A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula.

Q: A 4 cm tall object is placed perpendicular to the principal axis of a convex lens of focal length 24 cm. The distance of the object from the lens is 16 cm. Find the position, size and nature of the image formed, using the lens formula.

Using the lens formula, the position of the image is at **$48\text{ cm}$** on the same side as the object (in front of the lens). The size of the image is **$12\text{ cm}$**, and its nature is **virtual and erect**.

Q: 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.

Q: 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.

Q: What topic does this question cover in Science?

This question covers the topic "Refraction by Spherical Lenses" from CBSE Class 10 Science.

Q: 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.

Light — Reflection and Refraction

Refraction by Spherical Lenses

Official Answer

Using the lens formula, the position of the image is at $48\text{ cm}$ on the same side as the object (in front of the lens). The size of the image is $12\text{ cm}$ , and its nature is virtual and erect.

lens formulav = -48 cmmagnificationh' = 12 cmvirtual and erectenlarged

Marking Scheme

11.5 marks
21 mark
30.5 marks

Hint

Identify the signs of $u$ and $f$ carefully. For a convex lens, $f$ is positive. Since the object is placed closer than the focal length ( $16\text{ cm} < 24\text{ cm}$ ), expect a virtual, erect, and enlarged image.

Quick Oral Answer

The image is formed at $48\text{ cm}$ on the same side as the object. It is virtual, erect, and has a height of $12\text{ cm}$ , which is three times the size of the object.

Analysis & Explanation

When an object is placed within the focal length of a convex lens ( $|u| < |f|$ , here $16\text{ cm} < 24\text{ cm}$ ), the lens acts as a magnifying glass. The refracted rays diverge, and their backward extensions meet to form a virtual, erect, and magnified image on the same side as the object. The calculations perfectly match this physical behavior, yielding a negative image distance ( $v = -48\text{ cm}$ ) and a positive magnification ( $m = +3$ ).

Common Mistakes

1Using the mirror formula ( $\frac{1}{v} + \frac{1}{u} = \frac{1}{f}$ ) instead of the lens formula ( $\frac{1}{v} - \frac{1}{u} = \frac{1}{f}$ ).
2Using the wrong sign for magnification in lenses ( $m = -v/u$ is for mirrors; for lenses, it is $m = v/u$ ).
3Taking the focal length of a convex lens as negative.

Interesting Facts

This setup represents how a simple magnifying glass works! When you hold a magnifying glass very close to an object (closer than its focal length), you see a giant, upright virtual image of the object.

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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.