Class 10 Science Chapter 9 Previous Year Questions - Ray Optics - Light : Reflection and Refraction
Previous Year Questions 2024
Q1: At what distance from a convex lens should an object be placed to get an image of the same size as that of the object on a screen? (2024)
(a) Beyond twice the focal length of the lens.
(b) At the principal focus of the lens.
(c) At twice the focal length of the lens.
(d) Between the optical centre of the lens and its principal focus.
View AnswerAns: (c)
To get an image of the same size as the object using a convex lens, the object should be placed at twice the focal length of the lens. This distance is called "twice the focal length" because at this position, the light rays coming from the object will converge to form an image that is equal in size.
Q2: An object is placed at a distance of 10 cm from a convex mirror of focal length 15 cm. Find the position of the image formed by the mirror. (2024)
View AnswerAns: u = - 10cm; f = +15 cm
Image is formed behind the mirror.
Q3: Source-based/case-based questions with 2 to 3 short subparts. Internal choice is provided in one of these sub-parts: (2024)
Study the data given below showing the focal length of three concave mirrors A, B and C and the respective distances of objects placed in front of the mirrors :
(i) In which one of the above cases the mirror will form a diminished image of the object ? Justify your answer.
(ii) List two properties of the image formed in case 2.
(iii) (A) What is the nature and size of the image formed by mirror C?
Draw ray diagram to justify your answer.
OR
(iii) (B) An object is placed at a distance of 18 cm from the pole of a concave mirror of focal length 12 cm. Find the position of the image formed in this case.
View AnswerAns: (i)
- Mirror A.
- as the object is placed beyond the centre of curvature of the mirror.
(ii) Same size/ Real / Inverted
(iii) (A) Nature-Virtual and erect Size-magnified
OR
(iii) (B) Here ƒ = –12 cm, u = –18 cm, n = ?
v = - 36cm
In front of the mirror at a distance of 36 cm from the pole of the mirror.
Q4: (a) State two laws of refraction of light.
OR
(b) Define the term absolute refractive index of a medium. A ray of light enters from vacuum to glass of absolute refractive index 1.5. Find the speed of light in glass. The speed of light in vacuum is 3 x 108 m/s. (2024)
View AnswerAns: Laws of Refraction of light :
(i) The incident ray, the refracted ray and the normal to the interface of two transparent media at the point of incidence, all lie in the same plane.
(ii) The ratio of the sine of angle of incidence to the sine of angle of refraction is a constant, for the light of a given colour and for the given pair of media.
OR
Absolute refractive index of a medium is the ratio of speed of light in air/vacuum to the speed of light in the given medium.
Given:
Absolute refractive index of a medium (nm)
Q5: The Phenomena of light involved in the formation of a rainbow in the sky are (2024)
(a) Refraction, dispersion and reflection
(b) Refraction, dispersion and total internal reflection
(c) Dispersion, scattering and reflection
(d) Dispersion, refraction and internal reflection
View AnswerAns: (d)
A rainbow forms in the sky due to the phenomena of dispersion, refraction, and internal reflection of light. When sunlight enters a raindrop, it bends (refraction), separates into different colors (dispersion), and reflects off the inside surface of the drop before coming back out, creating a rainbow.
Q6: Absolute refractive index of glass and water is 3/2 and 4/3 respectively. If the speed of light in glass is 2 x 108 m/s, the speed of light in water is: (2024)
(a) 
(b) 
(c) 
(d) 
View AnswerAns: (a)
To find the speed of light in water, we can use the formula for refractive index, which is the ratio of the speed of light in a vacuum to the speed of light in a medium. Given the refractive indices, we can calculate the speed of light in water using the speed of light in glass and the ratios. When we do the calculation, we find that the speed of light in water is 
Q7: (a) The variation of image distance (v) with object distance (u) for a convex lens is given in the following observation table. Analyse it and answer the questions that follow: (2024)
(i) Without calculation, find the focal length of the convex lens. Justify your answer.
(ii) Which observation is not correct ? Why? Draw ray diagram to find the position of the image formed for this position of the object.
(iii) Find the approximate value of magnification for u = - 30 cm.
OR
(b) (i) Define principal axis of a lens. Draw a ray diagram to show what happens when a ray of light parallel to the principal axis of a concave lens passes through it.
(ii) The focal length of a concave lens is 20 cm. At what distance from the lens should a 5 cm tall object be placed so that its image is formed at a distance of 15 cm from the lens? Also calculate the size of the image formed.
View AnswerAns: (a) (i) S. No. 3, 2f is 50 cm. ∴ 2f = 50 cm, or f = 25 cm.
Justification: Object distance(u) and image distance (v) are same so it implies that object is placed at 2F.
(ii) S. No. 6, is not correct.
Reason: For u = −15 cm, sign of v must be – ve ( as the image is formed on the same side of the lens as the object)
(iii) Magnification: m = v/u
OR
(b) (i) Principal axis: It is an imaginary line passing through the two centres of curvatures of a lens. 
(ii) f = − 20 cm; h = 5 cm; v = −15 cm
or u = − 60 cm object is at a distance of 60 cm from the lens
Size of the image(magnification):
Q8: How will the image formed by a convex lens be affected, if the upper half of the lens is wrapped with black paper? (2024)
(a) The size of the image formed will be one-half of the size of the image due to the complete lens.
(b) The image of the upper half of the object will not be formed.
(c) The brightness of the image will reduce.
(d) The lower half of the inverted image will not be formed.
View AnswerAns: (c)
If the upper half of a convex lens is wrapped in black paper, it blocks some light from passing through. As a result, the brightness of the image formed will reduce, but the size and shape of the image will remain the same since the lower half of the lens can still focus the light.
Q9: The phenomena of light involved in the formation of a rainbow are: (2024)
(a) Refraction, reflection, and dispersion
(b) Refraction, dispersion, and internal reflection
(c) Reflection, dispersion, and internal reflection
(d) Refraction, dispersion, scattering, and total internal reflection
View AnswerAns: (b)
A rainbow forms due to the processes of refraction, dispersion, and internal reflection of light in raindrops. When sunlight enters a raindrop, it bends (refraction), spreads out into different colors (dispersion), and reflects off the inside surface of the drop, resulting in the colorful arc of a rainbow.
Q10: Case-based/data-based questions with 3 short sub-parts. Internal choice is provided in one of these sub-parts. (2024)
A highly polished surface such as a mirror reflects most of the light falling on it. In our daily life we use two types of mirrors plane and spherical. The reflecting surface of a spherical mirrors may be curved inwards or outwards. In concave mirrors, reflection takes place from the inner surface, while in convex mirrors reflection takes place from the outer surface.
(a) Define the principal axis of a concave mirror.
(b) A ray of light is incident on a concave mirror, parallel to its principal axis. If this ray after reflection from the mirror passes through the principal axis from a point at a distance of 10 cm from the pole of the mirror, find the radius of curvature of the mirror.
(c) (i) An object is placed at a distance of 10 cm from the pole of a convex mirror of focal length 15 cm. Find the position of the image.
OR
(c) (ii) A mirror forms a virtual, erect and diminished image of an object. Identify the type of this mirror. Draw a ray diagram to show the image formation in this case.
View AnswerAns: (a) It is straight line passing through the pole and centre of curvature of a concave mirror.
(b) Radius of curvature ,R= 20 cm
(c) (i) u = -10 cm, f = +15 cm
OR
(c) (ii) Convex mirror / Diverging mirror
Q11: (a) (i) Draw a ray diagram to show the path of the refracted ray in each of the following cases: (2024)
A ray of light incident on a concave lens
(1) parallel to its principal axis, and
(2) is directed towards its principal focus.
(ii) A 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 24 cm. The distance of object from the lens is 16 cm. Find the position and size of image formed.
OR
(b) (i) Draw a ray diagram to show the path of the reflected ray in each of the following cases: A ray of light incident on a convex mirror
(1) parallel to its principal axis, and
(2) is directed towards its principal focus
(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 mirror formula to determine the position and size of the image formed.
View AnswerAns: (a) (i)
(1)
(ii) Given u = –16 cm, f = + 24 cm, h = 4 cm
v = – 48 cm
Image is formed on the same side as the object
OR
(b) (i)
(1)
(ii) Here f = – 12 cm, u = – 18 cm, v = ?, h = 1·5 cm, h' = ? 
Q12: The color of light for which the refractive index of glass is minimum, is: (2024)
(a) Red
(b) Yellow
(c) Green
(d) Violet
View AnswerAns: (a)
The refractive index of glass is lowest for red light, meaning that red light travels through glass the fastest compared to other colors. As a result, red light bends the least when it enters the glass, which is why it has the minimum refractive index.
Previous Year Questions 2023
Q1: The magnification produced when an object is placed at a distance of 20 cm from a spherical mirror is +1/2. Where should the object be placed to reduce the magnification to +1/3? (2023)
View AnswerAns: The object needs to be moved closer to the spherical mirror so that the magnification can be decreased from +1/2 to +1/3.
Assume that the image distance is v and the object distance is u. Magnification is calculated as follows: magnification = -v/u
Since the initial magnification is +1/2, we have: +1/2 = -v/20 cm
Solving for v, we get: v = -10 cm
Substituting this value of v into the magnification equation, we get,
+1/3 = -(-10)/u
Simplifying, we get: u = 30 cm
Therefore, the object should be placed at a distance of 30 cm from the spherical mirror to reduce the magnification to +1/3.
Q2: Define the following terms in the context of a diverging mirror: (2023)
(i) Principal focus
(ii) Focal length
Draw a labelled ray diagram to illustrate your answer.
View AnswerAns: For Diverging mirror (Convex Mirror)
(i) Principal focus: It is the point on the principal axis where rays incident parallel to the principal axis appear to diverge after reflection.
(ii) Focal length: The distance between the pole of the mirror and the principal focus is called focal length.
Q3: An object of height 10 cm is placed 25 cm away from the optical centre of a converging lens of focal length 15 cm. Calculate the image distance and height of the image formed. (2023)
View AnswerAns: Object height, O = +10 cm
Focal length, f = +15 cm
Object distance, u = -25 cm
Image distance, v = ?
Image height, I = ?
Using lens formula,
I = (-1.5) × 10 = 15 cm
Height of image is 15 cm, inverted.
Q4: 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 center of this lens, what will be the sign of magnification and nature of image in this case? (2023)
View AnswerAns: The ability of a lens to converge or diverge the rays of light is called power of lens. It is the reciprocal of focal length of lens.
Its Sl unit is dioptre.
If f = - 10 cm
So, lens is concave in nature . u = - 20 cm, f = - 10 cm .
As the object placed beyond F image is virtual and thus magnification is +ve.
Q5: The ability of 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. (2023)
i. 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 × 108 m/s.
ii. 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 θ), then write the increasing order of the angle of refraction in these media.
iii. (A) The speed of light in glass is 2 × 108 m/s and is water is 2.25 × 108 m/s.
(a) Which one of the two optically denser and why?
(b) 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.
OR
(B) The absolute refractive indices of glass and water are 4/3 and 3/2, respectively. If the speed of light in glass is 2 × 108 m/s, calculate the speed of light in (i) vacuum (ii) water.
View AnswerAns: (i) Refractive index of diamond,
= 1.25 × 108 m/s.
(ii) rwater < rglass <
(iii) (A) (a) Since the speed of light is greater in water than in glass, glass is optically denser than water. This demonstrates that glass presents a greater barrier to light transmission than water.
(b) After refraction, the light will bend towards normal.
OR
(B) Refractive index of glass, ηg = 4/3
Speed of light in vacuum 
Refractive index of water 
Speed of light in water = 1.73 x 108 m/s
Because the information provided is wrong, ideally the speed of light in vacuum is 3 × 108 m/s and the speed of light in water is 2.25 × 108 m/s.
The correct solution is
Refractive index of glass 
Speed of light in vacuum
Speed of light in water 
Speed of light in water = 2.25 x 108 m/s
Q6: Many optical instruments consists of a number of lenses. They are combined to increase the magnification and sharpness of the image. The net power (P) of the lenses places 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 lens, widely used to design lens systems of cameras, microscopes and telescopes. These lens systems can have a combination of convex lenses and also concave lenses. (2023)
(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 and a concave lens form virtual image?
View AnswerAns: (a) Given: P1 = 4 D, P2 = -2 D
P = P1 + P2 = 4D - 2D
So the lens is convergent in nature.
(b) P = -2.5 D
(c) P = 0.1 D, u = -20 cm
Previous Year Questions 2022
Q1: An optical device forms an erect image of an object placed in front of it. If the size of the image is one half that of the object, the optical device is a (2022)
(a) concave mirror
(b) convex mirror
(c) plane mirror
(d) convex lens.
View AnswerAns: (b)
Sol: The image formed by a convex mirror is always erect and of smaller in size than object.
Q2: The relation R = 2f is valid (2022)
(a) for concave mirrors but not for convex mirrors
(b) for convex mirrors but not for concave mirrors
(c) neither for concave mirrors nor for convex mirrors
(d) for both concave and convex mirrors.
View AnswerAns: (d)
Sol: It is valid for both concave mirrors and convex mirrors.
Q3: In which of the following is a concave mirror used? (2022)
(a) A solar cooker
(b) A rear view mirror in vehicles
(c) A safety mirror in shopping malls
(d) In viewing full size image of distant tall buildings
View AnswerAns: (a)
Sol: Concave mirrors are the mirrors best suited ir solar cookers because concave mirrors are convergent mirror and they are reflect sunlight towards a single foca point.
Q4: For the diagram shown, according to the new Cartesian sign convention the magnification of the image formed will have the following specifications : (2022)
(a) Sign - Positive, Value - Less than 1
(b) Sign - Positive, Value - More than 1
(c) Sign - Negative, Value - Less than 1
(d) Sign - Negative, Value - More than 1
View AnswerAns: (b)
Sol: Magnification : Sign-positive, value-more the 1 because the object is placed between the focus and the pole. So, magnified image will be formed on other side of mirror. Hence, magnification of image formed will have i positive sign and value more than one.
Q5: The radius of curvature of a converging mirror is 30 cm. At what distance from the mirror should an object be placed so as to obtain a virtual image? (2022)
(a) Infinity
(b) 30 cm
(c) Between 15 cm and 30 cm
(d) Between 0 cm and 15 cm
View AnswerAns: (d)
Radius of curvature of a converging mirror, R = 30 cm
Focal Length, f = 30/2 cm = 15 cm
Thus, virtual image can be obtained from the mirror if an | object is placed between pole and focus, i.e., between 0 i cm and 15 cm.
Q6: Which of the following statements is not true in reference to the diagram shown above? (2022)
(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.
View AnswerAns: (b)
Sol: Image formed is enlarged is not true. When object is placed at C, image formed is real, inverted and of same size as object.
Q7: An object of height 4 cm is kept at a distance of 30 cm from the pole of a diverging mirror. If the focal length of the mirror is 10 cm, the height of the image formed is (2022)
(a) +3.0 cm
(b) +2.5 cm
(c) +1.0 cm
(d) +0.75 cm
View AnswerAns: (c)
Sol: Given, height of object (h) = +4 cm
Object distance (u) = -30 cm (object placed left side of the mirror)
Focal length, f = +10 cm
Mirror Formula,
1/f = 1/v + 1/u or v = uf/u - f
Now, Magnification (m) = 
or h' = 1 cm
Hence, height of the image formed is 1 cm.
Q8: If a lens and a spherical mirror both have a focal length of -15 cm, then it may be concluded that (2022)
(a) both are concave
(b) the lens is concave and the mirror is convex
(c) the lens is convex and the mirror is concave
(d) both are convex.
View AnswerAns: (a)
Sol: As the focal length of a concave mirror and a concave lens is taken as negative, both are concave in nature.
Q9: A student determines the focal length of a device’ A’ by focusing the image of a far off object on a screen placed on the opposite side of the object. The device 'A’ is (2022)
(a) concave lens
(b) concave mirror
(c) convex lens
(d) convex mirror.
View AnswerAns: (c)
Sol: If the rays are travelling from far off distance and focused on opposite side of the lens, this is only possible in convex lens.
Q10: When light is incident on a glass slab, the incident ray, refracted ray and the emergent ray are in three media A, B and C. If n1, n2 and n3 are the refractive indices of A, B and C respectively and the emergent ray is parallel to the incident ray, which of the following is true ? (2022)
(a) n1 < n2 < n3
(b) n1 > n2 > n3
(c) n1 < n2 = n3
(d) n3 = n3 < n2
View AnswerAns: (d)
Sol: Here, medium A and C is same and 8 is glass. n1 = n3 < n2
Q11: The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. According to new cartesian sign convention, if the image is three times the size of the flame, then the lens is (2022)
(a) concave and magnification is +3
(b) concave and magnification is -3
(c) convex and magnification is -3
(d) convex and magnification is +3.
View AnswerAns: (c)
Sol: The image of the candle flame is three times larger than the flame itself, which means the magnification is -3 according to the new Cartesian sign convention, where a negative sign indicates that the image is inverted. Since this type of magnification occurs with a convex lens, the correct answer is that the lens is convex and the magnification is -3.
Q12: The power of a combination of two lenses in contact is +1.0 D. If the focal length of one of the lenses of the combination is +20.0 cm, the focal length of the other lens would be (2022)
(a) -120.0 cm
(b) +80.0 cm
(c) -25.0 cm
(d) -20.0 cm
View AnswerAns: (c)
Sol: Given, combined power = +1 D
Focal length of one lens, f1 = 20 cm
Combined focal length, fcombined = 1 m = 100 cm
Solving this equation, we get f2 = -25 cm Focal length of other lens = -25 cm
Q13: Study the diagram given below and identify the type of the lens XX' and the position of the point on the principal axis OO' where the image of the object AB appears to be formed (2022)
(a) Concave; between O' and Y
(b) Concave : between O and Y
(c) Convex; between O' and Y
(d) Convex; between O and Y
View AnswerAns: (b)
Sol: As the ray after passing XX 'is diverging therefore, XX' is concave lens and image is formed between O and Y.
Q14: An object of height 3.0 cm is placed vertically on the principal axis of a convex lens. When the object i distance is -37.5 cm, an image of height -2.0 cm j is formed at a distance of 25.0 cm from the lens. I Next, the same object is placed vertically at 25.0 cm | from the lens. In this situation the image distance v and height h of the image is (according to the new j Cartesian sign convention) (2022)
(a) v = +37.5 cm; h = +4.5 cm
(b) v = -37.5 cm; h = +4.5 cm
(c) v = +37.5 cm; h = - 4.5 cm
(d) v = -37.5 cm; h = -4.5 cm
View AnswerAns: (c)
Sol: Given, object height = 3.0 cm
Let object distance is u and image distance is v.
Case-1: u = -37.5 cm and v = 25 cm
h = -2 cm (real and inverted)
From the lens formula, 
f = +15 cm ... (i)
Q15: An object is placed in front of a concave lens. For all positions of the object the image formed is always (2022)
(a) Real, diminished and inverted
(b) Virtual, diminished and erect
(c) Real, enlarged and erect
(d) Virtual, erect and enlarged.
View AnswerAns: (b)
Sol: The image formed by of a concave lens is always virtual, diminished and erect.
Q16: A ray of light starting from air passes through j medium A of refractive index 1.50, enters medium B of refractive index 1.33 and finally enters medium C of refractive index 2.42. If this ray emerges out in air from C, then for which of the following pairs of media the bending of light is least? (2022)
(a) air-A
(b) A-B
(c) B-C
(d) C-air
View AnswerAns: (b)
Sol: As refractive index of a medium increases, more bending of light takes place. For A - B, ratio of refractive indices of A and B is least, so least bending of light takes place for this pair of media.
Q17: A ray of light is incident as shown. If A, B and C are three different transparent media, then which among the following options is true for the given diagram? (2022)
(a) ∠1 > ∠4
(b) ∠1< ∠2
(c) ∠3 = ∠2
(d) ∠3 > ∠4
View AnswerAns: (c)
Sol: In A, B and C transparent media, ∠1 > ∠2 as light bends towards the normal and ∠3 < ∠4 as light bends away from the normal. ∠3 = ∠2 , because of alternate angles between two normals.
Q18: In the diagram shown above n1, n2 and n3 are refractive indices of the media 1, 2 and 3 respectively. Which one of the following is true in this case. (2022)
(a) n1 = n2
(b) n1 > n2
(c) n2 > n3
(d) n3 > n1
View AnswerAns: (d)
Sol:
In medium 2, the light ray bends towards the normal, so n2 > n1.
Similarly, in medium 3, the light ray bends more towards the normal which indicate that refractive index of medium 3 is greater than medium 2. So, n3 > n1
Q19: The refractive index of medium A is 1.5 and that of medium B is 1.33. If the speed of light in air is 3 x 108 m/s, what is the speed of light in medium A and B respectively? (2022)
(a) 2 x 108 m/s and 1.33 x 1 08 m/s
(b) 1.33 x 108 m/s and 2 x 108 m/s
(c) 2.25 x 108 m/s and 2 x 108 m/s
(d) 2 x 108 m/s and 2.25 x 108 m/s
View AnswerAns: (d)
Sol: Given, refractive index of medium A, μA = 1.5
Refractive index of medium B, μB = 1.33
Speed of light in air, c = 3 x 108 m/s
VA = 2 x 108 m/s
VB = 2.25 x 108 m/s
Hence, speed of light in medium A is 2 x 108 m/s and in medium B is 2.25 x 108 m/s.
Q20: A student wants to obtain magnified image of an object AB as on a Screen. Which one of the following arrangements shows the correct position of AB for him/her to be successful? (2022)
(a) 
(b) 
(c) 
(d) 
View AnswerAns: (c)
Sol: To get real & magnified image, object should be kept either between 2F and F or at F. If it is kept at F, image will form at infinity which cannot be taken on screen, so object should be kept between 2F and F.
Q21: The following diagram shows the use of an optical device to perform an experiment of light. As per the arrangement shown, the optical device is likely to be a (2022)
(a) concave mirror
(b) concave lens
(c) convex mirror
(d) convex lens
View AnswerAns: (b)
Sol: As per the arrangement, an optical device to perform an experiment of light is likely to be a concave lens.
Q22: If a lens can converge the sun rays at a point 20 cm j away from its optical centre, the power of this lens is (2022)
(a) +2D
(b) -2D
(c) +5D
(d) -5D
View AnswerAns: (c)
Sol: Converging point, f = 20 cm
Power P = ?
As we know that,
Hence, power of convex lens is +5 D.
Q23: A converging lens forms a three-times magnified image of an object, which can be taken on a screen. If the focal length of the lens is 30 cm, then the distance of the object from the lens is (2022)
(a) -55 cm
(b) -50 cm
(c) -45 cm
(d) -40 cm
View AnswerAns: (d)
Sol: –40 cm
Reason — Given, convex lens as converging lens
f = 30 cm
From formula,
m = v/u
-3 = v/u [-ve sign as real and inverted image is formed.]
v = -3u
Using formula,
1/f = 1/v - 1/u
We get,
Hence, distance of the object from the lens = -40 cm.
Previous Year Questions 2021
Q1: The refractive index of glass is 1.50. What is the meaning of this statement? (2021)
View AnswerAns: It gives us the idea about the speed of light in the air and in the glass. It means that speed of light is 1.5 time more in air than the speed of light in the glass.
Previous Year Questions 2020
Q1: Define pole of a spherical mirror. (2020)
View AnswerAns: The pole of a spherical mirror define the geometrical center of the spherical surface of the mirror. It is the center of reflecting surface of spherical mirror and lies on the surface of spherical mirror.
Q2: The refractive index of a medium 'x' with respect to a medium 'y' is 2/3 and the refractive index of medium 'y' with respect to medium 'z' is 4/3. Find the refractive index of medium 'z' with respect to medium 'x'. If the speed of light in medium 'x' is 3 x 108 m s-1, calculate the speed of light in medium 'y. (2020)
View AnswerAns: Given, refractive index of medium x with respect to y, 
Refractive index of medium y with respect to z,
∴ Refractive index of medium x with respect to z,
∴ Refractive index of medium z with respect to x,
Now speed of light in x = 3 × 108 m/s
Speed of light in y, vy = ?
∵ 
Q3: Study the ray diagram given below and answer the questions that follow: (2020)
(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?
View AnswerAns: (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.
Q4: Rishi went to a palmist to show his palm. The palmist used a special lens for this purpose. (2020)
(i) State the nature of the lens and reason for its use.
(ii) Where should the palmist place/hold the lens so as to have a real and magnified image of an object?
(iii) If the focal length of this lens is 10 cm, the lens is held at a distance of 5 cm from the palm, use lens formula to find the position and size of the image. (2020)
View AnswerAns: (i) The lens used here is a convex lens and it is used as a magnifying glass because at close range, i.e., when the object is placed between optical centre and principal focus it forms an enlarged, virtual and erect image of the object.
(ii) When this lens is placed such that the object is between the centre of curvature and the principal focus, the palmist obtain a real and magnified image.
(iii) Given, focal length, f = 10 cm and u = -5 cm According to lens formula,
∴
Thus, the image will be formed at 10 cm on the same side of the palm and the size of the image will be enlarged.
Previous Year Questions 2019
Q1: How far should an object be placed from a convex lens of focal length 20 cm to obtain its real image at a distance of 30 cm from the lens? Determine the height of the image if the object is 4 cm tall. (2019)
View AnswerAns: Focal length of convex lens (f) = 20 cm
Real image formed at a distance, (v) = 30 cm
Height of image (h1) = 4 cm
Let the object distance be u.
Lens formula, 
; u = - 60 cm .From,
Hence, height of image of object is 2 cm.
Q2: State laws of reflection of light. List four characteristics of the image formed by a plane mirror. (2019)
View AnswerAns: The laws of reflection of light state that:
- The incident ray, the reflected ray and the normal to the surface at the point of incidence all lie in the same plane.
- The angle of incidence (i) is always equal to the angle of reflection (∠r) i.e. ∠i = ∠r
Characteristics of the image formed by a plane mirror
- Image formed in a plane mirror is virtual and erect.
- It is the same size as the object.
- The image formed is laterally inverted.
- The image formed is at same distance behind the mirror as object is in front of mirror.
Q3: A student, holding a mirror in his hand, directed the reflecting surface of the mirror towards the sun. He then directed the reflected light on to a sheet of paper held close to the mirror. (2019)
(a) What should he do to burn the paper?
(b) Which type of mirror does he use?
(c) Will he be able to determine the approximate value of focal length of this mirror from this activity ? Give reason and draw ray diagram to justify your answer in this case.
View AnswerAns:
(a) The student should adjust the distance between the mirror and the paper so that solar rays are sharply focussed on the paper.
(b) The mirror is a concave mirror.
(c) The student can find the approximate focal length by measuring the distance between the paper and the mirror.
As shown in Fig. 10.29, parallel rays from the sun are focussed on the paper at point A' in focal plane of mirror such that PB' = f.
Q4: A real image, 2/3 rd of the size of an object, is formed by a convex lens when the object is at a distance of 12 cm from it. Find the focal length of the lens. (2019)
View AnswerAns: Here distance of the object from the lens u = - 12 cm and magnification of real image m = -2/3
As per relation m = v/u, we have v = mu =(-2/3) x (-12) = + 8 cm
So as per lens formula 
we have
cm
Q5: Draw a ray diagram to show refraction through a rectangular glass slab. How is the emergent ray related to incident ray ? What is its lateral displacement ? (2019)
View AnswerAns: A ray diagram showing refraction through a rectangular glass slab has been shown in adjoining Fig.
The emergent ray GH is exactly parallel to the incident ray EFNM. It means that ∠r2 = ∠i1.
However, the emergent ray is laterally (side ways) displaced as compared to the original path of light ray. In ray diagram, the lateral displacement is GN. Its value increases on increasing the width of glass slab.
Q6: An object is placed at a distance of 60 cm from a concave lens of focal length 30 cm. (2019)
(i) Use the lens formula to find the distance of the image from the lens.
(ii) List four characteristics of the image (nature, position, size, erect/inverted) formed by the lens in this case.
(iii) Draw a ray diagram to justify your answer of the part (ii).
View AnswerAns: We have, (i) Object distance, u= —60 cm Focal length of the concave lens, f= —30 cm Using lens formula,
v = -20 cm
The image will be formed at a distance of 20 cm in front of the lens.
(ii) Nature of the image is virtual. The position of the image is between F1 and optical center 0. Size of the image is diminished. The image is Erect.
(iii)
Q7: (a) List four characteristics of the image formed by a convex lens when an object is placed between its optical centre and principal focus.
(b) Size of the image of an object by a concave lens of focal length 20 cm is observed to be reduced to 1/3 rd of its size. Find the distance of the object from the lens. (2019)
View AnswerAns: (a) When an object is placed between the optical centre and principal focus of a convex lens, the image formed is virtual, erect and enlarged. Moreover, the image is formed on the same side of lens behind the object.
(b) Here magnification of given concave lens m = 1/3 and focal length of lens f = - 20 cm
As per relation m = v/u for a lens, we get
Therefore, as per sign convention followed, both u and v are -ve.
Using lens formula 
we have
So the object is placed at a distance of 40 cm from the lens.
Previous Year Questions 2018
Q1: State the laws of refraction of light. Explain the term ‘absolute refractive index of a medium’ and write an expression to relate it with the speed of light in vacuum. (2018)
View AnswerAns: Laws of refraction: a. The incident ray, refracted ray and normal to the point of incidence all lie in the same plane.
b. The ratio of sin of incident angle to sin of angle of refraction for a given pair of medium is constant.
Absolute refractive index of a medium is the ratio of speed of light in air or vacuum to speed of light in the medium.
Absolute refractive index
Q2: What is meant by power of a lens? Write its SI unit. A student uses a lens of focal length 40 cm and another of -20 cm. Write the nature and power of each lens. (2018)
View AnswerAns: Power of lens is the ability of a lens to converge or diverge light rays passing through it. It is the reciprocal of the focal length. S.I. Unit: If focal length is measured in metre then unit of power of a lens is Dioptre (D)
Power of first lens: Focal length=+40 cm. Focal length is positive hence it is a convex lens.
Power of second lens
Focal length = - 20 cm
Its focal length is negative hence it is a concave lens.
Q3: An object of height 4.0 cm is placed at a distance of 30 cm from the optical centre ‘O’ of a convex lens of focal length 20 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre ‘O’ and principal focus ‘F’ on the diagram. Also find the approximate ratio of size of the image to the size of the object. (2018)
View AnswerAns: Ray diagram: Position of O and F
hi = - 8 cm
Ratio = hi/ho is approximately 2:1
Previous Year Questions 2017
Q1: If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it ? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror? (2017)
View AnswerAns: Only a convex mirror always form an erect and diminished image behind the mirror between its pole and focus point for all positions of the object placed in front of the mirror.
A ray diagram showing the image formation of an object AB is shown here . A convex mirror is used as a rear view mirror in automobiles because it gives erect and diminished images of vehicles coming from behind. As a result, it helps the driver in having a much wider field of view.
Q2: An object 4 cm in height, is placed at 15 cm in front of a concave mirror of focal length 10 cm. At what distance from the mirror should a screen be placed to obtain a sharp image of the object. Calculate the height of the image. (2017)
View AnswerAns: Here distance of object u = - 15 cm, height of object h = +4 cm and the focal length of concave mirror f = - 10 cm.
As per mirror formula
we have
⇒ u = -30 Cm.
Thus, a screen be placed infront of mirror at a distance of 30 cm from it.
Thus, image is an inverted image of height 8 cm.
Q3: An object is placed at a distance of 15 cm from a concave lens of focal length 30 cm. List four characteristics (nature, position, etc.) of the image formed by the lens. (2017)
View AnswerAns: Here, object distance, u = -15 cm
Using lens formula,
Magnification, m = u/v = -10/-15 = +23
Four characteristics of the image formed by the concave lens are :
(i) Image formed is virtual
(ii) Image is erect
(iii) Image is formed on the same side of the lens as the object
(iv) Image is smaller than the object
Q4: Give any two applications of a concave and convex mirror. (2017)
View AnswerAns: (i) Concave mirrors:
- Concave mirrors are used while applying make–up or shaving, as they provide a magnified image.
- They are used in torches, search lights, and head lights as they direct the light to a long distance.
(ii) Convex mirrors:
- Convex mirrors are used in vehicles as rearview mirrors because they give an upright image and provide a wider field of view as they are curved outwards.
- They are found in the hallways of various buildings including hospitals, hotels, schools, and stores. They are usually mounted on a wall or ceiling where hallways make sharp turns.
Q5: Define power of a lens. (2017)
View AnswerAns: The ability of lens to converge or diverge the ray of light is called power of lens. It is equal to the reciprocal of focal length, i.e. P = 1/f
Q6: The magnification of an image formed by a lens is -1. If the distance of the image from the optical centre of the lens is 25 cm, where is the object placed? Find the nature and focal length of the lens. If the object is displaced 15 cm towards the optical centre of the lens, where would the image be formed? Draw a ray diagram to justify your answer. (2017)
View AnswerAns: For real image, m = - 1
Therefore, m = v/u = -1
or u = -v, v = -u, - u = v = — 25
Using lens formula,
1/f = 1/v - 1/u
= 1/25 - 1/-25 = 2/25 = 1/12.5
f = +12.5 cm
Thus, the positive focal length shows that the given lens is a convex lens of focal length 12.5 cm. If the object is now displaced 15 cm towards the optical centre of the lens i.e, object is now placed at a distance of 25 - 15 = 10 cm from the optical centre.Therefore u = - 10 cm , and f = +12.5 cm.
Using lens formula again,
1/f = 1/v - 1/u
or
1/v = 1/f + 1/u
= 2/25 + 1/-10
= 2/25 - 1/10 = -1/50
v = -50 So, in this case, virtual image is formed on the same side of the object at a distance of 50 cm from the optical centre o f the lens as shown in figure
Q7: If the image formed by a lens for all positions of an object placed in front of it is always erect and diminished, what is the nature of this lens? Draw a ray diagram to justify your answer. If the numerical value of the power of this lens is 10 D, what is its focal length in the Cartesian system? (2017)
View AnswerAns: It is a concave or diverging lens.
f = 1/P
P = - 10 D,
f = 1/-10D = -0.1 m
Or f = -10cm.
Q8: Define the term magnification as referred to spherical mirrors. If a concave mirror forms a real image 40 cm from the mirror, when the object is placed at a distance of 20 cm from its pole, find the focal length of the mirror. [Delhi 2017 C]
View AnswerAns: Magnification of spherical mirror (m): It is equal to the ratio of size (height) o f the image to the size (height) of the object. Thus,
m = Size of Image (h2)/Size of Object (h1)
Given: For concave mirror u = - 20 cm,v = - 40 cm,
Using mirror equation,
1/f = 1/v + 1/u
or
1/f = 1/-40 + 1/-20
= -(1/40 + 1/20)
= -3/40
⇒ f = - 40/3
Q9: State Snell's law of refraction of light. Express it mathematically. Write the relationship between absolute efractive index of a medium and speed of light in vacuum. [AI 2017 C]
View AnswerAns: Snell’s law: The ratio of sine of angle of incidence (i.e. sin i) to the sine of angle of refraction (i.e. sin r) is always constant for the light of given colour and for the given pair of media.
Mathematically, sin i/ sin r = constant = n21
The constant n21 is called refractive index of the second medium with respect to the first medium.
Absolute refractive index of the medium is given by nm = Speed of light in vacuum (c)/ Speed of light in medium (v) = c/v
Q10: (a) What is the minimum number of rays required for locating the image formed by a concave mirror for an object? Draw a ray diagram to show the formation of a virtual image by a concave mirror.
(b) The linear magnification produced by a spherical mirror is +3. Analyze this value and state the (i) type of mirror and (ii) position of the object with respect to the pole of the mirror. Draw ray diagram to show the formation of image in this cased
(c) An object is placed at a distance of 30 cm in front of a convex mirror of focal length 15 cm. Write four characteristics of the image formed by the mirror. (2017)
View AnswerAns: (a) Two rays are required.
(b) The linear magnification produced by a spherical mirror is +3. It shows that the size of image is three times the size of object, image is virtual and erect and formed behind the mirror.
Hence
(i) the mirror is a concave mirror, and (ii) the object is placed between the pole and the focus of a concave mirror.
(c) The four characteristics of the image formed by the convex mirror are virtual, erect, diminished and laterally inverted.
Q11: (a) To construct a ray diagram we use two rays which are so chosen that it is easy to know their directions after reflection from the mirror. List two such rays and state the path of these rays after reflection in case of concave mirrors. Use these two rays and draw ray diagram to locate the image of an object placed between pole and focus of a concave mirror.
(b) A concave mirror produces three times magnified image on a screen. If the object is placed 20 cm in front of the mirror, how far is the screen from the object? (2017)
View AnswerAns: (a) Rays which are chosen to construct a ray diagram for reflection are: O') A ray parallel to the principal axis and A ray passing through the centre of curvature of a concave mirror.
Path of these light rays after reflection:
(i) It will pass through the principal focus of a concave mirror
(ii) It gets reflected back along the same path. When an object is placed between the pole and the principal focus of a concave mirror, a virtual, erect and enlarged image is formed behind the concave mirror as shown in the figure.
(b) Given: u = — 20 cm and m = 3Magnification, m, is given by m = v/-u
∴ v = - m x u= - (- 3) (- 20 cm) = - 60 cm
Distance between the object and the screen is= - 60 cm - (- 20 cm)= - 40 cm.
Q12: 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: (2017) 
(a) What is the focal length of the convex lens? Give reason to justify your answer.
(b) Write the serial number of the observation which is not correct. On what basis have you arrived at this conclusion?
(c) Select an appropriate scale and draw a ray diagram for the observation at S.No. 2. Also find the approximate value of magnification.
View AnswerAns: (a) From the observation 3, the radius of curvature of the lens is 40 cm as distance of object and the distance of the image is same.
We know, focal length, f = R/2 = 40/2 = 20 cm
(b) S. No. 6 is not correct, because for this observation the object distance is between focus and pole and for such cases, the image formed is always virtual. But in this case real image is formed as the image distance is positive.
From the figure, object distance u = - 60 cm and image distance v = 30 cm.
We know, magnification = v/u = +30/-60 = -0.5
Q13: (a) If the image formed by a mirror for all positions of the object placed in front of it is always diminished, erect and virtual, state the type of the mirror and also draw a ray diagram to justify your answer. Write one use such mirrors are put to and why?
(b) Define the radius of curvature of spherical mirrors. Find the nature and focal length of a spherical mirror whose radius of curvature is +24 cm. (2017)
View AnswerAns: (a) Convex (diverging) mirror
Reason: (i) It always produces a virtual and erect image.
(ii) The size of image formed is smaller than the object.
Therefore, it enables the driver to see a wide field view of the traffic behind the vehicle in a small mirror.
(b) Radius of Curvature: The separation between the pole and the centre of curvature or the radius of the hollow sphere, of which the mirror is a part, is called radius of curvature (R), i.e., PC = R.
Since focal length of the mirror is +24 cm. It indicates that nature of the given spherical mirror is convex/diverging mirror.
As R = 2f = 24 cm
Therefore, f = +12 cm
Q14: (a) Draw labelled ray diagrams for each of the following cases to show the position, nature and size of the image formed by a convex lens when the object is placed. (2017)
(i) between its optical center (O) and principal focus (F)
(ii) between F and 2 F
(b) How will the nature and size of the image formed in the above two cases, (i) and (ii) change, if the convex lens is replaced by a concave lens of same focal length?
View AnswerAns: (a) A convex lens of focal length 'f' can form
(i) a magnified and erect image only when the object is placed between its focus 'F' and optical centre ‘O’ of the lens.
(ii) a magnified and inverted image when an object is placed in the following positions: Between F1 and 2F1
(b) Whatever be the position o f object as given in case (i) and (ii),the image formed by the concave lens is always virtual, erect and diminished.
Q15: State the laws that are followed when light is reflected by spherical mirrors. Draw a ray diagram to show the formation of image of an object placed in front of a convex mirror. List two characteristics of the image formed. Briefly explain one use of convex mirrors. (2017)
View AnswerAns: Laws of reflection
(i) The incident ray, reflected ray and normal to the reflecting surface at the incident point all lie in the same plane.
(ii) The angle of reflection is equal to the angle of incidence.
(1) The convex mirror always form virtual and erect images.
(2) The size of the image is always lesser than the object.
Use of convex mirror-
This mirror is used in vechiles mirror to see the nearer object to vechicle.
Previous Year Questions 2016
Q1: What is a prism? (2016)
View AnswerAns: A prism is an optical device with two triangular bases along with three rectangular lateral surfaces commonly inclined at an angle of 60°.
Q2: Define the term reflection. (2016)
View AnswerAns: The bouncing back of a ray of light in the same medium after striking on a surface of an object.
Q3: The nature, size and position of image of an object produced by a lens or mirror are as shown below. Identify the lens/ mirror (X) used in each case and draw the corresponding complete ray diagram, (size of the object about half of the image). (2016) 
View AnswerAns: a. Convex lens when object is in between F and C (2F).
b. Concave mirror when object is in between P and F its enlarged, erected and virtual image is formed.
Q4: (a) Calculate the distance at which an object should be placed in front of a convex lens of focal length 10 cm to obtain a virtual image of double its size.
(b) In the above given case, find the magnification, if image formed is real. Express it in terms of relation between v and u (2016)
View AnswerAns: Given f = + 10 cm, u = ?
For virtual image
m = + 2
As
m = v/u or v/u = 2
v = 2u ...(1)
1/v = 1/u = 1/f ...(2)
Substituting (1) in (2)
u = - 5 cm
For real image, f = 10 cm, m = - 2
v/u = - 2, v = - 2u
-3/2u = 1/10 or u = - 15 cm
Q5: One half of a convex lens is covered with a black paper. (2016)
a. Show the formation of image of an object placed at 2Fp of such covered lens with the help of ray diagram. Mention the position and nature of image.
b. Draw the fay diagram for same object at same position in front of the same lens, but now uncovered. Will there be any difference in the image obtained in the two cases? Give reason for your answer.
View AnswerAns: a. If the lower half of the lens is covered even then it will form a complete real, inverted image of same size at C2(2F2) with reduced intensity of image.
b. There will be no change in the nature and position of the object except in later case the image will be brighter.
Q6: State the relation between object distance, image distance and focal length of a concave or convex mirror. A concave mirror produces two times magnified real image of an object at 10 cm from it. Find the position of the image. (2016)
View AnswerAns: For concave or convex mirrors the relation between u, v and f is given by
mirror formula,
m = - 2
u = - 10 cm
m = - v/u = - 2 or v = 2u = -20 cm
v = - 20 cm
Previous Year Questions 2015
Q1: What is the magnification of the images formed by plane mirrors and why? (2015)
View AnswerAns: Magnification of images formed by plane mirrors is unity because for plane mirrors, the size of the image formed is equal to that of the object.
Q2: What is meant by power of a lens? (2015)
View AnswerAns: Power is the degree of convergence or divergence of light rays achieved by a lens.
It is defined as the reciprocal of its focal length. i.e, P = 1/f
Q3: Name the mirror that is used by a dentist in examining teeth. (2015)
View AnswerAns: Dentist uses a concave mirror to See large images of the teeth of patients.
Q4: What is lateral displacement of a light ray passing through a glass slab? (2015)
View AnswerAns: The shifting of the light ray sideways (though in the direction of original ray) on emergence from a rectangular glass slab is called “lateral displacement”.
Q5: Define power of a lens and write its SI unit. (2015)
View AnswerAns: Reciprocal of focal length of a lens, expressed in meter, is called the power of that lens. Its SI unit is 1 dioptre (1 D), where 1 D = 1 m-1.
Q6: Name the lens which can be used as a magnifying glass. (2015)
View AnswerAns: A convex lens can be used as a magnifying glass so as to form magnified image of a tiny object placed near it.
Q7: Which type of lens has a negative power? (2015)
View AnswerAns: A concave (diverging) lens has a negative power.
Q8: What is the difference between virtual image of an object formed by a convex lens and that formed by a concave lens? (2015)
View AnswerAns: 'Virtual image formed by a convex lens is always magnified but that formed by a concave lens is diminished one.
Q9: During its passage from one medium to another, where does a light ray change its path? (2015)
View AnswerAns: During its passage from one medium to another a light ray changes its path at the boundary face separating the two media.
Q10: The power of a lens is + 5 D. Find its focal length in metres. (2015)
View AnswerAns: Focal length f = 1/P = +1/5 m = 0.2m
Q11: What are the units of power of a lens? (2015)
View AnswerAns: If the focal length is measured in metre then the unit of power of a lens is dioptre.
Q12: If the image formed by a mirror for all positions of the object placed in front of it is always erect and diminished, what type of mirror is it? Draw a ray diagram to justify your answer. Where and why do we generally use this type of mirror? (2015)
View AnswerAns: Only in convex mirror, for all positions of the object placed in front of it is always virtual, erected and diminished. Hence this mirror is convex mirror.
Convex mirrors are used in automobiles as a rear view mirror because of wider field of view and formation of erect image.
Q13: Name the type of mirror used in the following: (2015)
a. Solar furnace
b. Side/rear - view mirror of a vehicle.
Draw a labelled ray diagram to show the formation of image in each of the above two cases.
Which of these mirrors could also form a magnified and virtual image of an object? Illustrate with the help of a ray diagram.
View AnswerAns: a. Concave mirror
b. Convex mirror
Concave mirror form magnified virtual image of an object.
Previous Year Questions 2014
Q1: Write down four important characteristics of image formed by a plane mirror. (2014)
View AnswerAns: Image is virtual, erect, laterally inverted and of same size as object.
Q2: Describe a spherical mirror. (2014)
View AnswerAns: Spherical mirror is a part of a sphere. If reflection takes place from inside, it is said to be concave mirror, and if the reflection takes place from outside surface it is a convex mirror.
Q3: Define the following terms in relation to concave spherical mirror: (2014)
(a) Pole
(b) Centre of curvature
(c) Radius of curvature
(d) Principal axis
(e) Principal focus
(f) Aperture
(g) Focal length
View AnswerAns: (a) The mid point of mirror is known as pole.
(b) The centre of curvature of a spherical mirror is the centre of that sphere of which mirror is a part,
(c) The distance between pole and centre of curvature is called radius of curvature of the mirror.
(d) The straight line joining the pole and centre of curvature is called principal axis.
(e) The point on the principal axis through which parallel rays to the principal axis passes or appear to pass after reflection.
(f) The diameter of the mirror or size of the mirror is called aperture.
(g) The distance between focus and pole of a mirror is the focal length of the mirror.
Q4: With the help of ray diagram show that angle of incidence is equal to the angle of reflection when a ray is incident on the concave/convex mirror. (2014)
View AnswerAns: 
Q5: A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of image and magnification. Describe what happens to the image as the needle is moved farther from the mirror. (2014)
View AnswerAns: Given: u = - 12 cm
f = + 15 cm
h1 = 4.5 cm
v = ?, m = ?
Mirror formula,
v = + 6.6 cm
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