Chapter 10 · Light - Reflection and Refraction

NCERT Exemplar Class 10 Science Chapter 10 Long Answer Questions

All 9 questions of NCERT Exemplar Class 10 Science Chapter 10 Long Answer Questions (Light - Reflection and Refraction), in the book's order, each with its answer and a step-by-step solution.

1. Q30
   Draw ray diagrams ​showing the ​image ​formation by a concave mirror when an object is placed

   1. (a)between pole and ​focus of the mirror
   2. (b)​between focus and centre ​of curvature of the mirror
   3. (c)at centre ​of curvature of the mirror
   4. (d)a little beyond centre of curvature of the mirror
   5. (e)at infinity
   Show answer

   The image is real and inverted for all positions beyond $F$, and virtual, erect and enlarged between $P$ and $F$.

   (a) Object between $P$ and $F$ ($u=-6$ cm): the image is virtual, erect and enlarged, $v=+15$ cm behind the mirror, $m=+2.5$. (b) Object between $F$ and $C$ ($u=-15$ cm): real, inverted and enlarged, $v=-30$ cm, beyond $C$, $m=-2$. (c) Object at $C$ ($u=-20$ cm): real, inverted and the same size, $v=-20$ cm at $C$ itself, $m=-1$. (d) Object a little beyond $C$ ($u=-30$ cm): real, inverted and diminished, $v=-15$ cm, between $F$ and $C$, $m=-0.5$. (e) Object at infinity: real, inverted and highly diminished to a point, formed at the focus, $v=f=-10$ cm.

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2. Q31
   Draw ray diagrams showing the image formation by a convex ​lens when an object is ​placed

   1. (a)between optical centre and focus of the lens
   2. (b)between focus and twice the focal length of the lens
   3. (c)at twice ​the focal ​length ​of the lens
   4. (d)at infinity
   5. (e)at the ​focus ​of the lens
   Show answer

   Virtual, erect and enlarged between O and F; real and inverted in the other cases, a point at F for an object at infinity.

   (a) Object between $O$ and $F$ ($u=-6$ cm): the image is virtual, erect and enlarged, $v=-15$ cm on the same side as the object, $m=+2.5$. (b) Object between $F$ and $2F$ ($u=-15$ cm): real, inverted and enlarged, $v=+30$ cm, beyond $2F$ on the other side, $m=-2$. (c) Object at $2F$ ($u=-20$ cm): real, inverted and the same size, $v=+20$ cm at $2F$ on the other side, $m=-1$. (d) Object at infinity: real, inverted and highly diminished to a point, formed at the focus, $v=f=+10$ cm. (e) Object at the focus ($u=-10$ cm): the emergent rays are parallel, so the image is formed at infinity and is real, inverted and highly enlarged.

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3. Q32
   Write laws ​of refraction. Explain the same ​with ​the help of ray diagram, when a ray of ​light passes through a rectangular glass ​slab.
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   Incident ray, refracted ray and normal are coplanar, and $\sin i/\sin r$ is constant; in a slab the emergent ray is parallel to the incident ray.

   Laws of refraction. First law: the incident ray, the refracted ray and the normal to the interface at the point of incidence all lie in the same plane. Second law (Snell's law): for light of a given colour passing between a given pair of media, $\frac{\sin i}{\sin r}=n_{21}$, a constant equal to the refractive index of the second medium with respect to the first. Applied to a rectangular glass slab with $n=1.5$ and $i=60^{\circ }$: at the first face the ray enters the denser glass and bends towards the normal, $\sin r=\frac{\sin 60^{\circ }}{1.5}=0.5774$, so $r=35.3^{\circ }$. Because the faces are parallel it meets the second face at the same $35.3^{\circ }$, and there it enters the rarer air and bends away from the normal, $\sin e=1.5\sin r=0.8660$, so $e=60^{\circ }=i$. The emergent ray is therefore parallel to the incident ray, displaced sideways by $d=\frac{t\sin (i-r)}{\cos r}=2.05$ cm for a $4$ cm slab.

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4. Q33
   Draw ray diagrams showing the image formation ​by a ​concave lens when an object is placed

   1. (a)​at the focus of the lens
   2. (b)between focus and twice the focal ​length of the lens
   3. (c)​beyond twice ​the ​focal length of the lens
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   In all three positions a concave lens forms a virtual, erect, diminished image on the object's side, between the optical centre and the focus.

   (a) Object at the focus ($u=-10$ cm): the image is virtual, erect and diminished to half, $v=-5$ cm on the same side as the object, $m=+0.5$. (b) Object between $F$ and $2F$ ($u=-15$ cm): virtual, erect and diminished, $v=-6$ cm on the same side, $m=+0.4$. (c) Object beyond $2F$ ($u=-30$ cm): virtual, erect and diminished to a quarter, $v=-7.5$ cm on the same side, $m=+0.25$. In every case the image lies between the optical centre and the focus on the object's own side, and a concave lens can form no other kind of image of a real object.

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5. Q34
   Draw ray diagrams ​showing the ​image formation by a ​convex mirror when an object is placed

   1. (a)at infinity
   2. (b)​at finite distance ​from the mirror
   Show answer

   At infinity the image is a point at $F$; at a finite distance it is virtual, erect, diminished, between $P$ and $F$.

   (a) Object at infinity: the image is virtual, erect and highly diminished to a point, formed at the focus $F$ behind the mirror, $v=f=+10$ cm, $m\to 0$. (b) Object at a finite distance (taking $u=-30$ cm): the image is virtual, erect and diminished, formed $v=+7.5$ cm behind the mirror, between the pole $P$ and the focus $F$, with $m=+0.25$. Whatever finite distance is chosen, $m=\frac{f}{f-u}$ lies between $0$ and $1$, so a convex mirror always gives an erect, diminished, virtual image between $P$ and $F$.

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6. Q35
   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?
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   The candle is 26.7 cm from the lens, which is convex ($f=+20\text{cm}$); the image is real, inverted and magnified three times.

   The candle should be placed $\frac{80}{3}=26.7$ cm from the lens (that is $u=-26.7$ cm, on the side away from the screen). The lens formula then gives $\frac{1}{f}=\frac{1}{80}+\frac{3}{80}=\frac{1}{20}$, so $f=+20$ cm: the lens is a convex (converging) lens. The image obtained $80$ cm from the lens is real, inverted and magnified three times.

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7. Q36
   Size of image of an object by a mirror having a focal length of $20$ cm is observed to be ​reduced ​to $1/3$rd of ​its size. At what ​distance the object ​has been ​placed ​from ​the mirror? What is the nature of the image and the mirror?
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   The object is 80 cm from the mirror; the image is real, inverted and one third the size, and the mirror is concave.

   Taking the mirror as concave, as the NCERT Exemplar answer list does: the object is placed $80$ cm in front of the mirror, $u=-80$ cm (that is $u=4f$, twice the radius of curvature). The image is then at $v=\frac{u}{3}=-26.7$ cm, in front of the mirror between $F$ and $C$, with $m=-\frac{1}{3}$: the image is real, inverted and diminished to one third, and the mirror is a concave mirror. If instead the printed focal length is read as $+20$ cm, the mirror is convex, the one-third reduction needs $u=-2f=-40$ cm and gives a virtual, erect image at $v=+13.3$ cm behind the mirror; the question does not print a sign, and the answer list takes the concave reading.

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8. Q37
   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?
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   Power is the reciprocal of focal length in metres, unit dioptre: the first lens is convex, $+2\text{D}$; the second concave, $-2\text{D}$.

   Power of a lens: the degree of convergence or divergence of light rays that a lens produces, measured as the reciprocal of its focal length in metres, $P=\frac{1}{f}$. Its unit is the dioptre (D); $1$ D is the power of a lens of focal length $1$ m, so $1\text{D}=1{\text{m}}^{-1}$. First student: $f=+50$ cm $=+0.5$ m, so $P=\frac{1}{0.5}=+2$ D, and the lens is a convex (converging) lens. Second student: $f=-50$ cm $=-0.5$ m, so $P=\frac{1}{-0.5}=-2$ D, and the lens is a concave (diverging) lens.

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9. Q38
   A student focussed ​the image of a ​candle flame on a white screen using a convex lens. He noted down the ​position of the candle screen and the lens as under Position of candle = $12.0$ cm Position of convex lens = $50.0$ cm Position of the screen = $88.0$ cm

   1. (i)What is the focal length of the convex lens?
   2. (ii)Where will the image be formed if he shifts the ​candle towards ​the lens at a position of $31.0$ cm?
   3. (iii)​What will be the nature of the image formed if he further shifts the candle towards the lens?
   4. (iv)Draw a ray ​diagram to show the formation of the image in case (iii) as said above.
   Show answer

   $f=+19.0\text{cm}$; the image is at infinity with the candle at the focus, and virtual, erect and magnified when it is closer.

   (i) The object and image distances are $u=12.0-50.0=-38.0$ cm and $v=88.0-50.0=+38.0$ cm, so $\frac{1}{f}=\frac{1}{38.0}+\frac{1}{38.0}=\frac{1}{19.0}$ and the focal length is $f=+19.0$ cm. (ii) At the $31.0$ cm mark the candle is at $u=-19.0$ cm, exactly at the focus, so $\frac{1}{v}=\frac{1}{19.0}-\frac{1}{19.0}=0$ and the image is formed at infinity: the refracted rays leave the lens parallel and no screen at any finite distance can catch it. (iii) Shifting the candle still closer puts it between the optical centre and the focus, and the image then becomes virtual, erect and magnified, formed on the same side of the lens as the candle. (iv) With the candle at $u=-9.5$ cm, for example, $v=-19.0$ cm and $m=+2$: the ray diagram shows the two emergent rays diverging, with their dashed backward extensions meeting $19.0$ cm from the lens to give an erect image twice the size of the flame.

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