Reflection of Light — ICSE Class 9 Physics Important Questions

Answer: (b) equal to the angle of reflection.

The first law of reflection states $\angle i=\angle r$i= r, both measured from the normal.

Answer: (c) virtual and erect.

A plane mirror forms a virtual, erect image of the same size, laterally inverted, and as far behind the mirror as the object is in front.

Answer: (b) $60^\circ$60^.

The angle of incidence is measured from the normal: $90^\circ-30^\circ=60^\circ$90^-30^=60^. By the law of reflection the angle of reflection is also $60^\circ$60^.

Answer: (b) gives an erect, diminished image and a wide field of view.

A convex mirror always forms a virtual, erect, diminished image and, being diverging, covers a large field of view, letting the driver see more traffic behind.

Answer: (a) Since the focus $F$F lies exactly midway between the pole $P$P and the centre of curvature $C$C, the focal length is half the radius of curvature, $f=\dfrac{R}{2}$f=R/2. R correctly explains A.

Lateral inversion is the apparent left-right reversal of an image formed by a plane mirror: the right side of the object appears as the left side of the image and vice versa.

Example: The word AMBULANCE is written laterally inverted on the front of the vehicle so that a driver ahead reads it correctly in the rear-view mirror. (The mirror image of the letter 'p' looks like 'q'.)

Centre of curvature $(C)$(C): the centre of the imaginary sphere of which the mirror forms a part. It lies in front of a concave mirror on the principal axis.

Principal focus $(F)$(F): the point on the principal axis at which rays of light coming parallel to the axis actually meet (converge) after reflection from a concave mirror.

Laws of reflection:

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

Here $\angle i$i (between incident ray and normal) equals $\angle r$r (between reflected ray and normal).

When the object is beyond $C$C, two standard rays are drawn: a ray parallel to the principal axis reflects through the focus $F$F, and a ray through $C$C retraces its path. They meet between $F$F and $C$C.

Nature of image: real, inverted and diminished, formed between $F$F and $C$C.

(a) Plane mirror: used as a looking glass (dressing mirror) because it forms an erect image of the same size as the object.

(b) Concave mirror: used as a shaving/make-up mirror because, when the face is placed within the focus, it gives a magnified, erect, virtual image. (Also used in torches and headlights as a reflector.)

(c) Convex mirror: used as a rear-view mirror in vehicles and as a security mirror because it gives an erect, diminished image with a wide field of view.

(a) When the object is between $P$P and $F$F, the reflected rays diverge and must be produced backward to meet behind the mirror, giving a virtual image.

(b) Nature: virtual and erect. Position: behind the mirror. Size: magnified (larger than the object).

(c) A shaving mirror or a dentist's mirror uses a concave mirror in this way to give an enlarged erect image.

(a) For $\theta=90^\circ$=90^:
$n=\frac{360^\circ}{90^\circ}-1=4-1=3.$n=360^/90^-1=4-1=3.
Three images are formed.

(b) For $\theta=60^\circ$=60^:
$n=\frac{360^\circ}{60^\circ}-1=6-1=5.$n=360^/60^-1=6-1=5.
Five images are formed.

(c) As the angle $\theta$ between the mirrors is decreased, $\dfrac{360^\circ}{\theta}$360^/ increases, so the number of images increases. When the mirrors become parallel $(\theta=0^\circ)$(=0^), the number of images becomes infinite (theoretically), as seen between two parallel mirrors in a barber's shop.

(i) The angle of incidence is measured from the normal, not the mirror. Angle of incidence $=90^\circ-35^\circ=55^\circ$=90^-35^=55^.

(ii) By the law of reflection, angle of reflection $=$= angle of incidence $=55^\circ$=55^.

(iii) Angle between incident and reflected rays $=\angle i+\angle r=55^\circ+55^\circ=110^\circ$= i+ r=55^+55^=110^.

(iv) When a plane mirror is rotated by an angle $\theta$ with the incident ray fixed, the reflected ray rotates by $2\theta$2. Here $\theta=10^\circ$=10^, so the reflected ray rotates by $2\times10^\circ=20^\circ$2×10^=20^.