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## Goyal Brothers Light Class-9 ICSE Physics Ch-8

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Exercise 2  Multiple Choice Questions.

Exercise 2 Subjective Questions

Unit I Exercise I

### Goyal Brothers Light Class-9 ICSE Physics Ch-8

Question 1.
(a) What do you understand by the following terms?

1. Light
2. Diffused light

(b) By giving one example and one use explain or define

1. Regular reflection
2. Irregular reflection.

(a)
(i) Light : Light is a form of energy which produces in us sensation of seeing.
(ii) Diffused light : Light obtained after reflection from rough surface is known as diffused light.
It is a soft light with neither the intensity nor the glare of direct light. It is scattered and comes from all directions. It does not cause harsh shadows.
(b)
(i) Regular reflection : The phenomenon due to which a parallel beam of light travelling through a certain medium, on striking some polished surface, bounces off from it, as parallel beam, in some other direction is called regular reflection.
For example : Reflection taking place from the objects like looking glass, still water, oil, highly polished metals is regular reflection.
Regular reflection is useful in the formation of images. We can see our face in a mirror only due to regular reflection.
(ii) Irregular reflection or Diffused reflection : The phenomenon due to which a parallel beam of light, travelling through some medium, gets reflected in various possible directions, on striking some rough surface is called irregular reflection.
For example : Reflection taking place from ground, walls, trees, suspended particles in air is irregular reflection.
Use : It helps in the general illumination of places and helps us to see things around us.

Question 2.
By drawing a neat diagram define the following :

1. Mirror
2. Incident ray
3. Reflected ray
4. Angle of incidence
5. Angle of reflection
6. Normal

1. Mirror is a highly polished and smooth surface which reflects almost the entire light falling on it. A plane mirror is made by silvering one side of a glass plate as shown in figure. 2. Incident ray : The light ray striking a reflecting surface is called the incident ray.
3. Reflected ray : The light ray obtained after reflection from the surface, in the same medium in which the incident ray is travelling, is called the reflected ray.
4. Angle of incidence : The angle which the incident ray makes with the normal at the point of incidence is called the angle of incidence. It is denoted by the letter “i”.
5. Angle of reflection : The angle which the reflected ray makes with the normal at point of incidence, is called the angle of reflection. It is denoted by the letter “r”. 6. Normal : The perpendicular drawn at the point of incidence, to the surface of mirror is called normal.

Question 3.    (Goyal Brothers Light Class-9 ICSE Physics Ch-8)

State the laws of reflection.
Laws of reflection :

1. The incident ray, the reflected ray and the normal ray at the point of incidence, lie in the same plane.
2. The angle of incidence i is equal to the angle of reflection r i.e. ∠i = ∠r.

Question 4.
A ray of light strikes a plane mirror, such that angle with the mirror is 20°. What is value of angle of reflection? What is the angle between the incident ray and the reflected ray?
∵ Light ray makes an angle of 20° with the mirror
∴ ∠ABM = 20°
Angle of incidence = ∠i = 90° – 20°
∠i = 70°
∵ ∠i = r ∴ ∠r = 70° Angle between incident ray and reflected ray = ∠i + ∠r = 70°+ 70°= 140°

Question 5.
Prove experimentally that images are formed as far behind in a plane mirror as the object is in front of it.
Consider an object‘O’ situated in front of a plane mirror MM,. A ray of light which starts from point ‘O’ perpendicularly, is reflected back along the same path (See figure).
However, another ray which moves along OB is reflected along BC, obeying the laws of reflection, such that BN is the normal. Produce OA and CB backward, such that they meet at point I. Then ‘I’ is image of ‘O’. Thus, in particular OA = IA.

Question 6.     (Goyal Brothers Light Class-9 ICSE Physics Ch-8)
Prove geometrically that when plane mirror turns through a certain angle, the reflected ray turns through twice the angle.
Consider a ray of light AB, incident on plane mirror in position MM’, such that BC is the reflected ray and BN is the normal.
∠ABM = ∠CBN = ∠i
∠ABC = 2 ∠i …(i)
Let the mirror be rotated through an angle ‘0’ about point B, such that M1M1 is the new position of the mirror and BN1 is the new position of normal. As the position of the incident ray remains same, therefore new angle of the incidence is ∠ABN1 whose magnitude is (i + θ). Let BD be the reflected ray, such that ∠DNB1 is the new angle of reflection. Thus, for a given incident ray, if the plane mirror is rotated through a certain angle, then the reflected ray rotates through twice the angle.

Question 7.

(a) What do you understand by the term lateral inversion?
(b) A printed card has letters PHYSICS. By drawing the diagram show the appearance of the letters. (No ray diagram is required).

(a) Lateral Inversion. The phenomenon due to which the image of an object turns through an angle of 180° through verticle axis rather the horizontal axis, such that right side of image appears as left or vice-versa is called Lateral Inversion. During lateral inversion the left side of object appears as right side of image and vice-versa. In a way the image turns through the angle of 180° about vertical axis. Question 8.

(a) State the mirror formula for the formation of total number of images formed in two plane mirrors, held at an angle.
(b) Calculate the number of images formed in two plane mirrors, when they are held at the angle of (i) 72° (ii) 36°.

(a) If θ = Angle of inclination between two mirrors
n = number of images formed Numeral one is subtracted because of the loss of one image due to overlapping of the images.
(b) (i) When θ = 72° Question 9.      (Goyal Brothers Light Class-9 ICSE Physics Ch-8)
Draw a neat two ray diagram for the formation of images in two plane mirrors, when mirrors are

1. at right angles to each other
2. facing each other.

(a) When two mirrors are inclined at right angles
‘O’ is an object placed in between two mirrors XY and
XZ, inclined at an angle of 90°. (See figure) Taking normal incidence, I1 and I2 are the images formed in the plane mirror XY and XZ respectively as far behind the mirrors, as point ‘O’ is in front of them.
However, image I1 acts as a virtual object for image mirror XZ1 and forms an image I3. Similarly, image I2 acts as a virtual object for the image mirror XY1 and forms the image I4. The images I3 and I4 overlap to form a very bright image. Thus, on the whole three images are seen. In order to draw two-ray diagrams, from the position FE of the eye, draw two rays meeting at I3,I4 such that these ray intersect the mirror XZ at D and C.
Now draw two rays from point I1 to join C and D intersecting mirror XY at A and B. Join O with A and B.
Similarly, in order to show image I2, draw two rays from I2 to the position of eye FE, such that they intersect at H and G Join H and G to ‘O’ so as to form incident beam.
(b) When two mirrors are parallel to each other Consider two plane mirrors XY and PQ facing each other and ‘A’ as an object situated anywhere between them (See figure).
It is clear that mirror XY forms its image in mirror PQ and vice versa. These images by themsleves will act as image mirror or virtual mirrors.
Let us consider the normal incidence towards the mirror XY for the object A. First of all an image X1 is formed as far behind, as the object is in front of it. This image X1 will fall in front of mirror PQ, and hence, forms an image X2. The image X2 falls in front of mirror XY and hence, forms an image X3. Thus, it continues and infinite images can be formed.
Similarly, taking normal incidence for mirror PQ image P1,P2,P3 etc. are formed.
In order to draw ray diagram, from point A, draw a divergent beam, meeting mirror PQ at points 1 and 2. With P1 as reference point, draw rays 1, 3 and 2, 4 meeting mirror XY. With P3 as reference point draw rays, such that they enter the eye.

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