Reflection of Light Class 9 ICSE Important Questions with Answers

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) $+20\ \text{cm}$+20 cm.

$f=\dfrac{R}{2}=\dfrac{40}{2}=20\ \text{cm}$f=R/2=40/2=20 cm. The focus of a convex mirror lies behind the mirror, on the side opposite to the incoming light, so its focal length is taken as positive. (A concave mirror's focal length is negative.)

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, take two rays from the top of the object: a ray parallel to the principal axis reflects through $F$F, and a ray through $C$C strikes the mirror normally and retraces its path. The reflected rays meet between $F$F and $C$C, where the image forms.

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

Sign convention: distances are measured from the pole; those in front of the mirror (on the object side) are negative. So $u=-20\ \text{cm}$u=-20 cm and, for a concave mirror, $f=-15\ \text{cm}$f=-15 cm.

Mirror formula: $\dfrac1v+\dfrac1u=\dfrac1f$1/v+1/u=1/f
$\frac1v=\frac1f-\frac1u=-\frac{1}{15}+\frac{1}{20}=\frac{-4+3}{60}=-\frac{1}{60}\ \Rightarrow\ v=-60\ \text{cm}.$1/v=1/f-1/u=-1/15+1/20=(-4+3)/60=-1/60 ⇒ v=-60 cm.
Magnification $m=-\dfrac{v}{u}=-\dfrac{-60}{-20}=-3$m=-v/u=--60/-20=-3.

The image is formed $60\ \text{cm}$60 cm in front of the mirror ($v$v negative), so it is real; $m=-3$m=-3 means it is inverted and three times the size of the object. (The object lies between $F$F and $C$C, so the image is beyond $C$C.)

(i) It passes through the principal focus $F$F.

(ii) It becomes parallel to the principal axis.

(iii) It strikes the mirror normally and retraces its path (comes back along itself).

(iv) The rays converge to a single point on the focal plane (the plane through $F$F perpendicular to the principal axis).

(For a convex mirror, ray (i) appears to come from $F$F behind the mirror, a ray directed towards $F$F becomes parallel to the axis, and a ray directed towards $C$C retraces its path.)

(a) When the object is between $P$P and $F$F, the reflected rays diverge. They never meet in front of the mirror, but when produced backwards (dashed) they appear to come from a point behind the mirror, where a virtual image forms.

(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) Each image formed in one mirror acts as an object for the other mirror, so reflection goes on back and forth and a long row of images forms (theoretically infinitely many). Some light is absorbed at every reflection and the later images are farther away, so they grow fainter.

(b) Take $A$A at $0$0 and $B$B at $30\text{ cm}$30 cm, object at $10\text{ cm}$10 cm.

• Image of the object in $A$A: $10\text{ cm}$10 cm behind $A$A.
• Object in $B$B: $20\text{ cm}$20 cm in front of $B$B, so its image is $20\text{ cm}$20 cm behind $B$B, i.e. $50\text{ cm}$50 cm from $A$A; its image in $A$A is $50\text{ cm}$50 cm behind $A$A.
• First image in $A$A ($10\text{ cm}$10 cm behind $A$A) is $40\text{ cm}$40 cm from $B$B; its image in $B$B is $70\text{ cm}$70 cm from $A$A; its image in $A$A is $70\text{ cm}$70 cm behind $A$A.

So the first three images in $A$A are $\boxed{10,\ 50\text{ and }70\text{ cm}}$10, 50 and 70 cm behind it.

(c) A periscope (two parallel plane mirrors, each at $45^\circ$45° to the tube); mirrors on opposite walls of a barber's shop or tailor's shop, so that a person can see the back of the head or clothes.

(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 $\theta$θ with the incident ray fixed, the reflected ray rotates by $2\theta$2θ. Here $\theta=10^\circ$θ=10°, so the reflected ray rotates by $2\times10^\circ=20^\circ$2×10°=20°.