Light class 10: CBSE previous question paper problems

Spherical mirrors

• Solution

  A1. ($\text{i}$‍) Pole: The centre of the reflecting surface of a spherical mirror.
  [$0.5$‍ marks]

  ($\text{ii}$‍) Centre of curvature: The reflecting surface of a spherical mirror is part of a sphere. The centre of that sphere is called centre of curvature.
  [$0.5$‍ marks]

  ($\text{iii}$‍) Principal axis: An imaginary line passing through the pole and the centre of curvature. The principal axis is normal to the mirror at the pole.
  [$0.5$‍ marks]

  ($\text{iv}$‍) Principal focus: The point on the principal axis, at which incident light rays parallel to the principal axis converge or appear to diverge from, is called the principal focus of the lens.
  [$0.5$‍ marks]

• Related video: Spherical mirrors, radius of curvature & focal length.

• Solution

  A2. ($\text{i}$‍) The headlights of a car use concave mirrors.
  [$2/3$‍ marks]

  Reason: When the headlight bulb is placed at the principal focus of the concave mirror, it produces a strong forward beam consisting of parallel rays of light.
  [$1$‍ mark]

  ($\text{ii}$‍) The rear-view mirrors use convex mirrors.
  [$2/3$‍ marks]

  Reason: A convex mirror, due to its curvature, directs light from a wider angle to the eye of the viewer. It also produces an erect image of the traffic behind.
  [$1$‍ mark]

  ($\text{iii}$‍) A solar furnace uses concave mirrors.
  [$2/3$‍ marks]

  Reason: The concave mirror focuses parallel rays of light coming from the sun at the principal focus. The concentration of sunlight at the principal focus produces a lot of useful heat.
  [$1$‍ mark]

• Related exercise: Applications of concave and convex mirrors.
• Related videos: ($\text{i}$‍) Concave mirror applications, ($\text{ii}$‍) Convex mirror & applications.

• Solution

  A3.
  ($\text{i}$‍)

  
  [$1$‍ mark]

  ($\text{ii}$‍)

  
  [$1$‍ mark]

  ($\text{iii}$‍)

  
  [$1$‍ mark]

• Related exercises: ($\text{i}$‍) Ray diagrams, ($\text{ii}$‍) Ray diagrams and curved mirrors.
• Related video: Convex & concave mirror ray diagrams.

• Solution

  A4. Only convex mirrors produce images that are always virtual and diminished.
  [$1$‍ mark]

  
  It is seen that the image ($\text{I}$‍) is virtual and diminished irrespective of the position of the object ($\text{O}$‍).
  [$2$‍ marks]

• Related videos: ($\text{i}$‍) Concave mirrors, ($\text{ii}$‍) Convex mirror & applications.

• Solution

  A5. ($\text{i}$‍) A magnification of $+1/3$‍ indicates an erect and diminished image. Only a convex mirror produces such an image.
  [$0.5$‍ marks]

  ($\text{ii}$‍) Position of object: Between infinity and the pole.
  [$0.5$‍ marks]

  The diagram below shows that for any position of the object between $\mathrm{\infty }$‍ and $P$‍, the convex mirror always forms an erect and diminished image.

  
  [$1$‍ mark]

• Solution

  A6. ($\text{i}$‍) The mirror is concave since it forms a real image.
  [$0.5$‍ marks]

  ($\text{ii}$‍) A magnification of $-1$‍ implies that both the object and image are located at the centre of curvature, $C$‍. The distance of the image from the object is zero.
  [$0.5$‍ marks]

  ($\text{iii}$‍) The radius of curvature is $50\text{cm}$‍. So, the focal length is $25\text{cm}$‍ ($1/2\times$‍ radius of curvature).
  [$1$‍ mark]

  ($\text{iv}$‍) The image formation is shown below.

  
  [$1$‍ mark]

• Related exercises: ($\text{i}$‍) Nature and size of images from magnification, ($\text{ii}$‍) Using the magnification formula for mirrors.
• Related video: Magnification formula for mirrors.

• Solution

  A7. Given:

  • object distance, $u=-4\text{m}$‍,
  • image distance, $v=+1\text{m}$‍.

  Magnification, $m\equiv -{\displaystyle \frac{v}{u}}=+{\displaystyle \frac{1}{4}}$‍.
  [$0.5$‍ marks]

  Since the magnification is positive and less than one, the mirror is a convex mirror.
  [$0.5$‍ marks]

• Solution

  A7.1. Data (in Cartesian sign convention):

  • object distance, $u=-20\text{cm}$‍
  • image distance, $v=?$‍
  • magnification, $m=-3$‍.

  From the magnification formula, we have,

  $\begin{array}{rl}m& =-{\displaystyle \frac{v}{u}}\\ \\ \Rightarrow v& =-mu\\ \\ & =-(-3)\times (-20\text{cm})\\ \\ & =-60\text{cm}\end{array}$‍
  [$2$‍ marks]

  Therefore, the distance between the screen (image) and the object,
  $|v-u|=|-40\text{cm}|=40\text{cm}$‍.
  [$1$‍ mark]

• Related exercises: ($\text{i}$‍) Using the magnification formula for mirrors, ($\text{ii}$‍) Nature and size of images from magnification.
• Related video: Magnification formula for mirrors.

• Solution

  A8. Since the image forms on a screen, it's a real image. The mirror is concave.
  [$0.5$‍ marks]

  We are given:

  • object distance, $u=-30\text{cm}$‍,
  • image distance, $v=-60\text{cm}$‍.

  The focal length, $f$‍, is calculated from the mirror formula,

  $\begin{array}{rl}{\displaystyle \frac{1}{f}}& ={\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}\\ \\ & =-{\displaystyle \frac{1}{60\text{cm}}}-{\displaystyle \frac{1}{30\text{cm}}}\\ \\ & =-{\displaystyle \frac{1+2}{60\text{cm}}}=-{\displaystyle \frac{\stackrel{1}{\cancel{3}}}{\underset{20}{\cancel{60}}\text{cm}}}\\ \\ f& =-20\text{cm}\end{array}$‍
  [$0.5$‍ marks]

• Related exercise: Using the mirror formula.
• Related videos: ($\text{i}$‍) Solve example: Mirror formula, ($\text{ii}$‍) Mirror formula.

• Solution

  A9. A virtual and diminished image implies that the mirror is convex.

  Data (in Cartesian sign convention):

  • object distance, $u=-18\text{cm}$‍
  • image distance, $v=?$‍
  • focal length, $f=?$‍
  • magnification, $m=+{\displaystyle \frac{1}{3}}$‍.

  We calculate $v$‍ using the magnification formula for mirrors, as follows,

  $\begin{array}{rl}m& =-{\displaystyle \frac{v}{u}}\\ \\ v& =-mu\\ \\ & =-(+{\displaystyle \frac{1}{\underset{1}{\cancel{3}}}})\times (-\stackrel{6}{\cancel{18}}\text{cm})\\ \\ v& =+6\text{cm}\end{array}$‍
  [$1$‍ mark]

  We calculate $f$‍ using the mirror formula.

  $\begin{array}{rl}{\displaystyle \frac{1}{f}}& ={\displaystyle \frac{1}{v}}+{\displaystyle \frac{1}{u}}\\ \\ & =[{\displaystyle \frac{1}{(+6)}}+{\displaystyle \frac{1}{(-18)}}]{\displaystyle \frac{1}{\text{cm}}}\\ \\ & ={\displaystyle \frac{\stackrel{1}{\cancel{3-1}}}{\underset{9}{\cancel{18}}}}{\displaystyle \frac{1}{\text{cm}}}\\ \\ f& =+9\text{cm}\end{array}$‍
  [$1$‍ mark]

  Ray diagram:

  
  [$1$‍ mark]

• Related exercise: Concave and convex mirrors.

• Solution

  A10. Range of object distance: The object must be between the pole and the focus of the concave mirror for an erect image. The range of the object distance, $u$‍, is, $0\text{cm}<u<20\text{cm}$‍.
  [$1$‍ mark]

  Two other characteristics of the image:
  ($\text{i}$‍) virtual,
  ($\text{ii}$‍) enlarged.
  [$1$‍ mark]

  Ray diagram:
  

  
  [$1$‍ mark]

Refraction of light and spherical lenses

• Solution

  A11. ($\text{i}$‍) The central point of a lens is called its optical centre. A ray of light passes undeviated through the optical centre of any lens.
  [$0.5$‍ marks]

  ($\text{ii}$‍) Each spherical surface of a spherical lens forms a part of a sphere. The centres of these spheres are called the centres of curvature.
  [$0.5$‍ marks]

  ($\text{iii}$‍) The principal axis of a lens is an imaginary line passing through its two centres of curvature.
  [$0.5$‍ marks]

  ($\text{iv}$‍) The aperture of a spherical lens is the diameter of its circular outline.
  [$0.5$‍ marks]

  ($\text{v}$‍) The point on the principal axis, at which incident light rays parallel to the principal axis converge or appear to diverge from, is called the principal focus of the lens.
  [$0.5$‍ marks]

  ($\text{vi}$‍) The distance between the optical centre and the principal focus of a spherical lens is called its focal length, denoted by the letter $f$‍.
  [$0.5$‍ marks]

• Related videos: ($\text{i}$‍) Convex lenses, ($\text{ii}$‍) Concave lenses.

• Solution

  A12. ($\text{a}$‍) The ray diagram is shown below. The object is placed between $O$‍ and $F_1$‍.

  
  [$1$‍ mark]

  ($\text{ii}$‍) The object distance ($u$‍) and the image distance ($v$‍) are both negative.
  [$1$‍ mark]

  These are related to the focal length of the lens, $f$‍, by the lens formula,

  $\boxed{{\displaystyle {\displaystyle \frac{1}{f}}={\displaystyle \frac{1}{v}}-{\displaystyle \frac{1}{u}}}}$‍.
  [$1$‍ mark]

• Related exercise: Lenses and ray diagrams.
• Related video: Sign convention for mirrors (& lenses).

• Solution

  A13. Lateral displacement: When a ray of light is incident obliquely on a parallel-sided glass slab, the emergent ray shifts laterally. The perpendicular distance between the incident ray and the emergent ray is called lateral displacement.
  [$1$‍ mark]

  Diagram:

  
  [$1$‍ mark]

  The factors on which lateral displacement depends are:
  ($\text{i}$‍) the thickness of the refracting material,
  ($\text{ii}$‍) the refractive index of the material.
  [$1$‍ mark]

• Related video: Refraction through glass slab.

• Solution

  A14. Snell's law: When a ray of light of a particular colour passing through a medium $1$‍ is incident on another medium $2$‍, the ratio of sine of the angle of incidence ($i$‍) and sine of the angle of refraction ($r$‍) is a constant.

  This constant is called the refractive index of medium $2$‍ with respect to medium $1$‍, denoted as $n_{21}$‍.
  [$1$‍ mark]

  Mathematically,

  $\begin{array}{rl}{n}_{21}& ={\displaystyle \frac{\sin i}{\sin r}}\end{array}$‍
  [$0.5$‍ mark]

  The refractive index of a medium $A$‍ with respect to another medium $B$‍, denoted by $n_{AB}$‍, is related to the speeds of light in the media as follows,

  $\begin{array}{rl}{n}_{AB}& ={\displaystyle \frac{\text{speed of light in medium }B}{\text{speed of light in medium }A}}.\end{array}$‍
  [$1$‍ mark]

  In the case above, if medium $B$‍ is vacuum or air, then $n_{AB}$‍ is called the absolute refractive index of medium $A$‍, and is written simply as $n_A$‍.
  [$0.5$‍ mark]

  Given:

  • absolute refractive index of glass, $n_G={\displaystyle \frac{3}{2}}$‍
  • absolute refractive index of water, $n_W={\displaystyle \frac{4}{3}}$‍
  • speed of light in glass, $v_G=2\times {10}^8\text{m/s}$‍.

  ($\text{i}$‍) Now,

  $\begin{array}{r}{n}_G={\displaystyle \frac{\text{speed of light in vacuum, }c}{v_G}}\end{array}$‍

  $\begin{array}{rl}\therefore c& =n_G\times v_G\\ \\ & ={\displaystyle \frac{3}{\cancel{2}}}\times \cancel{2}\times {10}^8\text{m/s}\\ \\ & =3\times {10}^8\text{m/s}\end{array}$‍
  [$1$‍ mark]

  ($\text{ii}$‍) The speed of light in water, $v_W$‍, is related to $n_W$‍ as,

  $\begin{array}{rl}{n}_W& ={\displaystyle \frac{c}{v_W}}\\ \\ \therefore v_W& ={\displaystyle \frac{c}{n_W}}\\ \\ & ={\displaystyle \frac{3\times {10}^8\text{m/s}}{4/3}}\\ \\ & ={\displaystyle \frac{9}{4}}\times {10}^8\text{m/s}\\ \\ & =2.25\times {10}^8\text{m/s}\end{array}$‍
  [$1$‍ mark]