Question 153 marksShort AnswerSection A

An object of height 6 cm is placed perpendicular to the principal axis of a concave lens of focal length 5 cm. Use lens formula to determine the position, size and nature of the image if the distance of the object from the lens is 10 cm.

Light — Reflection and Refraction
Lens Formula and Magnification

Official answer

Using the lens formula and magnification equations for a concave lens, we can determine the position, size, and nature of the image.


  • Position of the image (vv): −3.33 cm-3.33\text{ cm} (on the same side as the object, at a distance of 3.33 cm3.33\text{ cm} from the lens)
  • Size of the image (h′h'): +2 cm+2\text{ cm} (erect and diminished)
  • Nature of the image: Virtual, erect, and diminished
Answer keywords:lens formulaconcave lensvirtualerectdiminishedimage distancemagnification

Marking scheme

  1. 1.Writing the correct lens formula and substituting values with correct sign conventions (1 mark)
  2. 2.Calculating the correct image distance (v=−3.33 cmv = -3.33\text{ cm} or −10/3 cm-10/3\text{ cm}) (1 mark)
  3. 3.Calculating the correct image height (h′=+2 cmh' = +2\text{ cm}) and stating the nature (virtual, erect, diminished) (1 mark)

Hint

Remember that a concave lens always has a negative focal length (f<0f < 0) and always produces a virtual, erect, and diminished image (v<0v < 0 and 0<m<10 < m < 1).

Quick oral answer

For a concave lens, the focal length is always negative. The lens formula is 1/f = 1/v - 1/u, and the magnification is v/u. It always forms a virtual, erect, and diminished image.

Common mistakes

  1. 1.Using the mirror formula (with a '+' sign) instead of the lens formula.
  2. 2.Taking the focal length of the concave lens as positive.
  3. 3.Confusing the magnification formula of a lens (m=v/um = v/u) with that of a mirror (m=−v/um = -v/u).

Analysis and explanation

A concave lens is a diverging lens. It always forms a virtual, erect, and diminished image, regardless of the position of the object. The calculations confirm this: the image distance v=−3.33 cmv = -3.33\text{ cm} is negative (virtual), the magnification m=+1/3m = +1/3 is positive (erect) and less than 1 (diminished), and the final image height is 2 cm2\text{ cm} compared to the original 6 cm6\text{ cm} object height.

Interesting facts

No matter how far or close you place an object in front of a concave lens, the image will always be formed between the focus and the optical center, and it will always be smaller than the object.

Question details

Q15 of 40 questions

Board
CBSE
Class
Class 10
Subject
Science
Year
2013
Section
Section A
Marks
3
Type
Short Answer
Chapter
Light — Reflection and Refraction
Topic
Lens Formula and Magnification
Bloom's level
Applying
Syllabus topic
Lens formula, magnification, power of a lens

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Frequently asked questions

How many marks does this question carry in CBSE Class 10 Science 2013?

This question carries 3 marks in the CBSE Class 10 Science 2013 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 "Lens Formula and Magnification" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2013 paper?

This is a Short Answer question from Section A in the CBSE Class 10 Science 2013 paper.