Light - Reflection and Refraction Class 10 Competency Questions

20 Questions | 70 Marks

A group of students were experimenting with different mirrors and lenses. They observed the image formed by a concave mirror when the object was placed at different positions.(a) If the image formed is virtual, erect and diminished, what is the position of the object relative to the mirror? [1 mark](b) If the image formed is real, inverted and highly diminished, what is the position of the object relative to the mirror? [1 mark](c) The students then used a convex lens and placed an object at a distance equal to its focal length. Describe the nature, position, and size of the image formed. 3marks

A student is trying to read a book in a dimly lit room. He uses a magnifying glass to see the small print.(a) What type of lens is used as a magnifying glass? [1 mark](b) Explain why this lens helps in reading small print. 2marks

Assertion (A): A ray of light bends towards the normal when it travels from water to glass.Reason (R): Glass is optically denser than water. 2marks

A dentist uses a mirror to examine a patient's tooth. The image of the tooth appears larger than the actual tooth.(a) What type of mirror is used by the dentist? [1 mark](b) Draw a ray diagram to show how this mirror forms an enlarged image of the tooth. 3marks

A ray of light passes from medium A to medium B. The angle of incidence is

30^{\circ}

and the angle of refraction is

20^{\circ}

.(a) Which medium is optically denser? Justify your answer. [2 marks](b) Calculate the refractive index of medium B with respect to medium A. 2marks

A student observes the image of a distant tree in a small convex mirror fitted on a wall.(a) Explain why a convex mirror can show the full image of a tall tree even if the mirror is small. [2 marks](b) Mention one specific use of such mirrors in vehicles. 1mark

Assertion (A): A convex lens always forms a real image.Reason (R): When an object is placed between the optical centre and the focal point of a convex lens, a virtual and erect image is formed. 2marks

An object is placed at 20 cm in front of a concave mirror of focal length 15 cm.(a) Determine the position of the image formed. [2 marks](b) State whether the image is real or virtual. 1mark

A student conducts an experiment on refraction of light through a rectangular glass slab. They observe that the emergent ray is parallel to the incident ray.(a) Name the phenomenon responsible for this observation. [1 mark](b) Explain why the emergent ray is parallel to the incident ray, even though there is a lateral displacement. 2marks

A concave lens has a focal length of 20 cm. An object is placed at 30 cm from the lens.(a) Calculate the position of the image formed. [2 marks](b) What is the magnification of the image? 1mark

A ray of light enters from air into a medium with a refractive index of 1.5. If the speed of light in air is

3 \times 10^{8}

m/s, calculate the speed of light in the medium. 2marks

Assertion (A): A concave mirror can form a magnified, erect, virtual image.Reason (R): When the object is placed between the pole and the principal focus of a concave mirror, a virtual, erect and magnified image is formed behind the mirror. 2marks

A thin lens has a focal length of -10 cm.(a) Identify the type of lens. [1 mark](b) If an object is placed at 15 cm from this lens, find the position of the image. 3marks

Consider the situation where light travels from a denser medium to a rarer medium.(a) How does the light ray bend with respect to the normal? [1 mark](b) Explain the reason for this bending. 2marks

A convex mirror used as a rear-view mirror in a car has a radius of curvature of 2.0 m. If a truck is located 4.0 m from the mirror, find the position and nature of the image. 3marks

Assertion (A): The power of a convex lens is positive.Reason (R): A convex lens converges light rays and its focal length is taken as positive in the sign convention. 2marks

A student is looking at a coin at the bottom of a water tank. The coin appears to be raised.(a) Name the phenomenon responsible for this observation. [1 mark](b) Draw a ray diagram to illustrate why the coin appears raised. 3marks

An object of height 2 cm is placed at a distance of 25 cm from a concave mirror of focal length 15 cm.(a) Calculate the distance of the image from the mirror. [2 marks](b) Calculate the size of the image. 2marks

A ray of light passes through the optical centre of a convex lens.(a) What happens to the ray after refraction? [1 mark](b) Draw a ray diagram showing this and explain why it happens. 3marks

Assertion (A): A concave lens always forms a diminished image.Reason (R): The focal length of a concave lens is negative. 2marks

Answers

(a) The object is placed between the pole (P) and the principal focus (F) of the concave mirror. 1mark(b) The object is placed at infinity. [1 mark](c) When the object is placed at the focal length (F) of a convex lens, the rays emerging from it become parallel after refraction. Hence, a real, inverted image is formed at infinity. [3 marks]

(a) A convex lens. 1mark(b) A magnifying glass (convex lens) forms a virtual, erect, and magnified image when the object is placed within its focal length. This allows the small print to appear larger and thus easier to read. [2 marks]

(a) Assertion (A) is false, Reason (R) is true. 1mark(b) When light travels from water (rarer) to glass (denser), it bends towards the normal. The assertion is incorrect. Glass is optically denser than water because its refractive index is higher than that of water, causing light to slow down and bend towards the normal. [1 mark]

(a) A concave mirror. 1mark(b) [Ray diagram showing an object placed between P and F of a concave mirror, with rays parallel to the principal axis and passing through F after reflection, and a ray passing through P and reflecting back, intersecting behind the mirror to form a virtual, erect, and magnified image.] [3 marks]

(a) Medium B is optically denser than medium A because the angle of refraction (

20^{\circ}

) is less than the angle of incidence (

30^{\circ}

), indicating that the light has slowed down upon entering medium B. 2marks(b) Refractive index of medium B with respect to medium A (

n_{B A}

) =

\frac{\sin i}{\sin r} = \frac{\sin 30^{\circ}}{\sin 20^{\circ}} = \frac{0.5}{\sin 20^{\circ}}

. Using

\sin 20^{\circ} \approx 0.342

n_{B A} \approx \frac{0.5}{0.342} \approx 1.46

. [2 marks]

(a) A convex mirror has a wider field of view because it diverges light rays, allowing it to reflect light from a larger area and present a diminished, erect image of distant objects. 2marks(b) Rear-view or side-view mirrors in vehicles. [1 mark]

(a) Assertion (A) is false. 1mark(b) Reason (R) is true and explains why Assertion (A) is false. A convex lens can form both real and virtual images depending on the object's position. When the object is placed between F and O, a virtual, erect and magnified image is formed. [1 mark]

(a) Using the mirror formula:

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

. Here,

u = - 25

cm,

f = - 15

cm.

\frac{1}{v} + \frac{1}{- 25} = \frac{1}{- 15}

\frac{1}{v} = \frac{1}{- 15} + \frac{1}{25} = \frac{- 5 + 3}{75} = \frac{- 2}{75}

$v = - \frac{75}{2} = - 37.5$ cm. [2 marks]
(b) The image is virtual as the image distance $v$ is negative. [1 mark]

(a) Refraction of light. 1mark(b) When light enters from air to glass (denser medium), it bends towards the normal. When it emerges from glass to air (rarer medium), it bends away from the normal. Due to the parallel surfaces of the glass slab, the bending away from the normal at the second surface is such that the emergent ray becomes parallel to the incident ray, but is laterally displaced. [2 marks]

(a) Using the lens formula:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

. Here,

u = - 30

cm,

f = - 20

cm.

\frac{1}{v} - \frac{1}{- 30} = \frac{1}{- 20}

\frac{1}{v} + \frac{1}{30} = - \frac{1}{20}

\frac{1}{v} = - \frac{1}{20} - \frac{1}{30} = \frac{- 3 - 2}{60} = \frac{- 5}{60}

$v = - \frac{60}{5} = - 12$ cm. [2 marks]
(b) Magnification $m = \frac{v}{u} = \frac{- 12 cm}{- 30 cm} = + 0.4$ . [1 mark]

Absolute refractive index (

n

) =

\frac{Speed of light in vacuum (c)}{Speed of light in medium (v)}

1.5 = \frac{3 \times 10^{8} m/s}{v}

$v = \frac{3 \times 10^{8} m/s}{1.5} = 2 \times 10^{8}$ m/s. [2 marks]

(a) Assertion (A) is true. 1mark(b) Reason (R) is true and correctly explains the assertion. [1 mark]

(a) Concave lens (since the focal length is negative). 1mark(b) Using the lens formula:

\frac{1}{v} - \frac{1}{u} = \frac{1}{f}

. Here,

u = - 15

cm,

f = - 10

cm.

\frac{1}{v} - \frac{1}{- 15} = \frac{1}{- 10}

\frac{1}{v} + \frac{1}{15} = - \frac{1}{10}

\frac{1}{v} = - \frac{1}{10} - \frac{1}{15} = \frac{- 3 - 2}{30} = \frac{- 5}{30}

$v = - \frac{30}{5} = - 6$ cm. [3 marks]

(a) The light ray bends away from the normal. 1mark(b) When light travels from a denser medium to a rarer medium, its speed increases. This change in speed causes the light ray to bend away from the normal. [2 marks]

Radius of curvature

R = + 2.0

m. Focal length

f = R / 2 = + 1.0

m. Object distance

u = - 4.0

m.Using the mirror formula:

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

\frac{1}{v} + \frac{1}{- 4.0} = \frac{1}{1.0}

\frac{1}{v} = 1 + \frac{1}{4.0} = \frac{4 + 1}{4.0} = \frac{5}{4.0}

$v = \frac{4.0}{5} = + 0.8$ m.
The image is formed at 0.8 m behind the mirror. Since $v$ is positive, the image is virtual and erect. [3 marks]

(a) Assertion (A) is true. 1mark(b) Reason (R) is true and correctly explains Assertion (A). [1 mark]

(a) Refraction of light. 1mark(b) [Ray diagram showing light rays from the coin at the bottom of the water tank bending away from the normal as they emerge into air. The rays appear to originate from a point above the actual position of the coin, making it look raised.] [3 marks]

(a) Using the mirror formula:

\frac{1}{v} + \frac{1}{u} = \frac{1}{f}

. Here,

u = - 25

cm,

f = - 15

cm.

\frac{1}{v} + \frac{1}{- 25} = \frac{1}{- 15}

\frac{1}{v} = \frac{1}{- 15} + \frac{1}{25} = \frac{- 5 + 3}{75} = \frac{- 2}{75}

$v = - \frac{75}{2} = - 37.5$ cm. [2 marks]
(b) Height of image $h^{'} = h \times \frac{v}{u} = 2 cm \times \frac{- 37.5 cm}{- 25 cm} = 2 cm \times 1.5 = + 3.0$ cm. [2 marks]

(a) The ray of light passes through the optical centre without any deviation. 1mark(b) [Ray diagram showing a ray of light passing through the optical centre O of a convex lens and continuing in a straight line without bending.] This happens because the optical centre is the midpoint of the lens, and the surfaces on either side of it are symmetrically curved. At the optical centre, the deviation of light is negligible for a thin lens. [3 marks]

(a) Assertion (A) is true. 1mark(b) Reason (R) is true, but it does not directly explain why a concave lens always forms a diminished image. The fact that the focal length is negative is a consequence of its diverging nature, which leads to diminished images for real objects. [1 mark]

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Light - Reflection and Refraction — Competency Based Questions

Competency based questions

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