# Light - Reflection and Refraction : NCERT Exemplar

Short Answer Type Questions : -

Q.1    Identify the device used as a spherical mirror or lens in following cases, when the image formed is virtual and erect in each case.
(a) Object is placed between device and its focus, image formed is enlarged and behind it.
(b) Object is placed between the focus and device, image formed is enlarged and on the same side as that of the object.
(c) Object is placed between infinity and device, image formed is diminished and between focus and optical centre on the same side as that of the object.
(d) Object is placed between infinity and device, image formed is diminished and between pole and focus, behind it.
Sol.

(a) Concave mirror
(b) Convex lens
(c) Concave lens
(d) Convex mirror

Q.2    Why does a light ray incident on a rectangular glass slab immersed in any medium emerges parallel to itself? Explain using a diagram.

[Delhi 2013]

Sol. In a glass slab the opposite faces are parallel, so the angle of refraction by first interface, becomes
angle of incidence at the second interface i.e. $\angle \,{r_1} = \angle \,{r_2}$
By the principal of reversibility, the emergent ray bends such that $\angle \,e = \angle \,i$
Thus the incident ray and emergent ray become parallel to each other.

Q.3    A pencil when dipped in water in a glass tumbler appears to be bent at the interface of air and water. Will the pencil appears to be bent to the same extent, if instead of water we use liquids like, kerosene or turpentine. Support your answer with reason.
Sol.       The pencil dipped in water appears bent due to refraction of light. The angle of refraction depends on the refractive index of the medium. Hence it will appear to be bent to different extent in different media. Greater the refractive index, more the pencil will appear to be bent.

Q.4    Refractive index of diamond with respect to glass is 1.6 and absolute refractive index of glass is 1.5 . Find out the absolute refractive index of diamond.
Sol.

${n_{dg}} = 1.6$
${n_g} = 1.5$
${n_{dg}} = {{{n_d}} \over {{n_g}}}$
$\,{n_d} = {n_{dg}} \times {n_g}$ = 1.5 × 1.6 = 2.3

Q.5    A convex lens of focal length 20 cm can produce a magnified virtual as well as real image. Is this a correct statement? If yes, where shall the object be placed in each case for obtaining these images?
Sol.      Yes it is a correct statement.

Magnified, virtual image will be formed when object distance is less than 20 m.
Magnified, real image will be formed when object distance lies between 20 m to 40 m.

*Q.6    Sudha finds out that the sharp image of the window pane of her science laboratory is formed at a distance of 15 cm from the lens. She now tries to focus the building visible to her outside the window instead of the window pane without disturbing the lens. In which direction will she move the screen to obtain a sharp image of the building? What is the approximate focal length of this lens?
Sol.      As the object distance increases, image distance decreases. Sudha should try to move the screen towards the lens to obtain sharp image of buildings.

The approximate focal length is 15 cm.

Q.7    How are power and focal length of a lens related? You are provided with two lenses of focal length 20 cm and 40 cm respectively. Which lens will you use to obtain more convergent light?
Sol.       $P = {1 \over f}$

The lens having focal length 20 cm will give more convergent light.

Q.8    Under what condition in an arrangement of two plane mirrors, incident ray and reflected ray will always be parallel to each other, whatever may be angle of incidence. Show the same with the help of diagram.
Sol.        When the two plane mirror are placed perpendicular to each other. Angle between incident ray, reflected ray = 2i
Angle between second incident ray, reflected ray = 180 - 2i
Sum of the angle $= {180^ \circ }$
$\Rightarrow$ Incident ray is parallel to reflected ray.

Q.9    Draw a ray diagram showing the path of rays of light when it enters with oblique incidence.
(i) from air into water ; (ii) from water into air.

Sol.      (i) (ii) Long Answer Type Questions : -

Q.10    Draw ray diagrams showing the image formation by a convex mirror when an object is placed
(a) at infinity
(b) at finite distance from the mirror
Sol.        (a) (b) Q.11    The image of a candle flame formed by a lens is obtained on a screen placed on the other side of the lens. If the image is three times the size of the flame and the distance between lens and image is 80 cm, at what distance should the candle be placed from the lens? What is the nature of the image at a distance of 80 cm and the lens?
Sol.

m = – 3
v = 80 cm.
$m = {v \over u}$
$\,u\, = {{ - 80} \over 3}\,cm$
$=- 26.67\,cm.$
The image is real and lens is convex lens.

Q.12    Size of image of an object by a mirror having a focal length of 20 cm is observed to be reduced to 1/3rd of its size. At what distance the object has been placed from the mirror? What is the nature of the image and the mirror?
Sol.           As image could be real or virtual i.e inverted or erect.

Case I
$m = {{ - 1} \over 3}$ (In case of concave mirror)

f = – 20 cm.
$m = {{ - 1} \over 3} = {{ - v} \over u}$
$\,v = {u \over 3}$
${1 \over f} = {1 \over v} + {1 \over u}$
${{ - 1} \over {20}} = {3 \over u} + {1 \over u} = {4 \over u}$
$\,u\, = \, - 80\,cm.$
Case II
$m\, = {{ + 1} \over 3}$ (In case of convex mirror)
f  = + 20 cm.
$m\, = {{ + 1} \over 3} = {{ - v} \over u}$
$\,v\, = {{ - u} \over 3}$
${1 \over f} = {1 \over v} + {1 \over u}$
${1 \over {20}} = {{ - 3} \over 4} + {1 \over 4} = {{ - 2} \over u}$

Therefore,  u = – 40 cm.

Q.13    Define power of a lens. What is its unit? One student uses a lens of focal length 50 cm and another of 50 cm. What is the nature of the lens and its power used by each  of them?
Sol.

Power of a lens is the degree of its convergence or divergence. It is the reciprocal of it's focal length in meters. It's unit is dioptre.
${P_1} = {1 \over f} = {{100} \over {50}} =+ 2D\, - Convex lens.$
${P_2} = {1 \over f} = {{ - 100} \over {50}} =- 2D\, - Concave lens.$

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