Light: Reflection and Refraction — CBSE Class 10 Science Important Questions

Answer: (b) Convex mirror. A convex mirror forms a virtual, erect and diminished image for all object positions, which is why it gives a wide field of view and is used as a rear-view mirror in vehicles.

Answer: (b) virtual, erect and diminished. Whatever the position of the object, a concave lens forms a virtual, erect, diminished image on the same side as the object, between its focus $F_1$F₁ and the optical centre $O$O.

Answer: (b) $-50$-50 cm, concave. Since $P=\frac{1}{f}$P=1/f, $f=\frac{1}{P}=\frac{1}{-2}=-0.5$f=1/P=1/-2=-0.5 m $=-50$=-50 cm. A negative power / negative focal length means a concave (diverging) lens.

Answer: (b) at C, real, inverted and the same size as the object. When an object is at the centre of curvature of a concave mirror, the image is formed at C itself, is real, inverted and of the same size ($m=-1$m=-1).

Answer: (a) Both A and R are true and R is the correct explanation of A. A ray directed through the centre of curvature meets the mirror along the normal (since the radius is perpendicular to the surface), so the angle of incidence is $0°$0°; by the law of reflection it is reflected back along the same path.

(i) The incident ray, the refracted ray and the normal to the interface of the two media at the point of incidence all lie in the same plane.
(ii) The ratio of the sine of the angle of incidence to the sine of the angle of refraction is a constant for a given pair of media and a given colour of light: $\dfrac{\sin i}{\sin r}=\text{constant}$(sin i)/(sin r)=constant. This is Snell's law of refraction; the constant is the refractive index of the second medium with respect to the first.

For a convex mirror, $f=+15$f=+15 cm and $u=-30$u=-30 cm.

Using $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$1/v+1/u=1/f:

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{15}-\frac{1}{-30}=\frac{1}{15}+\frac{1}{30}=\frac{2+1}{30}=\frac{3}{30}=\frac{1}{10}$1/v=1/f-1/u=1/15-1/-30=1/15+1/30=(2+1)/30=3/30=1/10

$v=+10$v=+10 cm.

The image is formed 10 cm behind the mirror; the positive sign shows it is virtual and erect.

Given: $h=4$h=4 cm, $u=-15$u=-15 cm, $f=-10$f=-10 cm (concave).

Position: $\frac{1}{v}+\frac{1}{u}=\frac{1}{f}$1/v+1/u=1/f

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-10}-\frac{1}{-15}=-\frac{1}{10}+\frac{1}{15}=\frac{-3+2}{30}=-\frac{1}{30}$1/v=1/f-1/u=1/-10-1/-15=-1/10+1/15=(-3+2)/30=-1/30

$v=-30$v=-30 cm → the image is 30 cm in front of the mirror.

Nature: the negative $v$v shows the image is real and inverted.

Size: $m=-\frac{v}{u}=-\frac{-30}{-15}=-2$m=-v/u=--30/-15=-2.

$h'=m\times h=-2\times4=-8$h'=m× h=-2×4=-8 cm.

The image is real, inverted, magnified (twice) and 8 cm tall, formed 30 cm in front of the mirror.

Ray diagram / nature: when the object is placed beyond 2F of a convex lens, the image is formed between F2 and 2F2 on the other side and is real, inverted and diminished (as shown).

Numerical (object at 15 cm): $f=+10$f=+10 cm, $u=-15$u=-15 cm.

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

$\frac{1}{v}=\frac{1}{f}+\frac{1}{u}=\frac{1}{10}+\frac{1}{-15}=\frac{3-2}{30}=\frac{1}{30}$1/v=1/f+1/u=1/10+1/-15=(3-2)/30=1/30

$v=+30$v=+30 cm.

Magnification $m=\frac{v}{u}=\frac{30}{-15}=-2$m=v/u=30/-15=-2. So for the 15 cm position (object between F and 2F) the image is real, inverted and magnified, formed 30 cm from the lens on the opposite side.

When light enters the glass slab it bends towards the normal (air to denser glass); when it leaves the slab it bends away from the normal (glass to rarer air) by an equal amount.

Why emergent ∥ incident: the two refracting surfaces of the slab are parallel, so the bending at the first surface is exactly cancelled by the opposite bending at the second surface. Hence the angle of emergence equals the angle of incidence, and the emergent ray is parallel to the incident ray (only shifted sideways).

Lateral displacement: it is the perpendicular distance between the direction of the incident ray (produced) and the emergent ray. Its value increases with the thickness of the slab, the angle of incidence, and the refractive index of the glass.

For a concave lens, $f=-15$f=-15 cm, and its image is virtual, on the same side as the object, so $v=-10$v=-10 cm.
Lens formula: $\dfrac{1}{v}-\dfrac{1}{u}=\dfrac{1}{f}\Rightarrow\dfrac{1}{u}=\dfrac{1}{v}-\dfrac{1}{f}=-\dfrac{1}{10}+\dfrac{1}{15}=\dfrac{-3+2}{30}=-\dfrac{1}{30}$1/v-1/u=1/f⇒ 1/u=1/v-1/f=-1/10+1/15=(-3+2)/30=-1/30
So $u=-30$u=-30 cm: the object is $30$30 cm from the lens, on the same side as the image.
Magnification $m=\dfrac{v}{u}=\dfrac{-10}{-30}=+\dfrac{1}{3}$m=v/u=-10/-30=+1/3, so the image is virtual, erect and one-third the size of the object.

A ray parallel to the axis diverges after refraction and appears to come from $F_1$F₁ (dashed); a ray through the optical centre $O$O passes undeviated. Extended back, they meet at the image.

(a) When the object is beyond C of a concave mirror, the image is formed between F and C, and is real, inverted and diminished (smaller than the object), as shown in the ray diagram. One ray parallel to the axis reflects through F, and another ray through C retraces its path; the reflected rays meet to form the image.

(b) Object at C means $u=-2f$u=-2f. Here $f=-15$f=-15 cm, so $u=-30$u=-30 cm; $h=5$h=5 cm.

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

$\frac{1}{v}=\frac{1}{f}-\frac{1}{u}=\frac{1}{-15}-\frac{1}{-30}=-\frac{1}{15}+\frac{1}{30}=\frac{-2+1}{30}=-\frac{1}{30}$1/v=1/f-1/u=1/-15-1/-30=-1/15+1/30=(-2+1)/30=-1/30

$v=-30$v=-30 cm → image is 30 cm in front of the mirror (at C).

$m=-\frac{v}{u}=-\frac{-30}{-30}=-1$m=-v/u=--30/-30=-1, so $h'=-1\times5=-5$h'=-1×5=-5 cm.

The image is real, inverted, of the same size (5 cm) as the object and formed at the centre of curvature.

(a) The power of a lens is a measure of its ability to converge or diverge a beam of light; it is the reciprocal of its focal length in metres, $P=\frac{1}{f\,(\text{m})}$P=1/(f (m)). Its SI unit is the dioptre (D), where $1\,\text{D}=1\,\text{m}^{-1}$1 D=1 m^-1. Power is positive for a convex lens and negative for a concave lens.

(b) Convex lens: $f_1=+10$f₁=+10 cm $=+0.10$=+0.10 m → $P_1=\frac{1}{0.10}=+10$P₁=1/0.10=+10 D.

Concave lens: $f_2=-20$f₂=-20 cm $=-0.20$=-0.20 m → $P_2=\frac{1}{-0.20}=-5$P₂=1/-0.20=-5 D.

Net power: $P=P_1+P_2=(+10)+(-5)=+5$P=P₁+P₂=(+10)+(-5)=+5 D.

Focal length of combination: $f=\frac{1}{P}=\frac{1}{+5}=+0.20$f=1/P=1/(+5)=+0.20 m $=+20$=+20 cm.

Since the net power (and focal length) is positive, the combination behaves as a converging (convex) lens of focal length 20 cm.

(i) Light travels slowest in diamond because it has the highest refractive index (2.42); diamond is therefore the optically densest medium in the table.

(ii) $v_{water}=\frac{c}{n}=\frac{3\times10^{8}}{1.33}\approx2.26\times10^{8}$v_water=c/n=3×10⁸1.33≈2.26×10⁸ m/s.

(iii) It means the speed of light in vacuum is 1.50 times the speed of light in glass, i.e. $n=\frac{c}{v}=1.50$n=c/v=1.50, so light slows down to $\frac{1}{1.50}$1/1.50 of its vacuum speed on entering glass.

(iv) Light bends towards the normal. Glass ($n=1.50$n=1.50) is optically denser than water ($n=1.33$n=1.33), and light entering a denser medium slows down and bends towards the normal.