Question 352 marksShort AnswerSection B

An object of height 2.5 cm is placed at a distance of 15 cm from the optical centre ‘O’ of a convex lens of focal length 10 cm. Draw a ray diagram to find the position and size of the image formed. Mark optical centre ‘O’, principal focus F and height of the image on the diagram.

Also asked in:2016
Light — Reflection and Refraction
Refraction by Spherical Lenses

Official answer



The position of the image is at a distance of 30 cm30\text{ cm} on the other side of the lens (real and inverted), and the size (height) of the image is 5.0 cm5.0\text{ cm} (pointing downwards).


Below is the ray diagram representing the image formation:


``svg {"svg": "<svg viewBox=\"0 0 240 168\" xmlns=\"http://www.w3.org/2000/svg\"><text x=\"120\" y=\"13\" font-family=\"sans-serif\" font-size=\"9\" fill=\"#0f172a\" text-anchor=\"middle\" font-weight=\"bold\">Convex lens: object between F and 2F</text><line x1=\"24.0\" y1=\"95.0\" x2=\"208.0\" y2=\"95.0\" stroke=\"#94a3b8\" stroke-width=\"1\"/><polygon points=\"208.0,95.0 202.5,98.2 202.5,91.8\" fill=\"#94a3b8\"/><line x1=\"99.2\" y1=\"32.0\" x2=\"99.2\" y2=\"158.0\" stroke=\"#0b5cad\" stroke-width=\"2\"/><path d=\"M 93.2 32.0 Q 99.2 40.0 105.2 32.0\" fill=\"none\" stroke=\"#0b5cad\" stroke-width=\"2\"/><path d=\"M 93.2 158.0 Q 99.2 150.0 105.2 158.0\" fill=\"none\" stroke=\"#0b5cad\" stroke-width=\"2\"/><text x=\"103.2\" y=\"107.0\" font-family=\"sans-serif\" font-size=\"8\" fill=\"#0b5cad\" text-anchor=\"start\" font-weight=\"normal\" stroke=\"#fff\" stroke-width=\"2\" paint-order=\"stroke\" stroke-linejoin=\"round\">O</text><line x1=\"65.6\" y1=\"91.0\" x2=\"65.6\" y2=\"99.0\" stroke=\"#64748b\" stroke-width=\"1\"/><text x=\"65.6\" y=\"108.0\" font-family=\"sans-serif\" font-size=\"8\" fill=\"#334155\" text-anchor=\"middle\" font-weight=\"normal\" stroke=\"#fff\" stroke-width=\"2\" paint-order=\"stroke\" stroke-linejoin=\"round\">F</text><line x1=\"32.1\" y1=\"91.0\" x2=\"32.1\" y2=\"99.0\" stroke=\"#64748b\" stroke-width=\"1\"/><text x=\"32.1\" y=\"108.0\" font-family=\"sans-serif\" font-size=\"8\" fill=\"#334155\" text-anchor=\"middle\" font-weight=\"normal\" stroke=\"#fff\" stroke-width=\"2\" paint-order=\"stroke\" stroke-linejoin=\"round\">2F</text><line x1=\"132.8\" y1=\"91.0\" x2=\"132.8\" y2=\"99.0\" stroke=\"#64748b\" stroke-width=\"1\"/><text x=\"132.8\" y=\"108.0\" font-family=\"sans-serif\" font-size=\"8\" fill=\"#334155\" text-anchor=\"middle\" font-weight=\"normal\" stroke=\"#fff\" stroke-width=\"2\" paint-order=\"stroke\" stroke-linejoin=\"round\">F</text><line x1=\"166.4\" y1=\"91.0\" x2=\"166.4\" y2=\"99.0\" stroke=\"#64748b\" stroke-width=\"1\"/><text x=\"166.4\" y=\"108.0\" font-family=\"sans-serif\" font-size=\"8\" fill=\"#334155\" text-anchor=\"middle\" font-weight=\"normal\" stroke=\"#fff\" stroke-width=\"2\" paint-order=\"stroke\" stroke-linejoin=\"round\">2F</text><line x1=\"48.9\" y1=\"95.0\" x2=\"48.9\" y2=\"68.5\" stroke=\"#0a7d55\" stroke-width=\"2\"/><polygon points=\"48.9,68.5 52.1,74.0 45.7,74.0\" fill=\"#0a7d55\"/><text x=\"48.9\" y=\"63.5\" font-family=\"sans-serif\" font-size=\"8\" fill=\"#0a7d55\" text-anchor=\"middle\" font-weight=\"normal\" stroke=\"#fff\" stroke-width=\"2\" paint-order=\"stroke\" stroke-linejoin=\"round\">Object</text><line x1=\"199.9\" y1=\"95.0\" x2=\"199.9\" y2=\"148.0\" stroke=\"#c0392b\" stroke-width=\"2\"/><polygon points=\"199.9,148.0 196.7,142.5 203.1,142.5\" fill=\"#c0392b\"/><text x=\"199.9\" y=\"160.0\" font-family=\"sans-serif\" font-size=\"8\" fill=\"#c0392b\" text-anchor=\"middle\" font-weight=\"normal\" stroke=\"#fff\" stroke-width=\"2\" paint-order=\"stroke\" stroke-linejoin=\"round\">Image</text><polyline points=\"48.9,68.5 99.2,68.5 199.9,148.0\" fill=\"none\" stroke=\"#1d4ed8\" stroke-width=\"1.3\"/><polygon points=\"199.9,148.0 193.6,147.1 197.6,142.1\" fill=\"#1d4ed8\"/><polyline points=\"48.9,68.5 199.9,148.0\" fill=\"none\" stroke=\"#7c3aed\" stroke-width=\"1.3\"/><polygon points=\"199.9,148.0 193.6,148.3 196.6,142.6\" fill=\"#7c3aed\"/></svg>", "caption": "Convex lens: object between F and 2F: f=10 cm, u=15 cm, h=2.5 cm -> real, inverted image at v=30 cm, h'=5.0 cm"} ``



Answer keywords:lens formulaconvex lensimage distancemagnificationreal and invertedoptical centreprincipal focus

Marking scheme

  1. 1.0.5 Marks for using the correct lens formula and calculating v=+30 cmv = +30\text{ cm}.
  2. 2.0.5 Marks for calculating the height of the image h′=−5.0 cmh' = -5.0\text{ cm}.
  3. 3.1.0 Mark for drawing a neat, labeled ray diagram showing the optical centre 'O', principal focus 'F', object, and image with correct directions of arrows.

Hint

Use the lens formula 1v−1u=1f\frac{1}{v} - \frac{1}{u} = \frac{1}{f} to find the image distance, and then use m=vu=h′hm = \frac{v}{u} = \frac{h'}{h} to find the height of the image.

Quick oral answer

For a convex lens of focal length 10 cm, if an object is placed at 15 cm, it lies between F and 2F. The image is formed beyond 2F on the other side, which is at 30 cm, and it is real, inverted, and twice the size of the object.

Common mistakes

  1. 1.Using the mirror formula (with a plus sign) instead of the lens formula.
  2. 2.Incorrect sign convention for object distance (uu), taking it as positive.
  3. 3.Forgetting to draw arrows on the light rays in the ray diagram to show the direction of propagation.

Analysis and explanation

Since the focal length of the convex lens is 10 cm10\text{ cm}, the principal focus F1F_1 is at 10 cm10\text{ cm} and 2F12F_1 is at 20 cm20\text{ cm}. The object is placed at 15 cm15\text{ cm}, which is between F1F_1 and 2F12F_1.


According to the rules of image formation by a convex lens, when an object is placed between F1F_1 and 2F12F_1, its image is formed beyond 2F22F_2 on the other side of the lens. The image is real, inverted, and magnified. This is verified by our calculations where v=+30 cmv = +30\text{ cm} (which is beyond 2F2=20 cm2F_2 = 20\text{ cm}) and the magnification is −2-2.

Interesting facts

A convex lens acts as a magnifying glass when the object is placed within its focal length, but when the object is outside the focal length, it forms real images that can be projected onto a screen.

Question details

Q35 of 36 questions

Board
CBSE
Class
Class 10
Subject
Science
Year
2016
Section
Section B
Marks
2
Type
Short Answer
Chapter
Light — Reflection and Refraction
Topic
Refraction by Spherical Lenses
Also asked in
2016
Bloom's level
Creating
Syllabus topic
Image formation by lenses, Lens formula, Magnification

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Frequently asked questions

How many marks does this question carry in CBSE Class 10 Science 2016?

This question carries 2 marks in the CBSE Class 10 Science 2016 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.

Has this question appeared in other CBSE Science papers?

Yes, a similar question appeared in the 2016 CBSE Class 10 Science paper.

What type of question is this in the CBSE Class 10 Science 2016 paper?

This is a Short Answer question from Section B in the CBSE Class 10 Science 2016 paper.