# Light – Reflection and Refraction

### One Marks Questions

Answer: The bouncing back of light from any surface is called the reflection of light.

Answer: A spherical mirror whose reflecting surface is curved inwards is called a concave mirror.

Answer: A spherical mirror whose reflecting surface is curved outwards is called a concave mirror.

Answer: The focal length of a spherical mirror is the distance between its pole and principal focus.

i. Infront of the mirror
ii. Behind the mirror

The radius of curvature = 2×focal length
( R = 2f )

Answer: Concave mirror ( When an object is placed between the pole and principal focus of a mirror )

Answer: Medium A has a greater refractive index.

Answer: The size of the image formed increases as the object is brought closer towards the convex lens.

### Two Marks Questions

Answer: The laws of reflection are
i. The angle of incidence is equal to the angle of reflection.
ii. The incident ray, the normal to the mirror at the point of incidence and reflected ray, all lie in the same plane.

The mirrors whose reflecting surfaces are spherical are called spherical mirrors.
Types:
i. Concave mirror
ii. Convex mirror

• Used in torches, searchlights, vehicle headlights.
• Used in shaving mirrors to see a larger image of the face.
• Used to concentrate sunlight to produce heat in a solar furnace.

Answer: They form erect image also they have a wider field of view. They enable the driver to view a much larger area than would be possible with a plane mirror.

• The bottom of a tank containing water appears to be raised.
• A lemon kept in water in a glass tumbler appears to be bigger than its actual size.

The power of the lens is defined as reciprocal of its focal length. Its S I unit is dioptre (D)

### Three Marks Questions

Answer: The properties of the image formed by a plane mirror are
• The image is virtual and erect.
• The size of the image is equal to that of the object.
• The image formed is as far behind the mirror as the object is in front of it.
• The image is laterally inverted.

i. At infinity.
ii. Between pole and principal focus of the mirror.
iii. At the centre of the curvature of the mirror.

a. Concave mirrors are used in a vehicle to get the powerful parallel beam of light. When the light source is kept at the principal focus of a mirror, we get a parallel beam of light after reflection by the concave mirror.

b. When an object is kept between the pole and principal focus of a mirror, the image formed is enlarged, virtual and erect.

c. The parallel rays from the sun get converged by the concave mirror.

Answer: When light travels obliquely from one medium to another, the direction of propagation of light changes in the second medium. This phenomenon is known as refraction.

The laws of refraction are :
The incident ray refracted ray, and the normal to the interface of two media at the point of incidence all lie in the same plane.
The ratio of the sine of the angle of incidence to the sine of the angle of refraction is constant. This is also known as Snell's law of refraction.
sin ⁡i / sin ⁡r
= constant
Where i= angle of incidence and
r = angle of refraction

i. At focus F2 , highly diminished, real and inverted.
ii. Between F2 and 2F2 , Diminished, real and inverted.
iii. At 2F2 , same size, real and inverted.

### Four Marks Questions

i. Optical centre: The central point of a lens is its optical centre.

ii. Centre of curvature: The centre of curvature of the lens in the centre of the sphere of which refracting surface of the lens forms a part.

iii. Aperture: The effective diameter of the circular outline of a spherical lens is called its aperture.

iv. Principal Focus: Principal focus of a spherical lens is the point on a principal axis where the parallel rays after refraction converge or appear to diverge from.

A concave lens always forms a virtual, erect image on the same side of the object.
Image-distance v = –10 cm
Focal length f = –15 cm
Object-distance = ?
Since,
1 / v
-
1 / u
=
1 / f

1 / u
=
1 / v
-
1 / f

1 / u
=
1 / (-10)
-
1 / (-15)

1 / u
=
1 / (-10)
+
1 / 15

1 / u
=
(-3+2) / 30

1 / u
=
1 / (-30)

u = - 30 cm
Thus object distance is 30 cm
Magnification m =
v / u

m =
(-10) / (-30)

m =
1 / 3

m = + 0.33
The positive sign indicates that the image is erect and virtual. The image is one-third of the size of the object.