Q36

5 marksLong AnswerSection D

(a) (i) Draw a ray diagram to show the path of the refracted ray in each of the following cases: A ray of light incident on a concave lens (1) parallel to its principal axis, and (2) is directed towards its principal focus. (ii) A 4 cm tall object is placed perpendicular to the principal axis of convex lens of focal length 24 cm. The distance of object from the lens is 16 cm. Find the position and size of image formed. OR (b) (i) Draw a ray diagram to show the path of the reflected ray in each of the following cases: A ray of light incident on a convex mirror (1) parallel to its principal axis, and (2) is directed towards its principal focus. (ii) A 1.5 cm tall candle flame is placed perpendicular to the principal axis of a concave mirror of focal length 12 cm. If the distance of the flame from the pole of the mirror is 18 cm, use mirror formula to determine the position and size of the image formed.

Light — Reflection and Refraction

Image Formation by Spherical Mirrors and Lenses

Official Answer

Option (a)

#### (i) Ray Diagrams for Concave Lens

1. A ray of light incident parallel to the principal axis:

After refraction from a concave lens, the ray appears to diverge from the principal focus ( $F_1$ ) located on the same side of the lens as the incident light.


                  Concave Lens

  Incident Ray         ||
  ====================>|| \ Refracted Ray (Diverging)
  \                    ||
   \                   ||

     \                 ||

F1             ||

  (Appears to diverge  ||
     from Focus F1)    ||




2. A ray of light directed towards the principal focus ( $F_2$ ):
After refraction through a concave lens, the ray emerges parallel to the principal axis.


                  Concave Lens

  Incident Ray         ||
  ====================>||====================> Refracted Ray (Parallel)
  \                    ||
   \                   ||

     \                 ||
      v                ||
  (Directed towards    ||
     Focus F2)         ||




#### (ii) Numerical Calculation for Convex Lens

Given:
Height of the object,  $h = +4\text{ cm}$ 
Focal length of the convex lens,  $f = +24\text{ cm}$ 
Object distance,  $u = -16\text{ cm}$ 




Using the Lens Formula:

   $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ 
   $\frac{1}{24} = \frac{1}{v} - \frac{1}{-16}$ 
   $\frac{1}{24} = \frac{1}{v} + \frac{1}{16}$ 
   $\frac{1}{v} = \frac{1}{24} - \frac{1}{16}$ 
   $\frac{1}{v} = \frac{2 - 3}{48} = -\frac{1}{48}$ 
   $v = -48\text{ cm}$ 



Using the Magnification Formula:

   $m = \frac{v}{u} = \frac{h'}{h}$ 
   $m = \frac{-48}{-16} = +3$ 
   $h' = m \times h = 3 \times 4\text{ cm} = +12\text{ cm}$ 



Result:
Position of the image:  $48\text{ cm}$  in front of the lens (on the same side as the object).
Size of the image:  $12\text{ cm}$  tall.
Nature of the image: Virtual and erect.






Option (b)


#### (i) Ray Diagrams for Convex Mirror


1. A ray of light incident parallel to the principal axis:
After reflection, the ray appears to diverge from the principal focus ( $F$ ) behind the mirror.


                  Convex Mirror
                       |)
  Incident Ray         |)
  ====================>|)\ Reflected Ray (Diverging)
                       |) \
                       |)  \
-----------------------|)---\-------------------
                       |)    * F (Focus behind mirror)
                       |) (Appears to diverge from F)




2. A ray of light directed towards the principal focus ( $F$ ):
After reflection, the ray emerges parallel to the principal axis.


                  Convex Mirror
                       |)
  Incident Ray         |)
  ====================>|)
  \                    |)
   \                   |)

     \                 |)
      v                |)
  (Directed towards    |)    * F (Focus behind mirror)
     Focus F)          |)


  Reflected Ray (Parallel)
  <====================|)

#### (ii) Numerical Calculation for Concave Mirror

Given:
Height of the candle flame (object), $h = +1.5\text{ cm}$
Focal length of the concave mirror, $f = -12\text{ cm}$
Object distance, $u = -18\text{ cm}$

Using the Mirror Formula:

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

$\frac{1}{-12} = \frac{1}{v} + \frac{1}{-18}$

$-\frac{1}{12} = \frac{1}{v} - \frac{1}{18}$

$\frac{1}{v} = -\frac{1}{12} + \frac{1}{18}$

$\frac{1}{v} = \frac{-3 + 2}{36} = -\frac{1}{36}$

$v = -36\text{ cm}$

Using the Magnification Formula:

$m = -\frac{v}{u} = \frac{h'}{h}$

$m = -\frac{-36}{-18} = -2$

$h' = m \times h = -2 \times 1.5\text{ cm} = -3.0\text{ cm}$

Result:
Position of the image: $36\text{ cm}$ in front of the mirror (on the same side as the object).
Size of the image: $3.0\text{ cm}$ tall.
Nature of the image: Real and inverted.

lens formulamirror formulamagnificationvirtual and erectreal and invertedCartesian sign conventionprincipal focusconcave lensconvex mirrorconcave mirror

Marking Scheme

1Option (a) (i): Correctly drawing the two ray diagrams for a concave lens with proper arrows and labels. (2 marks)
2Option (a) (ii): Correct substitution in the lens formula and calculating $v = -48\text{ cm}$ . (2 marks)
3Option (a) (ii): Correct calculation of image height $h' = +12\text{ cm}$ and stating the nature (virtual and erect). (1 mark)
4Option (b) (i): Correctly drawing the two ray diagrams for a convex mirror with proper arrows and labels. (2 marks)
5Option (b) (ii): Correct substitution in the mirror formula and calculating $v = -36\text{ cm}$ . (2 marks)
6Option (b) (ii): Correct calculation of image height $h' = -3.0\text{ cm}$ and stating the nature (real and inverted). (1 mark)

Hint

For lenses, use the lens formula $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ and magnification $m = \frac{v}{u}$ . For mirrors, use the mirror formula $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ and magnification $m = -\frac{v}{u}$ . Remember to apply the Cartesian sign conventions carefully.

Quick Oral Answer

Q: What is the nature of the image formed by a concave mirror when the object is placed between its focus and pole? A: The image formed is virtual, erect, and magnified, and it is located behind the mirror.

Analysis & Explanation

The question tests the application of sign conventions, ray optics rules, and mathematical formulas for both spherical mirrors and lenses.

Sign Conventions (New Cartesian Sign Convention):

Distances measured in the direction of incident light are positive, while those against it are negative.
For a convex lens, the principal focus is real and lies on the opposite side of the incident light, making its focal length positive ( $f = +24\text{ cm}$ ).
For a concave mirror, the principal focus lies in front of the mirror, making its focal length negative ( $f = -12\text{ cm}$ ).
Object distance ( $u$ ) is always negative as the object is placed in front of the optical device.

Lens vs. Mirror Formulas:

The lens formula contains a negative sign: $\frac{1}{f} = \frac{1}{v} - \frac{1}{u}$ , and its magnification is $m = \frac{v}{u}$ .
The mirror formula contains a positive sign: $\frac{1}{f} = \frac{1}{v} + \frac{1}{u}$ , and its magnification is $m = -\frac{v}{u}$ .
Confusing these two formulas is a very common source of error for students.

Ray Diagrams:

The ray diagrams are based on fundamental laws of reflection and refraction. For a convex mirror, since the focus is virtual (behind the mirror), rays must be projected using dotted lines behind the mirror surface to show their virtual paths.

Common Mistakes

1Using the mirror formula instead of the lens formula for lens calculations, or vice versa.
2Forgetting to apply negative signs to object distance ( $u$ ) and concave mirror focal length ( $f$ ).
3Omitting arrows on ray diagrams, which leads to a loss of marks in CBSE exams.
4Confusing the magnification formulas: using $m = -\frac{v}{u}$ for lenses instead of $m = \frac{v}{u}$ .

Interesting Facts

A convex mirror always forms a virtual, erect, and diminished image, which is why it is preferred as a rear-view mirror in vehicles to provide a wider field of view.

The virtual image formed by a convex lens when the object is within its focal length is the principle behind a simple magnifying glass.

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Frequently Asked Questions

How many marks does this question carry in CBSE Class 10 Science 2024?

This question carries 5 marks in the CBSE Class 10 Science 2024 examination.

Which chapter does this question come from in Science?

This question is from the chapter "Light — Reflection and Refraction" in the CBSE Class 10 Science syllabus.

What topic does this question cover in Science?

This question covers the topic "Image Formation by Spherical Mirrors and Lenses" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2024 paper?

This is a Long Answer question from Section D in the CBSE Class 10 Science 2024 paper.