Direction : For question number 32, two statements are given – one labelled as Assertion
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(A) / Both Assertion (A) and Reason (R) both are true, and Reason (R) is the correct explanation of Assertion (1 Mark)
Light: Reflection and Refraction — Class 10 Science PYQs
Previous-year board questions (2026) with marking-scheme solutions, grouped by topic and marks.
Try each question first, then press ↓ (or tap Show Solution) to reveal the answer. Press ↓ again for the next question.
Reflection and spherical mirrors
1 Mark Questions
Pushkar was studying a light phenomenon and wrote down the following statements in that context :
(i) A concave mirror can be used as a shaving mirror because it can form an enlarged, virtual image when the object is placed between the pole and focus.
(ii) A convex lens always forms real, inverted images, regardless of the position of the object.
(iii) The relationship between object distance (u), image distance (v) and focal length (f) of the lens is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Choose the option enlisting all the correct statements :
(i) A concave mirror can be used as a shaving mirror because it can form an enlarged, virtual image when the object is placed between the pole and focus.
(ii) A convex lens always forms real, inverted images, regardless of the position of the object.
(iii) The relationship between object distance (u), image distance (v) and focal length (f) of the lens is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$.
Choose the option enlisting all the correct statements :
- (a)(i) and (ii)
- (b)(i) and (iii)
- (c)(ii) and (iii)
- (d)(i), (ii) and (iii)
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(B) (i) and (iii) (1)
2 Marks Questions
Write any two differences between the images formed by convex mirror and plane mirror.
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Images formed by convex mirror
1 Size of image is always smaller than object. (1 Mark)
2 Image distance is not equal to object distance. (1 Mark)
Images formed by plane mirror
Size of image is same as size of object.
Image distance is equal to object distance.
(any other suitable difference)
1 Size of image is always smaller than object. (1 Mark)
2 Image distance is not equal to object distance. (1 Mark)
Images formed by plane mirror
Size of image is same as size of object.
Image distance is equal to object distance.
(any other suitable difference)
Ray diagrams for mirrors
1 Mark Questions
Pushkar was studying a light phenomenon and wrote down the following statements in that context :
(i) A concave mirror can be used as a shaving mirror because it can form an enlarged, virtual image when the object is placed between the pole and focus.
(ii) A convex lens always forms real, inverted images, regardless of the position of the object.
(iii) The relationship between object distance (u), image distance (v) and focal length (f) of the lens is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Choose the option enlisting all the correct statements :
(i) A concave mirror can be used as a shaving mirror because it can form an enlarged, virtual image when the object is placed between the pole and focus.
(ii) A convex lens always forms real, inverted images, regardless of the position of the object.
(iii) The relationship between object distance (u), image distance (v) and focal length (f) of the lens is given by $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$
Choose the option enlisting all the correct statements :
- (a)(i) and (ii)
- (b)(i) and (iii)
- (c)(ii) and (iii)
- (d)(i), (ii) and (iii)
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B (i) and (iii)
1
1
Mirror formula and magnification
1 Mark Questions
From the following options choose the one which gives negative magnification greater than one :
- (a)Concave mirror; Nature of image – real
- (b)Concave mirror; Nature of image – virtual
- (c)Convex mirror; Nature of image – real
- (d)Convex mirror; Nature of image – virtual
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(A) / Concave mirror, Nature of image-real (1 Mark)
When a convergent beam of light is incident on a convex mirror, the reflected rays appear to be coming from a point $20\ \text{cm}$ behind the mirror and retrace their path. The focal length of the mirror is
- (a)$20\ \text{cm}$
- (b)$10\ \text{cm}$
- (c)$40\ \text{cm}$
- (d)$30\ \text{cm}$
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(B) / 10 cm (1 Mark)
Refraction: laws and refractive index
2 Marks Questions
(a) Relate the speed of light in the given medium with its optical density.
(b) Using the information given in the table below, arrange the medium A, B and C in the ascending order of their optical density.
begin{tabular}{|c|c|}
hline Medium & Speed of light (m/s)
hline A & $2.25 \times 10^8$
hline B & $2 \times 10^8$
hline C & $2.08 \times 10^8$
hline
end{tabular}
(b) Using the information given in the table below, arrange the medium A, B and C in the ascending order of their optical density.
begin{tabular}{|c|c|}
hline Medium & Speed of light (m/s)
hline A & $2.25 \times 10^8$
hline B & $2 \times 10^8$
hline C & $2.08 \times 10^8$
hline
end{tabular}
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(a) In optically rarer medium speed of light is more and in optically denser medium speed of light is less. (1 Mark)
(b) $A < C < B$ (1 Mark)
(b) $A < C < B$ (1 Mark)
(a) Define absolute refractive index of an optical medium.
(b)
begin{tabular}{|c|c|}
hline Material Medium & Refractive Index
hline A & 1.50
hline B & 1.46
hline C & 1.31
hline D & 1.77
hline
end{tabular}
Arrange these material mediums given in the table in increasing order of speed of light through them.
(b)
begin{tabular}{|c|c|}
hline Material Medium & Refractive Index
hline A & 1.50
hline B & 1.46
hline C & 1.31
hline D & 1.77
hline
end{tabular}
Arrange these material mediums given in the table in increasing order of speed of light through them.
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(a) Absolute refractive index of a medium is the ratio of speed of light in air to the speed of light in the given medium / Refractive index of a medium with respect to air / $n = \frac{c}{v}$ (1 Mark)
(b) $V_D < V_A < V_B < V_C$ (1 Mark)
(b) $V_D < V_A < V_B < V_C$ (1 Mark)
When an incident ray of light enters in a medium ‘X’ from medium ‘Y’, it bends away from the normal.
Comment about the following :
(a) Speed of light in medium ‘X’ with respect to the speed of light in medium of ‘Y’.
(b) Optical density of medium ‘X’ with respect to the optical density of medium ‘Y’.
Give reason for your answer in each case.
Comment about the following :
(a) Speed of light in medium ‘X’ with respect to the speed of light in medium of ‘Y’.
(b) Optical density of medium ‘X’ with respect to the optical density of medium ‘Y’.
Give reason for your answer in each case.
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(a) • Speed of light in medium 'X' is more than in medium 'Y'. (1/2)
• Because light ray bends away from the normal in optically rarer medium. (1/2)
(b) • Optical density of medium 'X' is less than medium 'Y'. (1/2)
• Because in optically rarer medium light bends away from the normal. (1/2)
• Because light ray bends away from the normal in optically rarer medium. (1/2)
(b) • Optical density of medium 'X' is less than medium 'Y'. (1/2)
• Because in optically rarer medium light bends away from the normal. (1/2)
10open ↗
Attempt either subpart (a) or (b) :
OR
(b) When a ray of light passes from water to air (I) will the angle of refraction in air (r) be greater than the angle of incidence in water (i) or less than it ? (II) and when we increase the angle of incidence in water, will the angle of refraction in air increase or decrease ? What is the limiting value for $\angle\text{r}$ ?
OR
(b) When a ray of light passes from water to air (I) will the angle of refraction in air (r) be greater than the angle of incidence in water (i) or less than it ? (II) and when we increase the angle of incidence in water, will the angle of refraction in air increase or decrease ? What is the limiting value for $\angle\text{r}$ ?
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(b) (I) $\angle r$ will be greater than $\angle i$. (1 Mark)
(II) • $\angle r$ will increase. (½ Mark)
• Limiting value of $\angle r$ will be $90^\circ$. (½ Mark)
(II) • $\angle r$ will increase. (½ Mark)
• Limiting value of $\angle r$ will be $90^\circ$. (½ Mark)
11open ↗
The values of refractive indices of two optical media X and Y are $1.44$ and $1.30$, respectively. Compare the values of the following quantities pertaining to them :
(a) Speed of light
(b) Angle of refraction for same angle of incidence
Justify your answer.
(a) Speed of light
(b) Angle of refraction for same angle of incidence
Justify your answer.
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(a) • Speed of light in medium X is less than the speed of light in medium Y. ($\frac{1}{2}$ Mark)
• Reason: Refractive index of medium X is more than medium Y hence it is optically denser than medium Y. ($\frac{1}{2}$ Mark)
(b) • Angle of refraction in medium X is less than in medium Y. ($\frac{1}{2}$ Mark)
• Reason: Refractive index of medium X is more than medium Y hence it is optically denser than medium Y. ($\frac{1}{2}$ Mark) (2 Marks)
• Reason: Refractive index of medium X is more than medium Y hence it is optically denser than medium Y. ($\frac{1}{2}$ Mark)
(b) • Angle of refraction in medium X is less than in medium Y. ($\frac{1}{2}$ Mark)
• Reason: Refractive index of medium X is more than medium Y hence it is optically denser than medium Y. ($\frac{1}{2}$ Mark) (2 Marks)
12open ↗
(b) Match the items given in Column I with the correct option given in Column II :
S.No. Column I Column II
1. Mirror Refraction of light
2. Lens Dispersion of light
3. Prism Tyndall effect
4. Scattering of light Reflection of light
S.No. Column I Column II
1. Mirror Refraction of light
2. Lens Dispersion of light
3. Prism Tyndall effect
4. Scattering of light Reflection of light
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(b) (1) Mirror Reflection of light ($\frac{1}{2}$x4 Marks)
(2) Lens Refraction of light
(3) Prism Dispersion of light
(4) Scattering of light Tyndall effect (2 Marks)
(2) Lens Refraction of light
(3) Prism Dispersion of light
(4) Scattering of light Tyndall effect (2 Marks)
4 Marks Questions
13open ↗
Read the following passage and answer the questions that follow :
Two students, A and B were performing an activity on refraction through glass slab and refraction through prism respectively. Student 'A' observed that the emergent light ray is parallel to the direction of incident ray, while student 'B' observed that the emergent ray is making a certain angle with respect to the direction of the incident ray in the prism, called angle of deviation. He also noted that the angle of deviation is different for different colours of light.
(a) When white light passes through a glass slab, it does not show dispersion, while it shows dispersion when it passes through a glass prism. Why ?
(b) Why do we get spectrum when white light passes through tiny water droplets present in air ?
(c) (i) For a given pair of mediums, if we change the angle of incidence gradually, will it change the speed of light in that medium ? Explain.
OR
(c) (ii) Calculate the speed of light in a given medium if angle of incidence in air is $60^\circ$ and angle of refraction in the medium is $30^\circ$. Given that speed of light in air is $3 \times 10^8\ \text{m/s}$.
Two students, A and B were performing an activity on refraction through glass slab and refraction through prism respectively. Student 'A' observed that the emergent light ray is parallel to the direction of incident ray, while student 'B' observed that the emergent ray is making a certain angle with respect to the direction of the incident ray in the prism, called angle of deviation. He also noted that the angle of deviation is different for different colours of light.
(a) When white light passes through a glass slab, it does not show dispersion, while it shows dispersion when it passes through a glass prism. Why ?
(b) Why do we get spectrum when white light passes through tiny water droplets present in air ?
(c) (i) For a given pair of mediums, if we change the angle of incidence gradually, will it change the speed of light in that medium ? Explain.
OR
(c) (ii) Calculate the speed of light in a given medium if angle of incidence in air is $60^\circ$ and angle of refraction in the medium is $30^\circ$. Given that speed of light in air is $3 \times 10^8\ \text{m/s}$.
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(a) Glass slab shows no dispersion because the colours dispersed by first
refracting surface recombine and emerge out as white light at the
second surface. (1)
In a glass prism different colours bend at different angles hence shows
dispersion
(b) Tiny water droplets act like prism. They refract and disperse the
sunlight to show spectrum. (1)
(c)
(i) With change in angle of incidence, angle of refraction will also change.
Hence refractive index remains constant. (1)
$\text{Refractive index} = \frac{\text{speed of light in air}}{\text{speed of light in given medium}}$
Hence speed of light in that medium will remain constant. (1)
OR
(c)
(ii) $\text{n} = \frac{\sin \text{i}}{\sin \text{r}}$
$\text{n} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$ ($\frac{1}{2}$)
$\text{n} = \frac{\text{speed of light in air}}{\text{speed of light in medium}}$
$\sqrt{3} = \frac{3 \times 10^8}{\text{v}}$ ($\frac{1}{2}$)
$\text{v} = \frac{3 \times 10^8}{\sqrt{3}}$
$\text{v} = \sqrt{3} \times 10^8 \text{ m/s}$ ($\frac{1}{2}$)
refracting surface recombine and emerge out as white light at the
second surface. (1)
In a glass prism different colours bend at different angles hence shows
dispersion
(b) Tiny water droplets act like prism. They refract and disperse the
sunlight to show spectrum. (1)
(c)
(i) With change in angle of incidence, angle of refraction will also change.
Hence refractive index remains constant. (1)
$\text{Refractive index} = \frac{\text{speed of light in air}}{\text{speed of light in given medium}}$
Hence speed of light in that medium will remain constant. (1)
OR
(c)
(ii) $\text{n} = \frac{\sin \text{i}}{\sin \text{r}}$
$\text{n} = \frac{\sin 60^{\circ}}{\sin 30^{\circ}} = \frac{\sqrt{3}/2}{1/2} = \sqrt{3}$ ($\frac{1}{2}$)
$\text{n} = \frac{\text{speed of light in air}}{\text{speed of light in medium}}$
$\sqrt{3} = \frac{3 \times 10^8}{\text{v}}$ ($\frac{1}{2}$)
$\text{v} = \frac{3 \times 10^8}{\sqrt{3}}$
$\text{v} = \sqrt{3} \times 10^8 \text{ m/s}$ ($\frac{1}{2}$)
Refraction by spherical lenses
2 Marks Questions
14open ↗
Read the following passage and answer the questions given below :
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a) A convex lens of focal length $20\ \text{cm}$ is used to form an image. If an object is placed at $40\ \text{cm}$ from the lens, what will be the position and nature of image ?
(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a) A convex lens of focal length $20\ \text{cm}$ is used to form an image. If an object is placed at $40\ \text{cm}$ from the lens, what will be the position and nature of image ?
(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
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$f_1 = 30 \text{ cm} = 0.3 \text{ m, } f_2=-15\text{cm} = -0.15 \text{ m}$
$P = \frac{1}{f}$
$P_1=\frac{+1}{0.3} \text{ D} ; P_2=\frac{-1}{0.15} \text{ D}$
$\frac{1}{2}$
Equivalent power, $P = P_1+P_2$
$P = - 3.33\text{D}$
$\frac{1}{2}$
Equivalent focal length, $f=\frac{1}{P}$
$\frac{1}{2}$
$f=\frac{-1}{3.33} = - 0.3 \text{ m} = - 30 \text{ cm}$
$\frac{1}{2}$
$P = \frac{1}{f}$
$P_1=\frac{+1}{0.3} \text{ D} ; P_2=\frac{-1}{0.15} \text{ D}$
$\frac{1}{2}$
Equivalent power, $P = P_1+P_2$
$P = - 3.33\text{D}$
$\frac{1}{2}$
Equivalent focal length, $f=\frac{1}{P}$
$\frac{1}{2}$
$f=\frac{-1}{3.33} = - 0.3 \text{ m} = - 30 \text{ cm}$
$\frac{1}{2}$
4 Marks Questions
15open ↗
Read the following passage and answer the questions given below :
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a) A convex lens of focal length $20\ \text{cm}$ is used to form an image. If an object is placed at $40\ \text{cm}$ from the lens, what will be the position and nature of image ?
(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
(c) (i) A lens combination consists of a convex lens of focal length $30\ \text{cm}$ and a concave lens of focal length $15\ \text{cm}$ placed together. Find the equivalent focal length and power of this lens combination.
OR
(c) (ii) Two lenses are placed in contact. One is a concave lens with focal length $2\ \text{m}$ and the other is a convex lens with focal length $1.5\ \text{m}$. What type of lens will the combination behave as (convex or concave) ? Give reason.
Lenses can form different types of images depending upon their focal length and position of object. A convex lens can create real, inverted or virtual, erect images, while a concave lens forms only virtual and diminished images. The focal length determines the power of lens. Convex lenses have positive focal length while concave lenses have negative focal length by convention. When lenses are placed together, their combined power is determined by the sum of their individual powers. Ray diagrams help to visualize how light converges or diverges through lens to form an image.
(a) A convex lens of focal length $20\ \text{cm}$ is used to form an image. If an object is placed at $40\ \text{cm}$ from the lens, what will be the position and nature of image ?
(b) Illustrate the formation of image with the help of ray diagram, when the object is placed between the optical centre and principal focus of concave lens.
(c) (i) A lens combination consists of a convex lens of focal length $30\ \text{cm}$ and a concave lens of focal length $15\ \text{cm}$ placed together. Find the equivalent focal length and power of this lens combination.
OR
(c) (ii) Two lenses are placed in contact. One is a concave lens with focal length $2\ \text{m}$ and the other is a convex lens with focal length $1.5\ \text{m}$. What type of lens will the combination behave as (convex or concave) ? Give reason.
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$f_1 = 30 \text{ cm} = 0.3 \text{ m}, f_2 = -15\text{cm} = -0.15 \text{ m}$ (½ Mark)
$P = \frac{1}{f}$
$P_1 = \frac{+1}{0.3} \text{ D} ; P_2 = \frac{-1}{0.15} \text{ D}$ (½ Mark)
Equivalent power, $P = P_1+P_2$
$P = -3.33 \text{D}$ (½ Mark)
Equivalent focal length, $f = \frac{1}{P}$
$f = \frac{-1}{3.33} = -0.3 \text{ m} = -30 \text{ cm}$ (½ Mark)
$P = \frac{1}{f}$
$P_1 = \frac{+1}{0.3} \text{ D} ; P_2 = \frac{-1}{0.15} \text{ D}$ (½ Mark)
Equivalent power, $P = P_1+P_2$
$P = -3.33 \text{D}$ (½ Mark)
Equivalent focal length, $f = \frac{1}{P}$
$f = \frac{-1}{3.33} = -0.3 \text{ m} = -30 \text{ cm}$ (½ Mark)
Lens formula, magnification, power of a lens
1 Mark Questions
16open ↗
A convex lens of focal length $15\ \text{cm}$, is forming a real image. If the size of image is same as the size of object, then position of object and position of image will be, respectively :
- (a)$15\ \text{cm}$ and $-15\ \text{cm}$ from lens
- (b)$-15\ \text{cm}$ and $+15\ \text{cm}$ from lens
- (c)$-30\ \text{cm}$ and $+30\ \text{cm}$ from lens
- (d)$-30\ \text{cm}$ and $-30\ \text{cm}$ from lens
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(C) / - 30 cm and + 30 cm from lens
1
1
1
1
17open ↗
The variation of image distance ($v$) with the object distance ($u$) for a convex lens is given in the following observation table. Analyse it and answer the questions given below :
(a) Without calculation, find out the focal length of the given convex lens. Justify your answer.
(b) Which one of the observations given in the table is not correct and why?
| S.No. | Object distance ($u$) cm | Image distance ($v$) cm |
|---|---|---|
| --- | --- | --- |
| 1 | $-150$ | $+30$ |
| 2 | $-75$ | $+37.5$ |
| 3 | $-50$ | $+50$ |
| 4 | $-37.5$ | $+75$ |
| 5 | $-30$ | $+150$ |
| 6 | $-15$ | $+37.5$ |
(b) Which one of the observations given in the table is not correct and why?
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(c) $m = \frac{v}{u} = \frac{150}{-30} = -5$
Nature of image: Image will be real and inverted
Reason: because the value of m is negative.
Nature of image: Image will be real and inverted
Reason: because the value of m is negative.
18open ↗
Rays from the sun converge at a point $25\ \text{cm}$ behind a convex lens. The distance at which an object be placed in front of the lens to get a virtual image, is :
- (a)20 cm
- (b)40 cm
- (c)50 cm
- (d)More than 50 cm
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(A) / 20 cm (1 Mark)
19open ↗
A convex lens and a concave lens having focal lengths $10\ \text{cm}$ and $20\ \text{cm}$ respectively are placed in contact with each other. Which of the following statements is correct about the lens combination ?
- (a)It will behave like a plane glass slab.
- (b)It will behave like a convex lens.
- (c)It will behave like a concave lens.
- (d)It will behave like a cylindrical lens.
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(B) It will behave like a convex lens. (1 Mark)
2 Marks Questions
20open ↗
(c) (i) A lens combination consists of a convex lens of focal length $30\ \text{cm}$ and a concave lens of focal length $15\ \text{cm}$ placed together. Find the equivalent focal length and power of this lens combination.
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Solution not available.
21open ↗
(c) (ii) Two lenses are placed in contact. One is a concave lens with focal length $2\ \text{m}$ and the other is a convex lens with focal length $1.5\ \text{m}$. What type of lens will the combination behave as (convex or concave) ? Give reason.
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Solution not available.
22open ↗
(c) (i) Find out the value of magnification for $u = -30\ \text{cm}$. Write the nature of the image formed. Give reason for your answer.
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Solution not available.
3 Marks Questions
23open ↗
(a) Write the expression for the magnification produced by a lens in terms of object distance and image distance.
(b) A $4\ \text{cm}$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $20\ \text{cm}$. Calculate the size of the image formed, if the distance of the object from the lens is $10\ \text{cm}$.
(b) A $4\ \text{cm}$ tall object is placed perpendicular to the principal axis of a convex lens of focal length $20\ \text{cm}$. Calculate the size of the image formed, if the distance of the object from the lens is $10\ \text{cm}$.
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(a) Magnification, $m = \frac{\text{Image distance}}{\text{Object distance}} / m = \frac{v}{u}$
(b)
$h_o= + 4 \text{ cm}$
$f = + 20 \text{ cm}$
$u = -10 \text{ cm}$
$h_i = ?$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{20} + \frac{1}{-10}$
$\frac{1}{v} = \frac{1-2}{20}$
$v = -20 \text{ cm}$
$m = \frac{v}{u}$
$m = \frac{-20}{-10}$
$m = 2$
$m = \frac{\text{height of image}}{\text{height of object}}$
height of image = $m \times$ height of object
height of image = $2 \times 4 \text{ cm} = 8 \text{ cm}$
(b)
$h_o= + 4 \text{ cm}$
$f = + 20 \text{ cm}$
$u = -10 \text{ cm}$
$h_i = ?$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{20} + \frac{1}{-10}$
$\frac{1}{v} = \frac{1-2}{20}$
$v = -20 \text{ cm}$
$m = \frac{v}{u}$
$m = \frac{-20}{-10}$
$m = 2$
$m = \frac{\text{height of image}}{\text{height of object}}$
height of image = $m \times$ height of object
height of image = $2 \times 4 \text{ cm} = 8 \text{ cm}$
24open ↗
(a) Write the name of the lens and position of the object in front of it so that a virtual and magnified image is formed.
(b) An object is placed perpendicular to the principal axis of a convex lens of focal length $20\ \text{cm}$. If object is located at $30\ \text{cm}$ from this lens, using lens formula find out the position and nature of the image formed.
(b) An object is placed perpendicular to the principal axis of a convex lens of focal length $20\ \text{cm}$. If object is located at $30\ \text{cm}$ from this lens, using lens formula find out the position and nature of the image formed.
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(a)
* Convex lens (½ Mark)
* When object is placed between focus and optical centre (½ Mark)
(b) $f = + 20 \text{ cm}$
$u = -30 \text{ cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{20} + \frac{1}{-30}$
$\frac{1}{v} = \frac{1}{60}$ (½ Mark)
$v = 60 \text{ cm}$
The image is formed at a distance of 60 cm on the other side of the optical centre.
* Nature of image – Real, inverted (1 Mark)
* Convex lens (½ Mark)
* When object is placed between focus and optical centre (½ Mark)
(b) $f = + 20 \text{ cm}$
$u = -30 \text{ cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{20} + \frac{1}{-30}$
$\frac{1}{v} = \frac{1}{60}$ (½ Mark)
$v = 60 \text{ cm}$
The image is formed at a distance of 60 cm on the other side of the optical centre.
* Nature of image – Real, inverted (1 Mark)
25open ↗
An object is placed at a distance of $30\,\text{cm}$ in front of a convex lens of focal length $15\,\text{cm}$. Use lens formula to determine the position of the image. What will be the size of the image in this case ?
Show SolutionHide Solution↓
• $u = -30\text{ cm}$, $f = 15\text{ cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{15} + \frac{1}{-30}$ (½ Mark)
$v = 30\text{ cm}$ (1 Mark)
• $m = \frac{v}{u} = +\frac{30}{-30} = -1$
Hence, size of image will be same as the object. / Size of image will be same as size of object because when object is kept at $2F_1$, image will be formed at $2F_2$. (1 Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{15} + \frac{1}{-30}$ (½ Mark)
$v = 30\text{ cm}$ (1 Mark)
• $m = \frac{v}{u} = +\frac{30}{-30} = -1$
Hence, size of image will be same as the object. / Size of image will be same as size of object because when object is kept at $2F_1$, image will be formed at $2F_2$. (1 Mark)
26open ↗
An object of height $6\ \text{cm}$ is placed at a distance of $30\ \text{cm}$ from the optical centre of a concave lens of focal length $15\ \text{cm}$.
Use lens formula to determine :
(a) The distance of image from the optical centre.
(b) The height of the image formed.
Use lens formula to determine :
(a) The distance of image from the optical centre.
(b) The height of the image formed.
Show SolutionHide Solution↓
(a) Height of object = $\text{6.0 cm}$
$u = \text{-30 cm}$
$f = \text{-15 cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30}$ (1/2)
(b) $v = \text{-10 cm}$
$m = \frac{v}{u}$
$m = \frac{-10}{-30}$
$m = \frac{1}{3}$
$\text{height of image} = \frac{h'}{\text{height of object}} = \frac{h'}{h}$
$h' = m \times h$
$h' = \frac{1}{3} \times 6$
$h' = \text{2 cm}$ (1/2)
$u = \text{-30 cm}$
$f = \text{-15 cm}$
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$
$\frac{1}{v} = \frac{1}{-15} + \frac{1}{-30}$ (1/2)
(b) $v = \text{-10 cm}$
$m = \frac{v}{u}$
$m = \frac{-10}{-30}$
$m = \frac{1}{3}$
$\text{height of image} = \frac{h'}{\text{height of object}} = \frac{h'}{h}$
$h' = m \times h$
$h' = \frac{1}{3} \times 6$
$h' = \text{2 cm}$ (1/2)
27open ↗
Optical device Object distance Focal length Height of object
(cm)
(cm)
(cm)
Convex lens 20 10 6
Concave mirror 30 10 6
By using the data given in the table, compare the properties of images formed by convex lens and concave mirror in terms of the positions and nature of images formed.
(cm)
(cm)
(cm)
Convex lens 20 10 6
Concave mirror 30 10 6
By using the data given in the table, compare the properties of images formed by convex lens and concave mirror in terms of the positions and nature of images formed.
Show SolutionHide Solution↓
For Convex lens
• Position:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (1 Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ (1 Mark)
$\frac{1}{v} = \frac{1}{10} + \frac{1}{-20}$ (1 Mark)
$v = +20cm$ (1 Mark)
• Nature: real, inverted (½ Mark)
For Concave mirror
• Position:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (1 Mark)
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ (1 Mark)
$\frac{1}{v} = \frac{1}{-10} - \frac{1}{-30}$ (1 Mark)
$v = -15cm$ (1 Mark)
• Nature: real, inverted (½ Mark)
/
Alternate answer
Convex lens: -
• When object is placed at centre of curvature, the image is formed at centre of curvature. (1 Mark)
• Nature of image is real and inverted. (½ Mark)
Concave Mirror: -
• When object is between infinity and centre of curvature the image is formed between focus and centre of curvature. (1 Mark)
• Nature of image is real and inverted. (½ Mark)
• Position:
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (1 Mark)
$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$ (1 Mark)
$\frac{1}{v} = \frac{1}{10} + \frac{1}{-20}$ (1 Mark)
$v = +20cm$ (1 Mark)
• Nature: real, inverted (½ Mark)
For Concave mirror
• Position:
$\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ (1 Mark)
$\frac{1}{v} = \frac{1}{f} - \frac{1}{u}$ (1 Mark)
$\frac{1}{v} = \frac{1}{-10} - \frac{1}{-30}$ (1 Mark)
$v = -15cm$ (1 Mark)
• Nature: real, inverted (½ Mark)
/
Alternate answer
Convex lens: -
• When object is placed at centre of curvature, the image is formed at centre of curvature. (1 Mark)
• Nature of image is real and inverted. (½ Mark)
Concave Mirror: -
• When object is between infinity and centre of curvature the image is formed between focus and centre of curvature. (1 Mark)
• Nature of image is real and inverted. (½ Mark)
28open ↗
The various positions of an object from a convex lens and the relative sizes of the images formed in each position are tabulated below. Look at the data carefully and answer the questions that follow the table :
S.No. Position of object from lens (cm) Relative size of image (cm)
1. 45 Smaller than object
2. 30 Same size as object
3. 20 Magnified
4. 10 Magnified
(a) What is the focal length of the lens ?
S.No. Position of object from lens (cm) Relative size of image (cm)
1. 45 Smaller than object
2. 30 Same size as object
3. 20 Magnified
4. 10 Magnified
(a) What is the focal length of the lens ?
Show SolutionHide Solution↓
(a) 15 cm (1 Mark)
(b) :
Observation 3: Real and inverted image is formed. (1 Mark)
Observation 4: Virtual and erect image is formed. (1 Mark) (3 Marks)
(b) :
Observation 3: Real and inverted image is formed. (1 Mark)
Observation 4: Virtual and erect image is formed. (1 Mark) (3 Marks)
29open ↗
The various positions of an object from a convex lens and the relative sizes of the images formed in each position are tabulated below. Look at the data carefully and answer the questions that follow the table :
S.No. Position of object from lens (cm) Relative size of image (cm)
1. 45 Smaller than object
2. 30 Same size as object
3. 20 Magnified
4. 10 Magnified
(b) What is the nature of the images formed in case of observation 3 and 4 ?
S.No. Position of object from lens (cm) Relative size of image (cm)
1. 45 Smaller than object
2. 30 Same size as object
3. 20 Magnified
4. 10 Magnified
(b) What is the nature of the images formed in case of observation 3 and 4 ?
Show SolutionHide Solution↓
(a) 15 cm (1 Mark)
(b) :
Observation 3: Real and inverted image is formed. (1 Mark)
Observation 4: Virtual and erect image is formed. (1 Mark) (3 Marks)
(b) :
Observation 3: Real and inverted image is formed. (1 Mark)
Observation 4: Virtual and erect image is formed. (1 Mark) (3 Marks)
5 Marks Questions
30open ↗
(i) Define the optical centre of a lens.
(ii) The light from an object placed on a meter scale at $8\ \text{cm}$ mark, was focused on a white screen placed at $92\ \text{cm}$ mark, with the help of a converging lens placed on the scale at $50\ \text{cm}$ mark :
(I) Find the focal length of converging lens.
(II) Find the position of image if the object is shifted towards the lens and kept at $29\ \text{cm}$ mark.
(III) What will be the nature of the image formed if the object is further shifted towards the lens ?
(ii) The light from an object placed on a meter scale at $8\ \text{cm}$ mark, was focused on a white screen placed at $92\ \text{cm}$ mark, with the help of a converging lens placed on the scale at $50\ \text{cm}$ mark :
(I) Find the focal length of converging lens.
(II) Find the position of image if the object is shifted towards the lens and kept at $29\ \text{cm}$ mark.
(III) What will be the nature of the image formed if the object is further shifted towards the lens ?
Show SolutionHide Solution↓
OR
The central point of a lens through which a ray of light passes without any deviation. (1 Mark)
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{42} - \frac{1}{-42} = \frac{1}{f}$
$f = 21 \text{ cm}$ (1 Mark)
Alternate answer for (I):
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
Since object distance is same as image distance. So, object is placed at $2F_1$.
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (1 Mark)
$u = -(50 - 29) = -21 \text{ cm}$
As now object is kept at focus $F_1$, image will be formed at infinity (1 Mark)
Now, object is kept between Optical centre and focus $F_1$,
Nature of image is virtual and erect. (1 Mark)
The central point of a lens through which a ray of light passes without any deviation. (1 Mark)
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$
$\frac{1}{42} - \frac{1}{-42} = \frac{1}{f}$
$f = 21 \text{ cm}$ (1 Mark)
Alternate answer for (I):
$u = -(50 - 8) = -42 \text{ cm}$ ($\frac{1}{2}$ Mark)
$v = (92 - 50) = 42 \text{ cm}$ ($\frac{1}{2}$ Mark)
Since object distance is same as image distance. So, object is placed at $2F_1$.
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (1 Mark)
$u = -(50 - 29) = -21 \text{ cm}$
As now object is kept at focus $F_1$, image will be formed at infinity (1 Mark)
Now, object is kept between Optical centre and focus $F_1$,
Nature of image is virtual and erect. (1 Mark)
31open ↗
(i) Define the focus of a convex lens.
(ii) An object placed on a meter scale at 8 $\text{cm}$ mark, was focused on a white screen placed at 92 $\text{cm}$ mark, by using a converging lens placed on the scale at 50 $\text{cm}$ mark.
(I) Find the focal length of converging lens.
(II) Find the position of object if the image forms at 71 $\text{cm}$ mark on meter scale.
(III) State the nature of image formed if the object is kept on 36 $\text{cm}$ mark of meter scale.
(ii) An object placed on a meter scale at 8 $\text{cm}$ mark, was focused on a white screen placed at 92 $\text{cm}$ mark, by using a converging lens placed on the scale at 50 $\text{cm}$ mark.
(I) Find the focal length of converging lens.
(II) Find the position of object if the image forms at 71 $\text{cm}$ mark on meter scale.
(III) State the nature of image formed if the object is kept on 36 $\text{cm}$ mark of meter scale.
Show SolutionHide Solution↓
It is a point on principal axis where light rays incident parallel to principal axis meet (converge) after refraction through a convex lens. (1 Mark)
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{42} - (\frac{1}{-42}) = \frac{1}{f}$ (½ Mark)
$f = 21 \text{ cm}$ (½ Mark)
Alternate answer for (I):
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
Since image distance is same as object distance. So, object is placed at $2F_1$. (½ Mark)
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (½ Mark)
$v = (71 - 50) \text{ cm} = 21 \text{ cm}$ (½ Mark)
Since, image is formed at focus $F_2$. So, object is at infinity. (1 Mark)
$u = - (50 - 36) \text{ cm} = -14 \text{ cm}$ (½ Mark)
Since, object is kept between focus $F_1$ and optical centre. So, nature of image formed will be virtual and erect. (1 Mark)
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
$\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ (½ Mark)
$\frac{1}{42} - (\frac{1}{-42}) = \frac{1}{f}$ (½ Mark)
$f = 21 \text{ cm}$ (½ Mark)
Alternate answer for (I):
$u = - (50 - 8) = -42 \text{ cm}$ (½ Mark)
$v = (92 - 50) = 42 \text{ cm}$ (½ Mark)
Since image distance is same as object distance. So, object is placed at $2F_1$. (½ Mark)
$2 \times f = 42 \text{ cm}$ or $f = 21 \text{ cm}$ (½ Mark)
$v = (71 - 50) \text{ cm} = 21 \text{ cm}$ (½ Mark)
Since, image is formed at focus $F_2$. So, object is at infinity. (1 Mark)
$u = - (50 - 36) \text{ cm} = -14 \text{ cm}$ (½ Mark)
Since, object is kept between focus $F_1$ and optical centre. So, nature of image formed will be virtual and erect. (1 Mark)
32open ↗
(i) What is meant by refraction of light? Write the laws of refraction.
(ii) If two lenses of focal length $+20\ \text{cm}$ and $-10\ \text{cm}$ are kept in contact, find out the net power of the lens combination. Will the combination behave as converging lens or a diverging lens ?
(ii) If two lenses of focal length $+20\ \text{cm}$ and $-10\ \text{cm}$ are kept in contact, find out the net power of the lens combination. Will the combination behave as converging lens or a diverging lens ?
Show SolutionHide Solution↓
(i)
$\bullet$ When a ray of light travels obliquely from one transparent medium to another, the direction of propagation of light in the second medium changes. This phenomenon is known as refraction of light. ($\frac{1}{2}$ Mark)
$\bullet$ Laws of refraction of light
(1) 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. (1 Mark)
(2) The ratio of 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. (1 Mark)
(ii)
$\text{P} = \frac{1}{\text{f}}$ ($\frac{1}{2}$ Mark)
$\text{P}_1 = \frac{1}{20 \text{ cm}} = \frac{100}{20 \text{ m}}$
$\text{P}_1 = +5 \text{ D}$ ($\frac{1}{2}$ Mark)
$\text{P}_2 = \frac{1}{-10 \text{ cm}} = \frac{-100}{10 \text{ m}}$
$\text{P}_2 = -10 \text{ D}$ ($\frac{1}{2}$ Mark)
Net Power of the lens combination = $\text{P} = \text{P}_1 + \text{P}_2$
$\text{P} = +5 \text{ D} - 10 \text{ D}$
$\text{P} = -5 \text{ D}$ ($\frac{1}{2}$ Mark)
The combination will behave as a diverging lens (concave lens). ($\frac{1}{2}$ Mark)
$\bullet$ When a ray of light travels obliquely from one transparent medium to another, the direction of propagation of light in the second medium changes. This phenomenon is known as refraction of light. ($\frac{1}{2}$ Mark)
$\bullet$ Laws of refraction of light
(1) 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. (1 Mark)
(2) The ratio of 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. (1 Mark)
(ii)
$\text{P} = \frac{1}{\text{f}}$ ($\frac{1}{2}$ Mark)
$\text{P}_1 = \frac{1}{20 \text{ cm}} = \frac{100}{20 \text{ m}}$
$\text{P}_1 = +5 \text{ D}$ ($\frac{1}{2}$ Mark)
$\text{P}_2 = \frac{1}{-10 \text{ cm}} = \frac{-100}{10 \text{ m}}$
$\text{P}_2 = -10 \text{ D}$ ($\frac{1}{2}$ Mark)
Net Power of the lens combination = $\text{P} = \text{P}_1 + \text{P}_2$
$\text{P} = +5 \text{ D} - 10 \text{ D}$
$\text{P} = -5 \text{ D}$ ($\frac{1}{2}$ Mark)
The combination will behave as a diverging lens (concave lens). ($\frac{1}{2}$ Mark)