CBSE Class 10 Science Light Reflection and Refraction VBQs Set 03

TOPIC : Combination, Power and Uses of Spherical Lens

Question. If the power of lens is –4.0D, then it means that the lens is a 
(a) concave lens of focal length –50 m
(b) convex lens of focal length +50 cm
(c) concave lens of focal length –25 cm
(d) convex lens of focal length –25 cm
Answer: (c) concave lens of focal length –25 cm

Question. In which of the following devices, convex lens is not used?
(a) Magnifying glass
(b) Microscope
(c) Flashlights
(d) Telescope
Answer: (c) Concave lens is used in flashlights to widen the beam produced by the bulb.

Question. Two convex lens P and Q have focal length 0.50 m and 0.40 m respectively. Which of the following is TRUE about the combined power of the two lenses?
(a) P is equal to 4.5 D.
(b) P is less than 4.5 D.
(c) P is more than 4.5 D.
(d) P cannot be determined from the information given.
Answer: (a) P is equal to 4.5 D.

Question. The combination of spherical lens of power +3D and –3D acts as a
(a) bifocal lens of power 6D
(b) plane glass sheet/slab
(c) convex lens having no power
(d) concave lens of power – 6D
Answer: (b) The combination will behave as a plane glass sheet/slab.

Question. The focal length of four convex lens P, Q, R and S are 20 cm, 15 cm, 5 cm and 10 cm respectively. The lens having lowest power is
(a) P
(b) Q
(c) R
(d) S
Answer: (a) P

Short Answer Type Questions

Question. (a) The power of a combination of two lenses XY is 5D if the focal length of lens X is 15 cm. State the nature and focal length of lens Y. 
(b) When object is placed between infinity and the pole of the convex mirror, where is the image formed?
Answer: (a) Power of a combination of two lenses X and Y = 5D
\( P = P_X + P_Y \)
\( \implies 5 = \frac{100}{15} + \frac{1}{f_Y} \) (Since \( f_X = 15 \text{ cm} \))
\( \implies \frac{1}{f_Y} = 5 - \frac{100}{15} = -\frac{25}{15} \)
\( \implies f_Y = -\frac{15}{25} = -0.6 \text{ m} = -60 \text{ cm} \)
Therefore, the focal length of given lens Y is 60 cm and it is a concave lens.
(b) Between P and F, behind the mirror.

Question. An object is kept at a distance of 18 cm, 20 cm, 22 cm and 30 cm respectively from a lens of power + 5D.
(a) In which case or cases would you get a magnified image?
(b) Which of the magnified image can be got on a screen? 
Answer: Since Power of lens = + 5 D
\( \implies f = \frac{1}{P} = \frac{1}{+5} = +0.2 \text{ m} = +20 \text{ cm} \)
(a) We would get a magnified image only when the object is kept at a distance of 18 cm, 22 cm and 30 cm respectively.
(b) The object at the positions of 22 cm and 30 cm will produce a magnified, real image on a screen.
Reason:
(i) A magnified virtual image is formed by a convex lens when the object lies between the focus and the optical centre of the lens.
(ii) A magnified real image is formed by a convex lens when the object lies between F and 2F.

Question. Define power of a lens. The focal length of a lens is – 10 cm. Write the nature of the lens and find its power. If an object is placed at a distance of 20 cm from the optical centre of this lens, according to the New Cartesian Sign Convention, what will be the sign of magnification in this case? 
Answer: • Power of a Lens: The ability of a lens to converge or diverge the ray of light after refraction, is called power (P) of the lens. It is defined as the reciprocal of the focal length, i.e. \( P = \frac{1}{f} \).
• Focal length of lens is –10 cm. Therefore, nature of lens = concave
Power of lens, \( P = \frac{100}{f(\text{cm})} = \frac{100}{-10 \text{ cm}} = -10\text{D} \)
• Magnification produced by concave lens is always positive as it forms only virtual and erect image on the same side of the object.

PRACTICE QUESTIONS

Question. A lens of power +2D and lens of power –1D are placed in close contact with each other. The combination will behave like
(a) a convergent lens of focal length 50 cm
(b) a convergent lens of focal length 100 cm
(c) a convergent lens of focal length 150 cm
(d) a divergent lens of focal length 100 cm
Answer: (b) a convergent lens of focal length 100 cm

Question. If a lens can converge the sun rays at a point 20 cm away from its optical centre, the power of the lens is
(a) +2D
(b) –2D
(c) +5D
(d) –5D
Answer: (c) +5D

Question. A convex lens of focal length 25 cm and a concave lens of focal length 20 cm are placed in close contact with one another.
(a) What is the power of the combination?
(b) Is this combination converging or diverging?
Answer: (a) \( P_1 = \frac{100}{25} = +4\text{D} \), \( P_2 = \frac{100}{-20} = -5\text{D} \). Power of combination \( P = P_1 + P_2 = +4 - 5 = -1\text{D} \).
(b) The combination is diverging because the total power is negative.

Question. A concave lens of power –2D is in contact with the convex lens of power +4D. Now a parallel beam of light is allowed to incident on this combination. At what distance from the combination will the beam get focused? State the nature of image formed.
Answer: Power of combination \( P = -2 + 4 = +2\text{D} \).
\( f = \frac{1}{P} = \frac{1}{+2} = 0.5 \text{ m} = 50 \text{ cm} \).
The beam will get focused at a distance of 50 cm from the combination. The image formed will be real.

Question. State three uses of (a) convex lens (b) concave lens in our daily life.
Answer: (a) Convex lens uses: (i) In magnifying glasses, (ii) In cameras, (iii) To correct hypermetropia (long-sightedness).
(b) Concave lens uses: (i) In peep holes of doors, (ii) In flashlights to spread the beam, (iii) To correct myopia (short-sightedness).

INTEGRATED (MIXED) QUESTIONS

Question. Distinguish between a convex and a concave mirror. 
Answer: A convex mirror has its reflecting surface bulging outwards and always forms a virtual, erect, and diminished image. A concave mirror has its reflecting surface curved inwards and can form both real and virtual, magnified or diminished images depending on the object's position.

Question. Why does a ray falling normally on a plane mirror, retrace its path? 
Answer: When a ray falls normally on a mirror, the angle of incidence is \( 0^\circ \). According to the laws of reflection, the angle of reflection must also be \( 0^\circ \). Therefore, the ray retraces its path.

Question. (a) Differentiate between reflection and refraction.
(b) A lemon kept in water in a glass tumbler appears to be bigger than its actual size, when viewed from the sides. Explain why it so appears. 
Answer: (a) Reflection is the bouncing back of light when it strikes a polished surface. Refraction is the bending of light when it passes from one transparent medium to another.
(b) The lemon appears bigger because light rays originating from the lemon in water refract as they enter the air. These rays bend away from the normal, making the lemon appear at a position closer and larger than its actual size.

Question. (a) “A convex lens of focal length ‘f’ can form a magnified erect as well as inverted image.” Justify this statement stating the position of the object with respect to the lens in each case for obtaining these images.
(b) A convex lens of focal length 25 cm and a concave lens of focal length 10 cm are placed in close contact with each other. Calculate the lens power of this combination. 
Answer: (a) Magnified erect image is formed when the object is placed between the optical centre and the principal focus (F). Magnified inverted image is formed when the object is placed between F and 2F.
(b) \( P_1 = \frac{100}{25} = +4\text{D} \), \( P_2 = \frac{100}{-10} = -10\text{D} \).
Combined Power \( P = P_1 + P_2 = +4 - 10 = -6\text{D} \).

Question. An object is kept at a distance of 18 cm, 20 cm, 22 cm and 30 cm respectively from a lens of power + 5D. 
(a) In which case or cases would you get a magnified image?
(b) Which of the magnified image can be got on a screen? 
Answer: (a) Magnified images are obtained for \( u = 18 \text{ cm} \) (virtual), \( 22 \text{ cm} \) (real), and \( 30 \text{ cm} \) (real).
(b) Magnified images for \( 22 \text{ cm} \) and \( 30 \text{ cm} \) can be caught on a screen as they are real images.

Question. (a) How can we differentiate between convex and concave lenses without touching them? 
(b) Two thin lenses of power +3.5 D and –2.5 D are placed in contact. Find the power and focal length of the lens combination. 
Answer: (a) Look at a distant object through them. A lens that forms a diminished and upright image is a concave lens. A lens that can form an inverted image of a distant building is a convex lens.
(b) \( P = +3.5 - 2.5 = +1.0\text{D} \).
\( f = \frac{1}{P} = \frac{1}{+1} = 1 \text{ m} = 100 \text{ cm} \).

ASSERTION AND REASON QUESTIONS

Question. Assertion: Incident light is reflected in only one direction from a smooth surface.
Reason: Since the angle of incidence and the angle of reflection are same, a beam of parallel rays of light falling on a smooth surface is reflected as a beam of parallel light rays in one direction only.
(a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.
(b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.
(c) Assertion is true but the Reason is false.
(d) The statement of the Assertion is false but the Reason is true.
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

Question. Assertion: The word AMBULANCE on the hospital vans is written in the form of its mirror as .
Reason: The image formed in a plane mirror is same size of the object.
(a) A
(b) B
(c) C
(d) D
Answer: (b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.

Question. Assertion: Concave mirrors are used as reflectors in torches, vehicle head-lights and in search lights.
Reason: When an object is placed beyond the centre of curvature of a concave mirror, the image formed is real and inverted.
(a) A
(b) B
(c) C
(d) D
Answer: (b) Both the Assertion and the Reason are correct but the Reason is not the correct explanation of the Assertion.

Question. Assertion: The light emerges from a parallel sided glass slab in a direction perpendicular to that in which it enters into the glass slab.
Reason: The perpendicular distance between the original path of incident ray and emergent ray coming out of glass slab is called lateral displacement of the emergent ray of light.
(a) A
(b) B
(c) C
(d) D
Answer: (d) The statement of the Assertion is false but the Reason is true.

Question. Assertion: When a pencil is partly immersed in water and held obliquely to the surface, the pencil appears to bend at the water surface.
Reason: The apparent bending of the pencil is due to the refraction of light when it passes from water to air.
(a) A
(b) B
(c) C
(d) D
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

Question. Assertion: 1.33 is the absolute refractive index of water.
Reason: Air is optically denser than water.
(a) A
(b) B
(c) C
(d) D
Answer: (c) Assertion is true but the Reason is false.

Question. Assertion: The value of f in a concave mirror is taken as –ve and in a convex mirror, it is taken as +ve.
Reason: All distances measured to the right of the origin are taken as +ve and those measured along the left of the origin are taken as –ve. 
(a) A
(b) B
(c) C
(d) D
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

Question. Assertion: A concave lens of very short focal length causes higher divergence than one with longer focal length.
Reason: The power of lens is directly proportional to its focal length. 
(a) A
(b) B
(c) C
(d) D
Answer: (c) Assertion is true but the Reason is false.

Question. Assertion: The SI unit of power of lens is ‘diopter’.
Reason: The power of concave lens is positive and that of convex lens is negative. 
(a) A
(b) B
(c) C
(d) D
Answer: (c) Assertion is true but the Reason is false.

Question. Assertion: Higher the refractive index of the medium, lesser will be the velocity of light in that media.
Reason: Refractive index is inversely proportional to the velocity of light in the medium.
(a) A
(b) B
(c) C
(d) D
Answer: (a) Both the Assertion and the Reason are correct and the Reason is the correct explanation of the Assertion.

CASE-BASED QUESTIONS

Question. Analyse the following observation table showing variation of image distance (v) with object distance (u) in case of a convex lens and answer the questions that follow, without doing any calculations:

(a) State the lens formula.
(b) Write the serial number of that observation which is not correct. How did you arrive at this conclusion?
(c) Take an appropriate scale to draw ray diagram for the observation at S. No. 2.
Answer: (a) Lens formula: \( \frac{1}{f} = \frac{1}{v} - \frac{1}{u} \)
(b) Observation S. No. 6 is incorrect. From observation S. No. 3, when \( u = -30 \text{ cm} \) and \( v = +30 \text{ cm} \), it indicates the object is at \( 2F \). Thus \( 2f = 30 \text{ cm} \)
\( \implies \) \( f = 15 \text{ cm} \). In observation 6, \( u = -10 \text{ cm} \), which is less than the focal length (\( f = 15 \text{ cm} \)). When an object is placed between the optical centre and focus of a convex lens, it forms a virtual image on the same side as the object, so \( v \) must be negative.
(c) [Ray diagram for \( u = -60 \text{ cm} \), \( f = 15 \text{ cm} \), \( v = +20 \text{ cm} \) showing object beyond \( 2F \) and image between \( F \) and \( 2F \)].

Question. Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid?
Answer: A biconvex lens acts as a plane glass sheet when it is immersed in a liquid having the same refractive index as that of the glass material of the lens. In this case, there is no bending of light at the interfaces.

Question. With the help of mirrors, we can form a variety of images. For example, in plane mirrors, images are the same size as the object and are located behind the mirror. Dental mirrors may produce a magnified image while security mirror in shops, on the other hand, form images that are smaller than the object. These images can be either real or virtual depending upon the position of object. The real image can be obtained on the screen only when the reflected rays meet actually. Virtual image does not form on the screen because after reflection, the reflected rays appear to meet.
(a) What are the advantages and disadvantages of using a convex mirror for seeing traffic at the rear?
(b) Name the mirror that can give an erect and enlarged image of the object.
(c) An object is placed at the distance of 10 cm, 20 cm, 30 cm and 40 cm respectively from a concave mirror of focal length 15 cm. Which position of the object will produce (i) virtual image and (ii) an image of same size?
Answer: (a) Advantage: It gives a wider field of view and always forms an erect image. Disadvantage: It forms diminished images, making objects appear further away than they actually are.
(b) Concave mirror (when the object is placed between the pole and the focus).
(c) (i) 10 cm (since it is less than the focal length of 15 cm). (ii) 30 cm (since it is at \( 2f \), the centre of curvature).

Question. What does the negative sign in the value of magnification produced by a mirror indicate about a image?
Answer: The negative sign in the value of magnification indicates that the image is real and inverted.

Question. The absolute refractive index of a medium is simply called its refractive index. The ability of a medium to refract light is also expressed in terms of its optical density. We have been using ‘rarer medium’ and ‘denser medium’ which actually means ‘optically rarer medium’ and ‘optically denser medium’. In comparing two media the one with larger refractive index is optically denser and vice-versa.

(a) Name the medium which have lowest and highest optical density.
(b) You are given water, kerosene, benzene and dense flint glass. In which of these media a ray of light incident obliquely at same angle would bend the most?
(c) How the absolute refractive index related to speed of light?
Answer: (a) Lowest optical density: Air (1.0003); Highest optical density: Diamond (2.42).
(b) Dense flint glass. It has the highest refractive index (1.65) among the given options, so light will bend the most towards the normal in it.
(c) Absolute refractive index \( n = \frac{c}{v} \), where \( c \) is the speed of light in vacuum and \( v \) is the speed of light in the medium. It is inversely proportional to the speed of light in the medium.

Question. A compound microscope is an instrument which consists of two lenses L1 and L2. The lens L1 called objective, forms a real, inverted and magnified image of the given object. This serves as the object for the second lens L2; the eye piece. The eye piece functions like a simple microscope or magnifier. It produces the final image, which is inverted with respect to the original object, enlarged and virtual.
(a) What types of lenses must be L1 and L2?
(b) What is the value and sign of magnification (according to the New Cartesian Sign convention) of the image formed by L1?
(c) If the power of the eye piece (L2) is 5 D and it forms an image at a distance 80 cm from its optical centre, at what distance should the object be?
Answer: (a) Both L1 and L2 must be convex lenses.
(b) The sign of magnification is negative (since the image is real and inverted) and the value of \( |m| \) is greater than 1.
(c) Given \( P_2 = 5\text{D} \)
\( \implies \) \( f_2 = \frac{100}{5} = 20 \text{

The ability of a medium to refract light is expressed in terms of its optical density. Optical density has a definite connotation. It is not the same as mass density. On comparing two media, the one with the large refractive index is optically denser medium than the other. The other medium with a lower refractive index is optically rarer. Also the speed of light through a given medium is inversely proportional to its optical density. 

Question. Determine the speed of light in diamond if the refractive index of diamond with respect to vacuum is 2.42. Speed of light in vacuum is \( 3 \times 10^8 \) m/s.
Answer: Given \( n = 2.42 \), \( c = 3 \times 10^8 \text{ m/s} \).
Refractive index \( n = \frac{c}{v} \)
\( \implies v = \frac{c}{n} = \frac{3 \times 10^8}{2.42} \approx 1.24 \times 10^8 \text{ m/s} \).

Question. Refractive indices of glass, water and carbon disulphide are 1.5, 1.33 and 1.62 respectively. If a ray of light is incident in these media at the same angle (say q), then write the increasing order of the angle of refraction in these media.
Answer: Higher the refractive index, more the light bends towards the normal, resulting in a smaller angle of refraction. The refractive indices are: water (1.33) < glass (1.5) < carbon disulphide (1.62).
Therefore, the increasing order of the angle of refraction is: carbon disulphide < glass < water.

Question. The speed of light in glass is \( 2 \times 10^8 \) m/s and in water is \( 2.25 \times 10^8 \) m/s. (i) Which one of the two is optically denser and why? (ii) A ray of light is incident normally at the water-glass interface when it enters a thick glass container filled with water. What will happen to the path of the ray after entering the glass? Give reason.
Answer: (i) Glass is optically denser because the speed of light is lower in glass (\( 2 \times 10^8 \text{ m/s} \)) compared to water (\( 2.25 \times 10^8 \text{ m/s} \)). Optical density is inversely proportional to the speed of light.
(ii) The ray of light will pass undeviated. This is because when a ray of light is incident normally (\( \angle i = 0^\circ \)) on the interface of two media, the angle of refraction is also zero (\( \angle r = 0^\circ \)).

Question. The absolute refractive indices of water and glass are 4/3 and 3/2 respectively. If the speed of light in glass is \( 2 \times 10^8 \) m/s, find the speed of light in (i) vacuum and (ii) water.
Answer: (i) For glass: \( n_g = \frac{c}{v_g} \)
\( \implies \frac{3}{2} = \frac{c}{2 \times 10^8} \)
\( \implies c = \frac{3}{2} \times 2 \times 10^8 = 3 \times 10^8 \text{ m/s} \).
(ii) For water: \( n_w = \frac{c}{v_w} \)
\( \implies \frac{4}{3} = \frac{3 \times 10^8}{v_w} \)
\( \implies v_w = \frac{3 \times 3 \times 10^8}{4} = 2.25 \times 10^8 \text{ m/s} \).