Class 10 light chapter prasant kirad, Exams of Physics

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Aarambh Chapterwise Problems

LIGHTLIGHT

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1. When a plane mirror is rotated through a certain angle, the reflected ray turns through twice as much and the size of the image: (a) is doubled (b) is halved (c) becomes infinite (d) remains the same
2. If an object is placed symmetrically between two plane mirrors, inclined at an angle of 72 degrees, then the total no. of images formed is: (a) 5 (b) 4 (c) 2 (d) infinite
3. Which statement is true for the reflection of light? (a) The angle of incidence and reflection are equal. (b) The reflected light is less bright than the incident light. (c) The sum of the angle of incidence and reflection is always greater than 90°. (d) The beams of the incident light, after reflection, diverge at unequal angles.
4. The focal length of a plane mirror is: (a) 0 (b) infinite (c) 25 cm (d) -25 cm
5. A ray of light passes through the optical centre of a convex lens. How does the ray emerge after refraction? (a) It bends towards the principal axis (b) It bends away from the principal axis (c) It passes undeviated (d) It gets reflected back
6. Which of the following statements is not true in reference to the diagram shown above? (a) Image formed is real. (b) Image formed is enlarged. (c) Image is formed at a distance equal to double the focal length. (d) Image formed is inverted.
7. An object is placed at a distance of 40 cm in front of a concave mirror of a focal length of 20 cm. The image produced is: (a) virtual and inverted

LIGHTLIGHTLIGHTLIGHT

PART 1 ( Basic MCQs)

Read the question properly!

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(c) is equal to one (d) can be more than or less than one, depending upon the position of the object in front of it

1. A student conducts an activity using a concave mirror with a focal length of 10 cm. He placed the object 15 cm from the mirror. Where is the image likely to form? (a) At 6 cm behind the mirror (b) At 30 cm behind the mirror (c) At 6 cm in front of the mirror (d) At 30 cm in front of the mirror
2. The image of an object placed in front of a convex mirror is formed at: (a) the object itself (b) twice the distance of the object in front of the mirror (c) half the distance of the object in front of the mirror (d) behind the mirror
3. A full-length image of a distant tall building can definitely be seen using: (a) a concave mirror (b) a convex mirror (c) a plane mirror (d) both concave as well as plane mirrors
4. A student conducts an activity using a flask of height 15 cm and a concave mirror. He finds that the image formed is 45 cm in height. What is the magnification of the image? (a) -3 times (b) -1/3 times (c) 1/3 times (d) 3 times
5. Which of the following can make a parallel beam of light from a point source incident on it? (a) Concave mirror as well as convex lens (b) Convex mirror as well as concave lens (c) Two plane mirrors placed at 90° to each other (d) Concave mirror as well as concave lens
6. A student studies that the speed of light in air is 300000 km/sec, whereas the speed in a glass slab is about 197000 km/sec. What causes the difference in the speed of light in these two media? (a) Difference in density (b) Difference in temperature (c) Difference in the amount of light (d) Difference in the direction of wind flow

ASSERTION AND REASON TYPE QUESTION (1 Marks)

In the following questions a statement of Assertion is followed by a statement of Reason. Mark the correct choice as two statements are given one labeled Assertion (A) and the other labeled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below: (a) Both A and R are true, and R is correct explanation of the assertion. (b) Both A and R are true, but R is not the correct explanation of the assertion.

(yaha Marks katate h)

(c) A is true, but R is false. (d) A is false, but R is true.

1. Assertion (A): The centre of curvature is not a part of the mirror. It lies outside its reflecting surface. Reason (R): The reflecting surface of a spherical mirror forms a part of a sphere. This sphere has a centre.
2. Assertion (A): A ray passing through the centre of curvature of a concave mirror after reflection, is reflected back along the same path. Reason (R): The incident rays fall on the mirror along the normal to the reflecting surface.
3. Assertion (A): A ray of light changes its direction when it travels obliquely from one transparent medium to another. Reason (R): The speed of light changes when it enters from one transparent medium to another.
4. Assertion (A): The emergent ray is parallel to the direction of the incident ray. Reason (R): The extent of bending of the ray of light at the opposite parallel faces (air-glass interface and glass-air interface) of the rectangular glass slab is equal and opposite.
5. Assertion (A): A ray of light travelling from a rarer medium to a denser medium slows down and bends towards the normal. When it travels from a denser medium to a rarer medium, it speeds up and bends away from the normal. Reason (R): The speed of light is higher in a rarer medium than in a denser medium.
6. Assertion (A): The mirrors used in search lights are concave spherical. Reason (R): In concave spherical mirror the image formed is always virtual.
7. Assertion (A): Light travels faster in glass than in air. Reason (R): Glass is denser than air.
8. The magnification produced by a spherical mirror is -3”. List four pieces of information you obtain from this statement about the mirror/ image.

29.The linear magnification produced by a spherical mirror is +3. Analyse this value and state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror. Draw a ray diagram to show the formation of image in this case

1. The linear magnification produced by a spherical mirror is -1/5. Analysing this value state the (i) type of spherical mirror and (ii) the position of the object with respect to the pole of the mirror. Draw ray diagram to justify your answer.
2. A concave mirror is used for image formation for different positions of an object. What inferences can be drawn about the following when an object is placed at a distance of 10 cm from the pole of a concave mirror of focal length 15 cm? (a) Position of the image, (b) Size of the image, (c) Nature of the image Draw a labelled ray diagram to justify your inferences

PART 2 (Subjective Questions)

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SHORT ANSWER TYPE QUESTIONS (2 and 3 Marks)

1. A convex lens and a concave lens have the same focal length. Compare their behaviour when parallel rays of light fall on them. Also state the type of image formed by each lens.
2. An object placed 24 cm from a concave lens produces a virtual image 8 cm from the lens on the same side as the object. Calculate the focal length and power of the lens.
3. A rear-view mirror in a vehicle forms an image of magnification +0.25. Identify the type of mirror used and explain why this mirror is preferred over a plane mirror.
4. A student claims that a concave mirror can produce both real and virtual images, whereas a convex mirror can produce only virtual images. Justify the statement with suitable reasoning.
5. A lens has a power of –2.5 D. (i) Identify the type of lens. (ii) Calculate its focal length. (iii) State the nature of image formed when an object is placed in front of it.
6. (A) Complete the following ray diagram to show the formation of image:

(B) Mention the nature, position and size of the image formed in this case. (C) State the sign of the image distance in this case using the Cartesian sign convention.

1. (a) (i) Draw a ray diagram to show the path of the refracted ray in each of the following cases: (1) A ray of light incident on a concave lens parallel to its principal axis. (2) A ray of light directed towards the principal focus of a concave lens. (ii) A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position and size of the image formed. OR (b) (i) Draw a ray diagram to show the path of the reflected ray in each of the following cases: (1) A ray of light incident on a convex mirror parallel to its principal axis. (2) A ray of light directed towards the principal focus of a convex mirror. (ii) A 1.5 cm tall candle flame is placed perpendicular to the principal axis of a concave mirror of focal length 12 cm. If the distance of the flame from the pole of the mirror is 18 cm, use the mirror formula to determine the position and size of the image formed

Q54. Study the ray diagram given below and answer the questions that follow: (a) Is the type of lens used converging or diverging? (b) List three characteristics of the image formed. (c) In which position of the object will the magnification be - 1?

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1. Read the source below and answer the questions that follow:

The images shown above are of a specialised slide projector. Slides are small transparencies mounted in sturdy frames that are ideally suited for magnification and projection because they possess very high resolution and excellent image quality. The projector contains a tray in which the slides are inserted in a particular orientation so that the audience can view enlarged erect images on the screen. For this reason, the slides are inserted upside down in the projector tray. To show her students the images of insects that she investigated in the laboratory, Mrs. Iyer brought a slide projector. Her slide projector produced a 500 times enlarged and inverted image of a slide on a screen placed 10 m away. 1 .Based on the information given above, identify the type of lens used in the slide projector. 2 .If v represents image distance and u represents object distance, state the sign convention for both in this case with one reason. 3 .A slide projector has a convex lens of focal length 20 cm. The slide is placed upside down at a distance of 21 cm from the lens. Calculate the distance at which the screen should be placed from the lens so that the image is obtained in clear focus.

Q56. (A) (a) Observe the following diagram and compare (i) speed of light and (ii) optical density of the three media A, B, and C. Also give justification for your answer of any one of the two cases in terms of refractive indices of A, B, and C.

(b) Redraw the path of a ray of light through the three media, if the ray of light starting from medium A falls on the medium B: (i) Obliquely and the optical density of medium B is made more than that of A and C. (ii) The ray falls normally from medium A to medium B.

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Case Study/Source Based Question (5 Marks)

PART 3 (Advanced Questions)

1. Many optical instruments consist of a number of lenses. They are combined to increase the magnification and sharpness of the image. The net power (P) of the lenses placed in contact is given by the algebraic sum of the powers of the individual lenses P₁, P₂, P₃ ... as: P = P₁ + P₂ + P₃ + ... This is also termed as the simple additive property of the power of lenses, widely used to design lens systems of cameras, microscopes and telescopes. These lens systems can have a combination of convex lenses as well as concave lenses. (a) What is the nature (convergent/divergent) of the combination of a convex lens of power +4 D and a concave lens of power −2 D? (b) Calculate the focal length of a lens of power −2.5 D. (c) Draw a ray diagram to show the nature and position of an image formed by a convex lens of power +0.1 D, when an object is placed at a distance of 20 cm from its optical centre. OR (c) How is a virtual image formed by a convex lens different from that formed by a concave lens? Under what conditions do a convex lens and a concave lens form virtual images?
2. Hold a concave mirror in your hand and direct its reflecting surface towards the Sun. Direct the light reflected by the mirror on to a white cardboard held close to the mirror. Move the cardboard back and forth gradually until you find a bright, sharp spot of light on the board. This spot of light is the image of the Sun on the sheet of paper, which is also termed as the “Principal Focus” of the concave mirror.

(A) List two applications of a concave mirror. (B) If the distance between the mirror and the principal focus is 15 cm, find the radius of curvature of the mirror. (C) Draw a ray diagram to show the type of image formed when an object is placed between the pole and focus of a concave mirror. (D) An object 10 cm in size is placed at 100 cm in front of a concave mirror. If its image is formed at the same point where the object is located, find: (i) Focal length of the mirror (ii) Magnification of the image formed with sign as per Cartesian sign convention.

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Reflection by Plane Mirrors – Q1, Q2, Q3, Q Refraction Through Lenses – Q5, Q8, Q9, Q13, Q34, Q39, Q41, Q44, Q47, Q48, Q51, Q52, Q53, Q54, Q55, Q57, Q Reflection by Spherical Mirrors – Q6, Q7, Q10, Q11, Q12, Q14, Q15, Q16, Q17, Q18, Q19, Q28, Q29, Q30, Q31, Q33, Q36, Q37, Q38, Q42, Q43, Q45, Q49, Q50, Q58, Q Refraction of Light – Q20, Q23, Q24, Q25, Q32, Q35, Q40, Q46, Q56, Q Assertion & Reason Based Questions – Q21, Q22, Q23, Q24, Q25, Q26, Q Concave Mirror – Q6, Q7, Q15, Q18, Q28, Q30, Q31, Q36, Q38, Q42, Q43, Q45, Q50, Q58, Q Convex Mirror – Q10, Q11, Q14, Q16, Q17, Q37, Q49, Q50, Q Convex Lens – Q5, Q8, Q9, Q34, Q39, Q41, Q44, Q47, Q48, Q52, Q53, Q54, Q55, Q57, Q Concave Lens – Q13, Q47, Q51, Q Magnification – Q14, Q18, Q28, Q29, Q30, Q42, Q49, Q57, Q Mirror Formula & Numerical Problems – Q7, Q12, Q15, Q30, Q31, Q42, Q58, Q Lens Formula & Numerical Problems – Q8, Q9, Q44, Q48, Q51, Q53, Q55, Q57, Q Power of Lens – Q8, Q48, Q51, Q Real and Virtual Images – Q6, Q10, Q11, Q13, Q15, Q16, Q28, Q29, Q30, Q31, Q36, Q39, Q41, Q50, Q60, Q Applications of Mirrors and Lenses – Q14, Q17, Q26, Q37, Q49, Q52, Q55, Q Competency & HOTS Based Questions – Q36, Q37, Q38, Q39, Q40, Q41, Q42, Q43, Q44, Q45, Q46, Q47, Q48, Q49, Q50, Q51, Q52, Q53, Q54, Q55, Q56, Q57, Q58, Q59, Q60, Q

MULTIPLE CHOICE QUESTIONS [MCQ's] 1 .(d) 2 .(b) 3 .(a) 4 .(b) 5 .(c) 6 .(b) 7 .(d) 8 .(c) 9 .(a) 10 .(d)

SOLUTIONS

11 .(a) 12 .(b) 13 .(c) 14 .(a) 15 .(d) 16 .(d) 17 .(b) 18 .(d) 19 .(a) 20 .(a)

21 .(a) 22 .(a) 23 .(a) 24 .(a) 25 .(a) 26 .(c) 27 .(d)

SHORT ANSWER TYPE QUESTIONS:

1. Negative sign of magnification indicates that the image is real and inverted. Since the image is real and inverted, the mirror is concave and magnification of -3 indicates that the image is magnified.
2. Positive value of the magnification indicates that image is virtual and erect. (i) Since the image is magnified, the mirror is concave. (ii) The object is between pole and focus of the mirror as shown
3. (i) Concave mirror (ii) Object is placed beyond C.

1. No, I do not agree. A convex lens forms real and inverted images when the object is placed beyond the focal length.
2. Light bends towards the normal when entering glass from air because it moves from a rarer medium to a denser medium and slows down. It bends away from the normal when moving from glass to air because it moves from a denser medium to a rarer medium and speeds up.
3. The image cannot be obtained on a screen because it is a virtual image formed by apparent intersection of rays. The rays do not actually meet.
4. Given: v = -36 cm m = - Calculation: m = -v/u -3 = -(-36)/u -3 = 36/u u = -12 cm Using mirror formula: 1/f = 1/v + 1/u 1/f = 1/(-36) + 1/(-12) 1/f = (-1 - 3)/ 1/f = -4/ 1/f = -1/ f = -9 cm Answer: Object distance, u = -12 cm Focal length, f = -9 cm
5. Image formed is: Real and inverted Enlarged Beyond the centre of curvature If the object moves towards the focus, the image moves farther away and becomes more enlarged.
6. Image distance, v = +80 cm Magnification, m = - Because the image formed by a convex lens on a screen is real and inverted. Using magnification formula: m = v/u -2 = 80/u u = -40 cm Now using lens formula: 1/f = 1/v - 1/u 1/f = 1/80 - 1/(-40) 1/f = 1/80 + 1/ 1/f = 1/80 + 2/ 1/f = 3/

LONG ANSWER TYPE QUESTIONS:

Therefore, f = 80/3 = 26.7 cm Final Answer: The focal length of the convex lens is 26.7 cm.

1. The object should be placed between the pole and focus of the concave mirror, i.e., at a distance less than 20 cm from the mirror. Reason: When the object is placed between the pole and focus, the concave mirror forms a virtual, erect, and highly enlarged image behind the mirror.
2. When light travels from glass to air, it goes from a denser medium to a rarer medium. Its speed increases, so the ray bends away from the normal. This bending of light is called refraction.
3. When parallel rays fall on a convex lens, the rays converge at the principal focus. When parallel rays fall on a concave lens, the rays diverge and appear to come from the principal focus. Image formed by: Convex lens: Can form real and inverted images. Concave lens: Always forms virtual, erect, and diminished images.
4. Given: Object distance, u = -24 cm Image distance, v = -8 cm Using lens formula: 1/f = 1/v - 1/u 1/f = 1/(-8) - 1/(-24) 1/f = -1/8 + 1/ 1/f = (-3 + 1)/ 1/f = -2/ 1/f = -1/ Therefore, f = -12 cm Now, power of lens: P = 100/f in cm P = 100/- P = -8.33 D Final Answer: Focal length = -12 cm Power = -8.33 D
5. The mirror used is a convex mirror. Reason: The magnification is positive and less than 1 (+0.25), which is the property of a convex mirror. It is preferred over a plane mirror because: it provides a wider field of view, forms erect and diminished images, helps the driver see more traffic behind.

Lens Formula: 1/f = 1/v − 1/u 1/24 = 1/v + 1/16 ⇒ 1/v = 1/24 − 1/ ⇒ 1/v = (2 − 3)/ ⇒ 1/v = −1/48 ⇒ v = −48 cm Magnification: m = v/u = (−48)/(−16) = 3 Image Height: h′ = m × h = 3 × 4 = 12 cm

1. (a) We have used a converging lens. (b) The characteristics of the image formed: (i) It is real. (ii) It is inverted (iii) It is enlarged. (c) We get the magnification of object, m = - 1 at the position 2F1.

CASE STUDY/SOURCE BASED QUESTION:

1. 1. Lens used: A convex lens is used because it forms a real, inverted and enlarged image on the screen.
2. Sign convention: u = negative because the object is on the left side of the lens. v = positive because the real image is formed on the right side of the lens.
3. Calculation: Given: f = +20 cm, u = -21 cm Using lens formula: 1/f = 1/v - 1/u 1/20 = 1/v - 1/(-21) 1/20 = 1/v + 1/ 1/v = 1/20 - 1/ 1/v = 1/ v = 420 cm
4. a) From the diagram, the ray bends towards the normal when it enters medium B from medium A. So, medium B is optically denser than medium A. When the ray goes from medium B to medium C, it bends away from the normal. So, medium C is optically rarer than medium B. Therefore: Optical density: B > A and B > C Speed of light: Speed of light is least in medium B. Refractive index: nB > nA and nB > nC

(b)(i) If medium B is made optically denser than both A and C, then the ray will bend towards the normal while entering from A to B, and it will bend away from the normal while going from B to C.

(b)(ii) If the ray falls normally from medium A to B, it will pass undeviated. Reason: At normal incidence, the angle of incidence is 0°, so the ray does not bend during refraction.

1. (a) Focal Length: From observation 4, u = −40 cm, v = +40 cm. Lens formula: 1/f = 1/v − 1/u = 1/40 − 1/(−40) = 1/40 + 1/ = 2/40 = 1/ ∴ f = 20 cm (positive, convex lens) Reason: The data is consistent with a convex lens of focal length 20 cm, as verified across observations. (b) Magnification for Observation 3: u = −30 cm, v = +60 cm. m = v/u = 60/(−30) = − The magnification is −2 (inverted, twice the size). (c) Differences:

• Observation 1: u = −15 cm, v = −60 cm. m = v/u = −60/(−15) = 4 (positive, virtual, erect image, same side)
• Observation 2: u = −25 cm, v = +100 cm. m = v/u = 100/(−25) = − (negative, real, inverted image, opposite side) Differences: 1 .Observation 1: Virtual, erect image; Observation 2: Real, inverted image. 2 .Observation 1: Image on same side as object; Observation 2: Image on opposite side.

1. (i) Diminished image Mirror A will form a diminished image. For mirror A: Focal length, f = 20 cm Centre of curvature, 2f = 40 cm Object distance, u = 45 cm Since the object is placed beyond the centre of curvature, the image formed will be real, inverted and diminished.

ii) Properties of image in case 2 For mirror B: Focal length, f = 15 cm Centre of curvature, 2f = 30 cm Object distance, u = 30 cm So, the object is placed at the centre of curvature. Two properties of the image: 1 .Image is real and inverted. 2 .Image is of the same size as the object.

(iii) (A) Image formed by mirror C Focal length, f = 30 cm Object distance, u = 20 cm Since u < f, the object is placed between the pole (P) and focus (F) of the concave mirror. So, the image formed is: Virtual, Erect and Magnified

P = 100/f (cm) ⇒ 0.1 = 100/f ⇒ f = 1000 cm 1/f = 1/v − 1/u 1/1000 = 1/v − 1/(−20) 1/1000 = 1/v + 1/ Step 3: Solve for v 1/v = 1/1000 − 1/ Taking LCM (1000): 1/v = (1 − 50)/ 1/v = −49/ v = −1000/ v ≈ −20.4 cm Nature of Image: Since v is negative, the image is formed on the same side as the object. Therefore, the image is virtual and erect.

OR

(c) Virtual images formed by convex and concave lenses differ in several ways:

• Convex Lens: Forms a virtual image when the object is within the focal length. The image is erect, magnified and on the same side as the object.
• Concave Lens: Always forms a virtual image. The image is erect, diminished and on the same side as the object.

1. (A) Applications of concave mirrors: (1) A concave mirror is used as a shaving mirror when the face is placed close to it so that it is within its focus and we get an erect and magnified image of the face. (2) Doctors use concave mirrors as head mirrors to concentrate parallel rays of light at its focus, which enables them to examine body parts such as eyes, throat, etc. (B) Given, f = 15 cm We know for a mirror, R = 2f

R = 2 × 15 cm R = 30 cm (C)

(D)(i) We can use the mirror formula to find the focal length of the mirror: 1/f = 1/v + 1/u where, f is the focal length of the mirror, v is the image distance and u is the object distance. Since the image is formed at the same point as the object, v = u = −100 cm (Distances to the left of the mirror are negative.)

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Substituting the values, we get: 1/f = 1/(−100) + 1/(−100) 1/f = −2/ f = −50 cm So the focal length of the mirror is −50 cm. (Negative sign indicates that it is a concave mirror.) (ii) The magnification of the image is: m = −v/u where, m is the magnification of the image. Substituting the values, we get: m = −(−100/−100) m = − So the magnification of the image is −1. (Negative sign indicates that the image is real and inverted.)