Case Based Questions for Ch 9 Light β Reflection and Refraction Class 10 Science
Important Questions1
A ray of light travelling from a rarer medium to a denser medium slows down and bends towards the normal. When it travels from denser medium to a rarer medium, it speeds up and bends away from the normal.
Consider an analogy to assist in our understanding of these two important principles. Suppose that a fast car is travelling across the road towards a thick mud at an angle, the mud slows down one side of the car, and the path of the car bends.
The more it is slowed, the more it bends.
Upon exiting the thick mud on the opposite side, the car speeds up and achieves its original speed. In effect, this analogy would be representative of light wave crossing two boundaries.
At the first boundary (the road to thick mud boundary), the light ray (or the car) would be slowing down; and at the second boundary (the mud to road boundary), the light ray (or the car) would be speeding up. We can apply our two important principles listed above and predict the direction of bending and the path of the car as it travels through the thick mud. As indicated in the diagram, upon entering the mud, the car slows down and the path of the car bends towards the normal (perpendicular line drawn to the surface). Upon exiting the mud, the car speeds up and the path of the car bends away from the normal. The path of the car is closer to the normal in the slower medium and farther from the normal in the faster medium.
This analogy can be extended to the path of a light ray as it passes from air into and out of a rectangular block of glass.
(i) A student studies that when a ray of light travels from air into the glass slab, it bends towards the normal. But as refracted ray emerges out of the glass slab to the vacuum, it bends away from the normal, as shown.
Which option explains the law of refraction of light through the glass slab?
(a) Light always bends towards the normal in a glass slab.
(b) Ray of light travelling in the air is always considered as the incident ray, and the one in the glass is the refracted ray.
(c) The incident ray, the refracted ray, and the normal to the interface always lie on the same plane.
(d) Ray of light always travels in a straight path irrespective of change in medium.
(ii) A student studies that speed of light in air is 300000 km/s whereas that of speed in a glass slab is about 197000 km/s. What causes the difference in speed of light in these two media?
(a) Difference in density
(b) Difference in amount of light
(c) Difference in direction of wind flow
(d) Difference in temperature
(iii) Rahul conducts an experiment using an object of height 10 cm and a concave lens with focal length 20 cm. The object is placed at a distance of 25 cm from the lens. Can the image be formed on a screen?
(a) Yes, as the image formed will be real.
(b) No, as the image formed will be inverted.
(c) No, as the image formed will be virtual.
(d) Yes, as the image formed will be erect.
(iv) A ray of light continues moving along the same path while passing through air β glass interface. The angle of incidence for the ray is
(a) zero
(b) 90Β°
(c) less than 90Β°
(d) greater than 90Β°
Answer
(i) β (c); All the refracted rays follow the first law of refraction i.e., the incident ray, the refracted ray and the normal to the interface of two transparent media at a point of incidence, all lie in the same plane.
(ii) β (a); The speed of light varies with density as the medium with higher density decreases the speed of light and medium with lower density increases the speed of light.
(iii) β (c); Here f = -20 cm [β΅ concave lens]
u = -25cm ; v = ?
Using lens formula
Β 
The negative sign shows that the image is formed in front of the lens and the image is virtual. So, it cannot be obtained on a screen.
(iv) β (a); No bending of light occurs when light is incident normally on a boundary of two media since angle of incidence and angle of refraction both are zero.
2
We know that the characteristics of image formed by a concave mirror depend on the position of the object with respect to the mirror.
When an object is placed between F and infinity, the image formed is real and inverted. But when the object is placed between F and mirror it cannot be obtained on the screen. The image formed in this case is virtual, erect and magnified. Such image may be seen by looking in the mirror directly.
When the object is moved from focus towards infinity, the image moves from infinity towards focus and its size decreases.
When object is placed at 2F image of the same size is formed at 2F, itself.
(i) If an object is placed 10 cm in front of a concave mirror of focal length 20 cm, the image will be
(a) real, erect, magnified
(b) real, inverted, diminished
(c) virtual, erect and magnified
(d) virtual erect and diminished
(ii) The minimum distance between the object and its real image for concave mirror is
(a) Zero
(b) F
(c) 2F
(d) 4F
(iii) A candle flame 3 cm high is placed at a distance of 3m from a wall. How far from the wall must a concave mirror be placed in order that it may form an image of the flame 9 cm high on the wall ?
(a) 1.5 m
(b) 2.5 m
(c) 3 m
(d) 4.5 m
(iv) An object is placed near a concave mirror at a distance of one β fourth the radius of curvature of the concave mirror. Which ray diagram shows the incident rays, reflected rays, and the position and nature of the image formed ? 
Answer
(i) β (c); Concave mirror forms erect and enlarged image when held closer to the cavity.(ii) β (a); Concave mirror forms a real and inverted image at 2F of the object kept at 2F.
(iii) β (d);
Clearly, the image distance = u + 3
m = -v/u = I/O ; β -(u + 3)/u = -9/3
Distance of wall from the mirror = u + 3 = 1.5 + 3 = 4.5 m
(iv) β (a).
3
If rays parallel to the axis fall on thin lens, they will be focused to a point called the focal point, F. This will not be precisely true for a lens with spherical surfaces. But it will be very nearly true; i.e., parallel rays will be focused to a tiny region that is nearly a point, if the diameter of the lens is small compared to radii of curvature of the two lens surface. This condition is satisfied by a thin lens. 
By drawing the same three rays we can determine the image position for diverging lens. 
To find the image point by drawing rays would be difficult if we had to determine all the refractive angles. So, to find the image point, we need to consider only three rays which shows an arrow as the object and a converging lens forming an image to the right. The three rays are drawn as follows:
(i) Ray 1 is drawn parallel to the axis; therefore it is refracted by the lens so that it passes along a line through the local point F.
(ii) Ray 2 is drawn on a line, passing through the other focal point F and emerges from the lens parallel to the axis.
(iii) Ray 3 is directed towards the very centre of the lens where the two surfaces are essentially parallel to each other; this ray therefore emerges from the lens at the same angle as it entered.
Any two of these rays will suffice to locate the image point, which is the point where they intersect. In this way, we can find the image point for one point of the object. The image points for all other points on the object can be found similarly to determine complete image of the object.
(i) How will the image formed by convex lens be affected if the upper half of the lens is wrapped with the black paper ?
(ii) What will be the object distance to form an image twice the size of object, using a convex lens of focal length 20 cm?
(iii) The refractive index of glass with respect to air is 3/2 and the refractive index of water with respect to air is 4/3. What will be the refractive index of glass with respect to water?
Answer
(i) Brightness of the image will be reduced.(ii) For virtual image, u < f and for real image, u lies between f and 2f. so, object will be between 20 cm and 40cm.
(iii) nga = refractive index of glass with respect to air
nga = 3/2
Now, nag = 1/(3/2)
nwa = refractive index of water w.r.t air = 4/3
and ngw = refractive index of glass w.r.t water
Then, nwa . ngw .nag = 1
ngw = (1Γ3Γ3)/(4Γ2) = 9/8
ngw = 1.125
4
The curved surface of a spoon can be considered as a spherical mirror. A highly smooth polished surface is called mirror. The mirror whose reflecting surface is curved inwards or outwards is called a spherical mirror. Inner part works as a concave mirror and the outer bulging part acts as a convex mirror. The center of the reflecting surface of a spherical mirror is called pole and the radius of the sphere of which the mirror is formed is called radius of curvature.
(i) When a concave mirror is held towards the sun and its sharp image is formed on a piece of carbon paper for some time, a hole is burnt in the carbon paper. What is the name given to the distance between the mirror and carbon paper?
(a) Radius of curvature
(b) Focal length
(c) Principal focus
(d) Principal axis
(ii) The distance between pole and focal point of a spherical mirror is equal to the distance between
(a) pole and center of curvature
(b) focus point and center of curvature
(c) pole and object
(d) object and image.
(iii) The focal length of a mirror is 15 cm. The radius of curvature is
(a) 15 cm
(b) 30 cm
(c) 45 cm
(d) 60 cm
(iv) The normal at any point on the mirror passes through
(a) focus
(b) pole
(c) center of curvature
(d) any point
(v) In a convex spherical mirror, reflection of light takes place at
(a) a flat surface
(b) a bent β in surface
(c) a bulging β out surface
(d) an uneven surface
Answer
(i) β (b); The focal length of a concave mirror is the distance between its pole and principal focus.(ii) β (b)
(iii) β (b); Given that, f = 15 cm
Radius of curvature of a spherical mirror is given as
R = 2F
β΄ R = 2 Γ 15 = 30 cm
(iv) β (c); In a spherical mirror, normal drawn at any point passes through the centre of curvature.
(v) β (c)
5
The spherical mirror forms different types of images when the object is placed at different locations.
When the image is formed on screen, the image is real and when the image does not form on screen, the image is virtual. When the two reflected rays meet actually, the image is real and when they appear to meet, the image is virtual.
A concave mirror always forms a real and inverted image for different positions of the object. But if the object is placed between the focus and pole, the image formed is virtual and erect.
A convex mirror always forms a virtual, erect and diminished image. A concave mirror is used as doctor's head mirror to focus light on body parts like eyes, ears, nose etc., to be examined because it can form erect and magnified image of the object. The convex mirror is used as a rear view mirrors in automobiles because it can form an small and erect image of an object.
(i) When an object is placed at the centre of curvature of a concave mirror, the image formed is
(a) larger than the object
(b) smaller than the object
(c) same size as that of the object
(d) highly enlarged.
(ii) No matter how far you stand from a mirror, your image appears erect. The mirror is likely to be
(a) plane
(b) concave
(c) convex
(d) either plane or convex.
(iii) A child is standing in front of a magic mirror. She finds the image of her head bigger, the middle portion of her body of the same size and that of the legs smaller. The following is the order of combinations for the magic mirror from the top.
(a) Plane, convex and concave
(b) Convex, concave and plane
(c) Concave, plane and convex
(d) Convex, plane and concave
(iv) To get an image larger than the object, one can use
(a) convex mirror but not a concave mirror
(b) a concave mirror but not a convex mirror
(c) either a convex mirror or a concave mirror
(d) a plane mirror.
(v) A convex mirror has wider field of view because
(a) the image formed is much smaller than the object and larger number of images can be seen.
(b) the image formed is much closer to the mirror
(c) both (a) and (b)
(d) none of these.
Answer
(i) β (c);
When the object is placed at the centre of curvature of concave mirror, the image formed is real, inverted and of the same size as that of the object.
(ii) β (d); The image is erect in a plane mirror and also in a convex mirror, for all positions of the object.
(iii) β (c); As the image of head is bigger, the upper portion of magic mirror is concave. The middle portion of the image is of same size, so, middle portion of magic mirror is plane. Now, the image of legs looks smaller, therefore, the lower portion of magic mirror is convex.
(iv) β (b)
(v) β (c)
6
The refraction of light on going from one medium to another takes place according to two laws which are known as the laws of refraction of light. These laws are
(1) The ratio of sine of angle of incidence to the sine of angle of refraction is always constant for the pair of media in contact.
sini/sinr =ΞΌ = constant
This constant is called refractive index of the second medium with respect to the first medium.
Refractive index is also defined as the ratio of speed of light in vacuum to the speed of light in medium.
(2) The incident ray, refracted ray and normal all lie in the same plane.
This law is called Snell's law of refraction.
(i) When light travels from air to glass,
(a) angle of incidence > angle of refraction
(b) angle of incidence < angle of refraction
(c) angle of incidence = angle of refraction
(d) can't say
(ii) When light travels from air to medium, the angle of incidence is 45Β° and angle of refraction is 30Β°. The refractive index of second medium with respect to the first medium is
(a) 1.41
(b) 1.50
(c) 1.23
(d) 1
(iii) In which medium, the speed of light is minimum?
(a) Air
(b) Glass
(c) Water
(d) Diamond
(iv) If the refractive index of glass is 1.5 and speed of light in air is 3 Γ 108 m/s. The speed of light in glass is
(a) 2 Γ 108 m/s
(b) 2.9 Γ 108 m/s
(c) 4.5 Γ 108 m/s
(d) 3 Γ 108 m/s
(v) Refractive index of a with respect to b is 2. Find the refractive index of b with respect to a.
(a) 0.4
(b) 0.5
(c) 0.25
(d) 2
Answer
(i) β (a); According to Snellβs law of refraction,sini/sinr > 1 or sini > sinr
or i > r.
(ii) β (a); As, 1ΞΌ2 = sini/sinr
sin 45Β°/sin 30Β° = (1/β2)/(1/2) = 1.41
(iii) β (d); As diamond has maximum value of refractive index, therefore it has minimum speed of light in medium.
(iv) β (a); As ΞΌglass = 1.5, c = 3 Γ 108 m/s
β΄ ΞΌ = c/v or 1.5 = (3 Γ 108)/v
v = 2 Γ 108 m/s
(v) β (b); Given refractive index of a with respect to b is bΞΌa = 2
β΄ Refractive index of b with respect to a is
1/ bΞΌa = aΞΌb = Β½ = 0.5