) Between the optical centre of the lens and its principal focus.<\/h2>\n

Answer:
\n(b) At twice the focal length.<\/h3>\n

Question 4
\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 :
\n(a) Both concave.
\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.<\/h2>\n

Answer:
\n(a) Both concave<\/h3>\n

Question 5
\nNo matter how far you stand from mirror, your image appears erect. The mirror is likely to be
\n(a) plane
\n(b) concave
\n(c) convex
\n(d) either plane or convex.<\/h2>\n

Answer:
\n(d) Either plane or convex.<\/h3>\n

Question 6
\nWhich of the following lenses would you prefer to use while reading small letters found in a dictionary ?
\n(a) A convex lens of focal length 50 cm.
\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.<\/h2>\n

Answer:
\n(c) A convex lens of focal length 5 cm.<\/h3>\n

Question 7
\nWe wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. What should be the range of distance of the object from the mirror ? What is the nature of the image ? Is the image larger or smaller than the object ? Draw a ray diagram to show the image formation in this case.<\/h2>\n

Answer:
\nA concave mirror gives an erect image when the object is placed between the focus F and the pole P of the concave mirror, i.e., between 0 and 15 cm from the mirror. The image thus formed will be virtual, erect and larger than the object.<\/h3>\n

\"\"
\nQuestion 8
\nName the type of mirror used in the following situations.
\n(a) Headlights of a car.
\n(b) Side\/rear-view mirror of a vehicle.
\n(c) Solar furnace.
\nSupport your answer with reason.<\/h2>\n

Answer:
\n(a) Concave mirrors are used as reflectors in headlights of cars. When a bulb is located at the focus of the concave mirror, the light rays after reflection from the mirror travel over a large distance as a parallel beam of high intensity.
\n(b) A convex mirror is used as a side\/rear-view mirror of a vehicle because
\n\u2022 A convex mirror always forms an erect, virtual and diminished image of an object placed anywhere in front it.
\n\u2022 A convex mirror has a wider field of view than a plane mirror of the same size.
\n(c) Large concave mirrors are used to concentrate sunlight to produce heat in solar furnaces.<\/h3>\n

Question 9
\nOne-half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object ? Verify your answer experimentally. Explain your observations.<\/h2>\n

Answer:
\nA convex lens forms complete image of an object, even if its one half is covered with black paper. It can be explained by considering following two cases.
\nCase I : When the upper half of the lens is covered
\nIn this case, a ray of light coming from the object will be refracted by the lower half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the following figure.<\/h3>\n

\"\"
\nCase II: When the lower half of the lens Is covered
\nIn this case, a ray of light coming from the object is refracted by the upper half of the lens. These rays meet at the other side of the lens to form the image of the given object, as shown in the given figure.<\/h3>\n

\"\"
\nQuestion 10
\nAn 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.<\/h2>\n

Answer:
\nHere : Object distance, u= -25 cm,
\nObject height, h = 5 cm,
\nFocal length, f = +10 cm
\nAccording to the lens formula, 1f=1\u03bd\u22121u , we have
\n\u21d2 1\u03bd=1f\u22121u=110\u2212125=15250or\u03bd=25015=16.66cm
\nThe positive value of v shows that the image is formed at the other side of the lens.<\/h3>\n

\"\"
\nThe negative value of image height indicates that the image formed is inverted.
\nThe position, size, and nature of image are shown alongside in the ray diagram.<\/h3>\n

\"\"
\nQuestion 11
\nA concave lens of focal length 15 cm forms an image 10 cm from the lens. How far is the object placed from the lens ? Draw the ray diagram.<\/h2>\n

Solution:
\nFocal length, f = -15 cm, Image distance, \u03bd = -10 cm (as concave lens forms the image on the same side of the lens)
\nFrom the lens formula 1f=1\u03bd\u22121u , we have<\/h3>\n

\"\"
\nObject distance, u = -30 cm
\nThe negative value of u indicates that the object is placed in front of the lens.<\/h3>\n

Question 12
\nAn object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the image.<\/h2>\n

Solution:
\nObject distance, u = -10 cm, Focal length, f = +15 cm, Image distance, \u03bd = ?<\/h3>\n

\"\"
\nThus, image distance, \u03bd = + 6 cm
\nBecause \u03bd is +ve, so a virtual image is formed at a distance of 6 cm behind the mirror.
\nMagnification, m=\u2212\u03c5u=\u22126\u221230=15 (i.e. < 1)
\nThe positive value of m shows that image erect and its value, which is less than 1, shows that image is smaller than the object. Thus, image is virtual, erect and diminished.<\/h3>\n

Question 13
\nThe magnification produced by a plane mirror is +1. What does this mean ?<\/h2>\n

Answer:
\nSince magnification, m=h\u2018h=\u2212\u03bdu. Given, m = +1, so h\u2019 = h and \u03bd = -u
\n(i) m = 1 indicates the size of image is same as that of object.
\n(ii) positive sign of m indicates that an erect image is formed.
\nThe opposite signs of \u03bd and u indicate that image is formed on the other side of the mirror from where the object is placed i.e., image is formed behind the mirror and thus image formed is virtual.<\/h3>\n

Question 14
\nAn object 5.0 cm in length is placed at a distance of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image, its nature and size.<\/h2>\n

Solution:
\nSince object size, h = +5 cm,
\nobject distance, u = -20 cm
\nand radius of curvature, R = +30 cm<\/h3>\n

\"\"
\nA virtual, erect image of height 2.2 cm is formed behind the mirror at a distance of 8.6 cm from the mirror.<\/h3>\n

Question 15
\nAn object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focussed image can be obtained ? Find the size and the nature of the image.<\/h2>\n

Answer:
\nHere, object size, h = +7.0 cm,
\nobject distance, u = -27 cm
\nand focal length, f = -18 cm
\nImage distance, \u03bd = ?
\nand image size, h\u2019 = ?
\nFrom the mirror formula, 1f=1\u03bd\u22121u, we have<\/h3>\n

\"\"
\nThe screen should be placed at a distance of 54 cm on the object side of the mirror to obtain a sharp image.<\/h3>\n

\"\"
\nThe image is real, inverted and enlarged in size.<\/h3>\n

Question 16
\nFind the focal length of a lens of power -2.0 D. What type of lens is this ?<\/h2>\n

Answer:
\nHere, P = -2.0 D
\nThe type of lens is concave because the focal length is negative.<\/h3>\n

\"\"
\nQuestion 17
\nA doctor has prescribed a corrective lens of power +1.5 D. Find the focal length of the lens. Is the prescribed lens diverging or converging ?<\/h2>\n

Answer:
\nHere, P = +1.5 D<\/h3>\n

\"\"
\nBecause the focal length is positive, the prescribed lens is converging.<\/h3>\n

Question 1.
\nDefine the principal focus of a concave mirror?<\/h2>\n

Answer:
\nLight rays that are parallel to the principal axis of a concave mirror converge at a specific point on its principal axis after reflecting from the mirror. This point is known as the principal focus of the concave mirror.<\/h3>\n

Question 2.
\nThe radius of curvature of a spherical mirror is 20 cm. What is its focal length?<\/h2>\n

Answer:
\nRadius of curvature, R = 20 cm
\nRadius of curvature of a spherical mirror = 2 x Focal length (f)
\nf = R\/2 = 20\/2 =10cm<\/h3>\n

Question 3.
\nName the mirror that can give an erect and enlarged image of an object.<\/h2>\n

Answer:
\nWhen an object is placed between the pole and the principal focus of a concave mirror, the image formed is virtual, erect, and enlarged.<\/h3>\n

Question 4.
\nWhy do we prefer a convex mirror as a rear-view mirror in vehicles?<\/h2>\n

Answer:
\nConvex mirrors give a virtual, erect, and diminished image of the objects placed in front of them. They are preferred as a rear-view mirror in vehicles because they give a wider field of view, which allows the driver to see most of the traffic behind him.<\/h3>\n

Question 1. Find the focal length of a convex mirror whose radius of curvature is 32 cm.<\/h2>\n

Answer: Radius of curvature, R = 32 cm
\nRadius of curvature = 2 x Focal length (f)
\nR = 2f
\nf = R\/2 = 32\/2 = 16cm
\nHence, the focal length of the given convex mirror is 16 cm.<\/h3>\n

Question 2.
\nA concave mirror produces three times magnified (enlarged) real image of object placed at 10 cm in front of it. Where is the image located?<\/h2>\n

Answer:
\nGiven, u = \u2013 10 cm
\nSince image is real inverted so, m = -3
\nm = -v \/ u
\n\u21d2 -3 = -v\/ -10
\nv= \u2013 30 cm
\nNegative sign indicates the image will be real and image is formed at 30 cm in front of the mirror.<\/h3>\n

Question 1.
\nDefine one dioptre of power of a lens?<\/h2>\n

Answer:
\nOne dioptre is the power Of a lens Of focal length 1m.
\nPower of lens is defined as the reciprocal of its focal length. If P is the power of a lens of focal length F in metres, then
\nP = 1\/ f (in meters)
\nThe S.I. unit of power of a lens is Dioptre. It is denoted by D.
\n1 dioptre is defined as the power of a lens of focal length 1 metre.
\n1 D = 1 m\u22121<\/h3>\n

Question 2.
\nA convex lens forms a real and inverted image of a needle at a distance of 50 cm from it. Where is the needle placed in front of the lens if the image is equal to the size of the object? Also find the power of the lens.<\/h2>\n

Answer:
\nv = + 50 cm
\nSince image is real and of same size. The position of image should be double the focal length.
\nHence, the object should be at 2f.
\nV = 2f = 50, f = 25 cm.
\nPower = 1\/f = 100\/25 = 4D<\/h3>\n

Question 3.
\nFind the power of a concave lens of focal length 2 m.<\/h2>\n

Answer:<\/h3>\n

\"\"
\nQuestion 1.
\nWhich one of the following materials cannot be used to make a lens?
\n(a) Water (b) Glass
\n(c) Plastic (d) Clay<\/h2>\n

Answer: (d) Clay<\/h3>\n

Question 2.
\nThe \u00a1mage formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
\n(a) Between the principal focus and the centre of Curvature
\n(b) At the centre of curvature
\n(c) Beyond the centre of curvature
\n(d) Between the pole of the mirror and Its principal focus.<\/h2>\n

Answer: (d) Between the pole of the mirror and its principal focus.<\/h3>\n

Question 3.
\nWhere should an object b. placed In front of a convex lens to get a real
\nimage of the size of the object?
\n(a) At the principal focus of the lens
\n(b) At twice the focal length
\n(c) At infinity
\n(d) Between the optical centre of the lens and its principal focus<\/h2>\n

Answer:
\n(b) At twice the focal length<\/h3>\n

Question 4.
\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:
\n(a) both concave
\n(b) both convex
\n(c) the mirror is concave, but the lens is convex
\n(d) the mirror is convex, but the lens is concave<\/h2>\n

Answer:
\n(a) Both concave.<\/h3>\n

Question 5.
\nNo matter how far you stand from a mirror, your Image appears erect. The mirror is likely to be
\n(a) plane
\n(b) concave
\n(c) convex
\n(d) Either plane or convex<\/h2>\n

Answer:
\n(d) Either plane or convex.<\/h3>\n

Question 6.
\nWhich of the following lenses would you prefer to use while reading small letters found \u00a1n a dictionary?
\n(a) A convex lens of focal length 50cm
\n(b) A concave lens of focal length 50cm
\n(c) A convex lens of focal length 5 cm
\n(d) A concave lens of focal length 5 cm.<\/h2>\n

Answer:
\n(c) A convex lens of focal length 5 cm<\/h3>\n

Question 7.
\nWe wish to obtain an erect image of an object, using a concave mirror of focal length 15 cm. what should be the range of distance of the object from the mirror? What is the nature of the image? Is the image larger or smaller than the object? Draw a ray diagram to show the image formation in this
\ncase.<\/h2>\n

Answer:
\nWe are given the focal length cf the concave mirror as f = -15cm.
\nFor getting an erect image using a concave mirror, the object should be placed at a distance less than the focal length.
\ni.e. 15 cm from the pole. The image formed will be virtual, enlarged and erect.<\/h3>\n

\"\"
\nQuestion 8.
\nName the type of mirror used in the following situations.
\n(a) Headlights of a car
\n(b) Side\/rear-view mirror of a vehicle
\n(c) Solar furnace
\nSupport your answer with reason.<\/h2>\n

Answer:
\n(a) Concave mirror, to get powerful and parallel beams of light.
\n(b) Convex mirror because it always gives an erect image and enables the driver to view much larger area.
\n(c) Concave or parabolic mirror because it can concentrate sunlight at the focus to produce heat in the solar furnace.<\/h3>\n

Question 9.
\nOne half of a convex lens is covered with a black paper. Will this lens produce a complete image of the object? Verify your answer experimentally. Explain your observations.<\/h2>\n

Answer:
\nYes, even when one half of the lens is covered with a black paper, complete image of the object will be formed. Take a convex lens and focus the light from a distant object onto a screen. As expected an image (sharp) is formed at a distance equal to the focal length Cover the lower or the upper half of the lens and focus the light from the same object onto the same screen. You will be able to get a sharp image again; however the brightness of the image will be less in the second case. The same effect w,ll be seen even if the lens is half covered with black strips.<\/h3>\n

Question 10.
\nAn object 5cm in length is held 25cm away from a converging lens of focal length 10 cm. Draw a ray diagram and find the position, size and the nature of the image formed.<\/h2>\n

Answer:<\/h3>\n

\"\"
\nTherefore, the mage \u00a1s formed between F2 and 2F2 on the other side of the lens. It is real and inverted, and smaller in size than the object.<\/h3>\n

Question 11.
\nA concave lens of focal length 15 cm forms an \u00a1mage 10 cm from the lens. How far is the object placed from the Pens? Draw the ray diagram.<\/h2>\n

Answer:<\/h3>\n

\"\"
\nQuestion 12.
\nAn object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position and nature of the \u00a1mage.<\/h2>\n

Answer:
\nf = +15 cm. u = -1o cm
\nFor mirror, we have<\/h3>\n

The image must be virtual and erect.<\/h3>\n

\"\"
\nQuestion 13.
\nThe magnification produced by a plane mirror is +1. What does this mean?<\/h2>\n

Answer:
\nThis means that size of the image is equal to the size of the object.<\/h3>\n

Question 14.
\nAn object 5.0 cm in length Is placedat a distanc, of 20 cm in front of a convex mirror of radius of curvature 30 cm. Find the position of the image nature and size.<\/h2>\n

Answer:<\/h3>\n

\"\"
\nQuestion 15.
\nAn object of size 7.0 cm is placed at 27 cm in front of a concave mirror of focal length 18 cm. At what distance from the mirror should a screen be placed, so that a sharp focused image can be obtained? Find the size and the nature of the image.<\/h2>\n

Answer:<\/h3>\n

\"\"
\nQuestion 16.
\nFind the focal length of a lens of power -2.0 D. What type of lens is this?<\/h2>\n

Answer:<\/h3>\n

\"\"
\nQuestion 17.
\nA doctor has prescribed a corrective lens of power +1.5 D. find the focal length of the lens. Is the prescribed lens diverging or converging?<\/h2>\n

Answer:<\/h3>\n

\"\"
\nQuestion 1.
\nHold a highly polished steel spoon curved inwards close to your face and move it slowly away from your face. What will you observe?
\n(a) Enlarged and erect image of your face
\n(b) Smaller and inverted image of your face
\n(c) Smaller and erect image of your face
\n(d) Enlarged and inverted image of your face<\/h2>\n

Answer:
\n(b) The inner curved surface of a highly polished steel spoon acts as a concave mirror. When the spoon is at a small distance from the face such that, the object lies between pole and focus of concave mirror, so an enlarged and erect image of your face will be observed but as the spoon is slowly moved away from the face, the image becomes smaller and appears inverted.<\/h3>\n

Question 2.
\nWhich one of the following materials cannot be used to make a lens?
\n(a) Water
\n(b) Glass
\n(c) Plastic
\n(d) Clay<\/h2>\n

Answer:
\n(d) Clay can never be transparent, so it cannot be used to make lens.<\/h3>\n

Question 3.
\nNo matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
\n(a) plane
\n(b) concave
\n(c) convex
\n(d) either plane or convex<\/h2>\n

Answer:
\n(d) Plane mirrors and convex mirrors always form the erect images.<\/h3>\n

Question 4.
\nThe image formed by a concave mirror is observed to be virtual, erect and larger than the object. Where should be the position of the object?
\n(a) Between principal focus and centre of curvature
\n(b) At centre of curvature
\n(c) Beyond centre of curvature
\n(d) Between pole of the mirror and its principal focus<\/h2>\n

Answer:
\n(d)<\/h3>\n

Question 5.
\nAn object AB is placed in front of a convex lens at its centre of curvature as shown in figure below.
\nFour students traced the path of light ray after refraction through the lens. Which one of them is correct?<\/h2>\n

\"\"<\/p>\n

\"\"
\n(a) Only I
\n(b) Only II
\n(c) Only III
\n(d) Only IV<\/h2>\n

Answer:
\n(d) When the object is placed at centre of curvature (2Fx) of convex lens, the same sized image is formed at 2F2. The image formed is real and inverted.<\/h3>\n

Question 6.
\nA spherical mirror and a thin spherical lens have each of a focal length -15 cm. The mirror and lens are likely to be
\n(a) both concave
\n(b) both convex
\n(c) mirror is concave and lens is convex
\n(d) mirror is convex and lens is concave<\/h2>\n

Answer:
\n(a) The focal length is taken as negative for both concave mirror and concave lens.<\/h3>\n

Question 7.
\nWhich of the following can make a parallel beam of light when light from a point source is incident on it?
\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<\/h2>\n

Answer:
\n(a) A ray passing through the principal focus of a concave mirror or convex lens, after reflection\/refraction, will emerge parallel to the principal axis.<\/h3>\n

\"\"
\nQuestion 8.
\nUnder which of the following conditions, a concave mirror can form an image larger than the actual object?
\n(a) When an object is kept at a distance equal to its radius of curvature
\n(b) When an object is kept at a distance less than its focal length
\n(c) When an object is placed between the focus and centre of curvature
\n(d) When an object is kept at a distance greater than its radius of curvature<\/h2>\n

Answer:
\n(c) A concave mirror can form an image enlarged, real and inverted than the actual object, beyond centre of curvature (C) when object is placed between the focus (F) and centre of curvature.<\/h3>\n

Question 9.
\nA light ray enters from medium A to medium Bas shown in the figure. The refractive index of medium B relative to A will be<\/h2>\n

\"\"
\n(a) greater than unity
\n(b) less than unity
\n(c) equal to unity
\n(d) zero<\/h2>\n

Answer:
\n(a) Since, light rays in the medium B goes towards normal. So, it has greater refractive index and lesser velocity of light w.r.t. medium A. So, refractive index of medium B w.r.t. medium A is greater than unity.<\/h3>\n

Question 10.
\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

\"\"
\nAnswer:
\n(a) Given, angle of incidence, i = 60\u00b0, angle of refraction, r = 45\u00b0
\nRefractive index of the medium B relative to medium A,<\/h3>\n

\"\"
\nQuestion 11.
\nBeams of light are incident through the holes A and B and emerge out of box through the holes C \u2013 and D respectively, as Box shown in the figure.<\/h2>\n

\"\"
\nWhich of the following could be inside the box?
\n(a) A rectangular glass slab
\n(b) A convex lens
\n(c) A concave lens
\n(d) A prism<\/h2>\n

Answer:
\n(a) Here, the emergent rays are parallel to the direction of the incident ray. Therefore, a rectangular glass slab could be inside the box as the extent of bending of light ray at the opposite parallel faces AB (air-glass interface) and CD (glass-air interface) of the rectangular glass slab are equal and opposite. This is why the ray emerges parallel to the incident ray.<\/h3>\n

Question 12.
\nA beam of light is incident through the holes on side A and emerges out of the holes on the other face of the box as shown in the figure. Which of the following could be inside the box?<\/h2>\n

\"\"
\n(a) Concave lens
\n(b) Rectangular glass slab
\n(c) Prism
\n(d) Convex lens<\/h2>\n

Answer:
\n(d) Since, in the figure all the parallel rays converge at a point. So, inside the box there must be a convex lens.<\/h3>\n

\"\"
\nQuestion 13.
\nWhich of the following statement is true?
\n(a) A convex lens has 4D power having a focal length 0.25 m
\n(b) A convex lens has 4D power having a focal length -0.25 m
\n(c) A concave lens has 4D power having a focal length 0.25 m
\n(d) A concave lens has 4D power having a focal length -0.25 m<\/h2>\n

Answer:
\n(a) The power P of a lens of focal length f is given by
\nP = 1\/f, where f is the focal length in metre and P is the power in dioptre.
\nP= 1\/f or f = 1\/P = 1\/4 = 0.25 m<\/h3>\n

Question 14.
\nMagnification produced by a rear view mirror fitted in vehicles [NCERT Exemplar]
\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.<\/h2>\n

Answer:
\n(a) The convex mirror forms virtual, erect and diminished image of the object and rear view mirror also form same type of image. Therefore, magnification (m) produced by a rear view mirror fitted in vehicles is less than one, i.e. m < 1.<\/h3>\n

Question 15.
\nRays from the Sun converge at a point 15 cm in front of a concave mirror. Where should an object be placed, so that size of its image is equal to the size of the object?
\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<\/h2>\n

Answer:
\n(b) The rays from the Sun, i.e. from infinity, are parallel to principal axis after reflection converge at a point is known as focus. Therefore, focal length if) of concave mirror is 15 cm. And we know that, same size, real and inverted image is formed by concave mirror when object is placed at focus 2 A or centre of curvature, so to form same size of image, object will be placed at 15 x 2 =30 cm.<\/h3>\n

Question 15.
\nRays from the Sun converge at a point 15 cm<\/h2>\n

in front of a concave mirror. Where should an object be placed, so that size of its image is equal to the size of the object?
\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<\/h2>\n

Answer:
\n(b) The rays from the Sun, i.e. from infinity, are parallel to principal axis after reflection converge at a point is known as focus. Therefore, focal length if) of concave mirror is 15 cm. And we know that, same size, real and inverted image is formed by concave mirror when object is placed at focus 2 A or centre of curvature, so to form same size of image, object will be placed at 15 x 2 =30 cm.<\/h3>\n

\"\"
\n(a) Only I
\n(b) Only II
\n(c) Only III
\n(d) Only IV<\/h2>\n

Answer:
\n(b) In a rectangular glass slab, the emergent rays are parallel to the direction of the incident ray, because the lateral deviation of bending of the ray of light at the opposite parallel faces (air-glass interface) and (glass-air interface) of the rectangular glass slab are equal and opposite. This is why the ray emerges are parallel to the incident ray.<\/h3>\n

Question 17.
\nYou are given water, mustard oil, glycerine and kerosene. In which of these media, a ray of light incident obliquely at same angle would bend the most?
\n(a) Kerosene
\n(b) Water
\n(c) Mustard oil
\n(d) Glycerine<\/h2>\n

Answer:
\n(d) The given material having their refractive index as kerosene is 1.44, water is 1.33, mustard oil is 1.46 and glycerine is 1.74. Thus, glycerine is most optically denser and hence have the largest refractive index. Therefore, ray of light bend most in glycerine.<\/h3>\n

Question 18.
\nA student placed a light bulb in midway between the two plane mirrors inclined at an angle of 60\u00b0. How many images will be observed by him?
\n(a) 4
\n(b) 6
\n(c) 5
\n(d) 8<\/h2>\n

Answer:
\n(c) Number of images formed by two plane mirrors inclined at an angle 60\u00b0 when a light bulb is placed in midway between them is
\nN = 360\u00b0\/60\u00b0 \u2013 1 = 6 \u2013 1 = 5<\/h3>\n

Question 19.
\nWhere should an object be placed in front of a convex lens to get a real image of the size of the object?
\n(a) At the principal focus of the lens
\n(b) At twice the focal length
\n(c) At infinity
\n(d) Between the optical centre of the lens and its principal focus<\/h2>\n

Answer:
\n(b) To set the real image of the size of the object, it should be placed at twice the focal length of a convex lens.<\/h3>\n

Question 20.
\nWhich of the following lenses would you prefer to use while reading small letters found in dictionary?
\n(a) A convex lens of focal length 50 cm
\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<\/h2>\n

Answer:
\n(c) Convex lens is used as magnifying glass. For better performance its focal length should be small.<\/h3>\n","protected":false},"excerpt":{"rendered":"

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