Question 213 marksShort AnswerSection A

The image of a candle flame placed at a distance of 45 cm from a spherical lens is formed on a screen placed at a distance of 90 cm from the lens. Identify the type of lens and calculate its focal length. If the height of the flame is 2 cm, find the height of its image.

Light — Reflection and Refraction
Refraction by Spherical Lenses

Official answer

The lens is a convex lens.

The focal length of the lens is +30 cm+30\text{ cm} (or 30 cm30\text{ cm}).

The height of the image is −4 cm-4\text{ cm} (the negative sign indicates that the image is real and inverted).

Answer keywords:convex lensreal imagelens formulafocal length 30 cmmagnificationheight of image -4 cm

Marking scheme

  1. 1.1.0 mark for identifying the lens as a convex lens with justification.
  2. 2.1.0 mark for correctly calculating the focal length (f=+30 cmf = +30\text{ cm}) using the lens formula.
  3. 3.1.0 mark for correctly calculating the height of the image (h′=−4 cmh' = -4\text{ cm}) using the magnification formula.

Hint

Since the image is caught on a screen, it must be real. Use the lens formula 1f=1v−1u\frac{1}{f} = \frac{1}{v} - \frac{1}{u} with proper sign conventions, and then use the magnification formula m=h′h=vum = \frac{h'}{h} = \frac{v}{u}.

Quick oral answer

A convex lens is used because only it can form a real image on a screen. The focal length is calculated using the lens formula to be 30 cm30\text{ cm}, and the image height is −4 cm-4\text{ cm}, indicating a real, inverted, and magnified image.

Common mistakes

  1. 1.Using the mirror formula instead of the lens formula (i.e., using ++ instead of −-).
  2. 2.Forgetting to apply the sign convention for object distance (u=−45 cmu = -45\text{ cm}).
  3. 3.Using m=−vum = -\frac{v}{u} (which is for mirrors) instead of m=vum = \frac{v}{u} (for lenses).

Analysis and explanation

A real image is always inverted, which is mathematically confirmed by the negative sign of the image height (h′=−4 cmh' = -4\text{ cm}). Since ∣v∣>∣u∣|v| > |u| (90 cm>45 cm90\text{ cm} > 45\text{ cm}), the image is magnified, which is also confirmed by the magnitude of the height (4 cm>2 cm4\text{ cm} > 2\text{ cm}).

Interesting facts

When an object is placed between FF and 2F2F of a convex lens, its real, inverted, and magnified image is formed beyond 2F2F on the other side. Here, f=30 cmf = 30\text{ cm}, so 2f=60 cm2f = 60\text{ cm}. The object at 45 cm45\text{ cm} is indeed between FF and 2F2F, and the image at 90 cm90\text{ cm} is beyond 2F2F (60 cm60\text{ cm}), perfectly matching theory!

Question details

Q21 of 41 questions

Board
CBSE
Class
Class 10
Subject
Science
Year
2012
Section
Section A
Marks
3
Type
Short Answer
Chapter
Light — Reflection and Refraction
Topic
Refraction by Spherical Lenses
Bloom's level
Applying
Syllabus topic
Lens formula, magnification, and numerical problems

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Frequently asked questions

How many marks does this question carry in CBSE Class 10 Science 2012?

This question carries 3 marks in the CBSE Class 10 Science 2012 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 "Refraction by Spherical Lenses" from CBSE Class 10 Science.

What type of question is this in the CBSE Class 10 Science 2012 paper?

This is a Short Answer question from Section A in the CBSE Class 10 Science 2012 paper.